Matrix
Subject: Optional Mathematics
Overview
A rectangular array of numbers arranged in horizontal and vertical lines enclosed between round or square brackets is called a matrix.
Matrix
Introduction
A rectangular array of numbers arranged in horizontal and vertical lines enclosed between round or square brackets is called a matrix. Some example of the matrix is given below:
X=\(\begin{bmatrix}1 &2\\ 3 &4\\ \end{bmatrix}\)
Y=\(\begin{bmatrix}1&2\\5&1\\7&4\\ \end{bmatrix}\)
Z=\(\begin{bmatrix}3&2&5\\4&6&5\\ \end{bmatrix}\)
Each member in the array is called an element. An element appearing in the ith row and jth column of a matrix called its (i,j)th element.
In the above examples, the order of matrix X is 2×2, Y is 3×2 and Z is 2×3. So, if the matrix contains rows and b column, then it is of order a×b.
Multiplication of Matrices
Let A and B be two matrices and product of A and B is denoted by AB.The products of AB can be defined or not according to its order. Matrices A and B are said to be compatible for product AB if and only if numbers of the column in A is equal to a number of rows in B. If this condition is not satisfied then the product of A and B cannot be performed.
if A and B are conformable for the product AB, then the number of rows if A followed by the number of columns in B gives the order of product AB.
If two matrices A and B are conformable for the product AB, it is not necessary that they are also conformable for the product BA. If A and Bare square matrices of the same order, then they are compatible for the product AB as well as BA.
The element in row i and column j of the product AB are obtained by multiplying the elements in the ith row of A by the corresponding elements in the jth column of B and adding up the resulting products.
(i, j)th element of AB = Sum of the products of the elements of the ith row of A with the corresponding elements of the jth column of B.
Let A=\(\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\\ \end{bmatrix}\) and B =\(\begin{bmatrix}b_{11}&b_{12}\\b_{21}&b_{22}\\ \end{bmatrix}\)
Here, order of AB will be 2×2. If cij denotes the elements of AB,then
AB=\(\begin{bmatrix}c_{11}&c_{12}\\c_{21}&c_{22}\\ \end{bmatrix}\)
where, c11 = 1st row of A×1st column of B = a11b11 + a12a21
c12 = 1st row of A×2nd column of B = a11b12 + a12b22
c21 = 2nd row of A×1st column of B = a21b11 + a22b21
c22 = 2nd row of A×2nd column of B = a21b12 + a22b22
∴ AB = \(\begin{bmatrix}a_{11}b_{11} +a_{12}b_{21}&a_{11}b_{12} +a_{12}b_{22}\\a_{21}b_{11} +a_{22}b_{21}&a_{21}b_{12} +a_{22}b_{22}\\ \end{bmatrix}\)
Examples:
(1) Let A=\(\begin{bmatrix}3&5\\−1&3\\ \end{bmatrix}\) and B=\(\begin{bmatrix}3\\8\\ \end{bmatrix}\) be the two martices.
Since A is order 2×2 and B is order 2×1, therefore, AB is defined and it is a matrix of order 2×1.
AB =\(\begin{bmatrix}3&5\\−1&3\\ \end{bmatrix}\) \(\begin{bmatrix}3\\8\\ \end{bmatrix}\) = \(\begin{bmatrix}3.3 +5.8\\−1.3 +3.8\\ \end{bmatrix}\) = \(\begin{bmatrix}49\\21\\ \end{bmatrix}\)
Note that b is of order 2×1 and A is of order 2×2, therefore, BA is not defined.
(2) Let A =\(\begin{bmatrix}1&3\\5&0\\ \end{bmatrix}\) and B =\(\begin{bmatrix}3&2\\6&3\\ \end{bmatrix}\) be two matrices.
Since A is of order 2×2 and B is of order 2×2, therefore, AB is defined ad it is a matrix of order 2×2.
AB =\(\begin{bmatrix}1&3\\5&0\\ \end{bmatrix}\) \(\begin{bmatrix}3&2\\6&3\\ \end{bmatrix}\) = \(\begin{bmatrix}1.3+3.6&1.2+3.3\\5.3+0.6&5.2+0.3\\ \end{bmatrix}\) = \(\begin{bmatrix}21&11\\15&10\\ \end{bmatrix}\)
Also b is of order 2×2 and A is of 2×2, therefore, BA is defined and it is a matrix of order 2×2.
BA =\(\begin{bmatrix}3&2\\6&3\\ \end{bmatrix}\) \(\begin{bmatrix}1&3\\5&0\\ \end{bmatrix}\) = \(\begin{bmatrix}3.1+2.5&3.3+2.0\\6.1+3.5&6.3+3.0\\ \end{bmatrix}\) = \(\begin{bmatrix}13&9\\21&18\\ \end{bmatrix}\)
Observe that AB ≠ BA
Some properties of matrix multiplication
(a) Multiplication of matrices is, in general, not commutative, i.e. AB ≠ BA, in general.
(i) When the matrix AB is defined, it is not always necessary that BA can also be defined. For example, if the matrix A is m × n and the matrix B is in n × p, AB exists whereas BA does not exist because p ≠ m.
(ii) When both the matrices AB and BA are defined, it is not always necessary that they should be of the same type. For example, if the matrix A is m × n and the matrix B is n × m, both AB and BA exist but the matrix AB is m × m while the matrix BA is n × n.
(iii) When A and B are square matrices of the same order, both AB and BA exist, but they are not necessarily equal.
Let A =\(\begin{bmatrix}1&2\\3&4\\ \end{bmatrix}\) and B =\(\begin{bmatrix}1&2\\4&3\\ \end{bmatrix}\)
Then AB =\(\begin{bmatrix}1&2\\3&4\\ \end{bmatrix}\) \(\begin{bmatrix}1&2\\4&3\\ \end{bmatrix}\) =\(\begin{bmatrix}1+8&2+6\\3+16&6+12\\ \end{bmatrix}\) =\(\begin{bmatrix}9&8\\19&18\\ \end{bmatrix}\)
BA =\(\begin{bmatrix}1&2\\4&3\\ \end{bmatrix}\) \(\begin{bmatrix}1&2\\3&4\\ \end{bmatrix}\) =\(\begin{bmatrix}1+6&2+8\\4+9&8+12\\ \end{bmatrix}\) =\(\begin{bmatrix}7&10\\13&20\\ \end{bmatrix}\)
Thus AB ≠ BA.
(b) Multipication of matrices is associative i.e if A, B and C are matrices conformable for multiplication, then (AB)C = A(BC)
Let A =\(\begin{bmatrix}1&3\\−2&0\\ \end{bmatrix}\), B=\(\begin{bmatrix}2&4\\5&1\\ \end{bmatrix}\) and C=\(\begin{bmatrix}−3&4\\6&0\\ \end{bmatrix}\)
Then AB =\(\begin{bmatrix}1&3\\−2&0\\ \end{bmatrix}\) \(\begin{bmatrix}2&4\\5&1\\ \end{bmatrix}\) =\(\begin{bmatrix}2+15&4+3\\−4+0&−8+0\\ \end{bmatrix}\) =\(\begin{bmatrix}17&7\\−4&−8\\ \end{bmatrix}\)
(AB)C =\(\begin{bmatrix}17&7\\−4&−8\\ \end{bmatrix}\) \(\begin{bmatrix}−3&4\\6&0\\ \end{bmatrix}\) =\(\begin{bmatrix}17&7\\−4&−8\\ \end{bmatrix}\) \(\begin{bmatrix}−51+42&68+0\\12−48&−16+0\\ \end{bmatrix}\) \(\begin{bmatrix}−9&68\\−36&−16\\ \end{bmatrix}\)
BC =\(\begin{bmatrix}2&4\\5&1\\ \end{bmatrix}\) \(\begin{bmatrix}−3&4\\6&0\\ \end{bmatrix}\) =\(\begin{bmatrix}−6+24&8+0\\−15+6&20+0\\ \end{bmatrix}\) =\(\begin{bmatrix}18&8\\−9&20\\ \end{bmatrix}\)
A(BC) =\(\begin{bmatrix}1&3\\−2&0\\ \end{bmatrix}\) \(\begin{bmatrix}18&8\\−9&20\\ \end{bmatrix}\) =\(\begin{bmatrix}1&3\\−2&0\\ \end{bmatrix}\) \(\begin{bmatrix}18−27&8+60\\−36+0&−16+0\\ \end{bmatrix}\) = \(\begin{bmatrix}−9&68\\−36&−16\\ \end{bmatrix}\)
(c) Multiplication of matrices is distributive with respect to addition i.e. if A, B and C are matrices conformable for the requisite addition and multiplication, then A(B+C) = AB+AC and (A + B)C = AC+ BC.
Let A =\(\begin{bmatrix}1&1\\2&1\\1&2\\ \end{bmatrix}\), B =\(\begin{bmatrix}0&−1\\1&2\\ \end{bmatrix}\) and C =\(\begin{bmatrix}1&0\\2&1\\ \end{bmatrix}\)
Then B + C =\(\begin{bmatrix}0&−1\\1&2\\ \end{bmatrix}\) + \(\begin{bmatrix}1&0\\2&1\\ \end{bmatrix}\) =\(\begin{bmatrix}1&−1\\3&3\\ \end{bmatrix}\)
A(B+C) =\(\begin{bmatrix}1&1\\2&1\\1&2\\ \end{bmatrix}\) \(\begin{bmatrix}1&−1\\3&3\\ \end{bmatrix}\) =\(\begin{bmatrix}1+3&−1+3\\2+3&−2+3\\1+6&−1+6\\ \end{bmatrix}\) =\(\begin{bmatrix}4&2\\5&1\\7&5\\ \end{bmatrix}\) ........(i)
AC =\(\begin{bmatrix}1&1\\2&1\\1&2\\ \end{bmatrix}\) \(\begin{bmatrix}1&0\\2&1\\ \end{bmatrix}\) =\(\begin{bmatrix}1+2&0+1\\2+2&0+1\\1+4&0+2\\ \end{bmatrix}\) =\(\begin{bmatrix}3&1\\4&\\5&2\\ \end{bmatrix}\)
AB =\(\begin{bmatrix}1&1\\2&1\\1&2\\ \end{bmatrix}\) \(\begin{bmatrix}o&−1\\1&2\\ \end{bmatrix}\) =\(\begin{bmatrix}o+1&−1+2\\0+1&−2+2\\0+2&−1+4\\ \end{bmatrix}\) =\(\begin{bmatrix}1&1\\1&0\\2&3\\ \end{bmatrix}\)
AB + AC =\(\begin{bmatrix}1&1\\1&0\\2&3\\ \end{bmatrix}\) + \(\begin{bmatrix}3&1\\4&1\\5&2\\ \end{bmatrix}\) =\(\begin{bmatrix}4&2\\5&1\\7&5\\ \end{bmatrix}\) ...........(ii)
From (i) and (ii) A(B+C) = AB + AC.
Similarly we can verify (A+B)C = AC + BC.
(d) If A is a square matrix and I is a null matrix of the same order, then AI = IA = A
Let A =\(\begin{bmatrix}1&2\\3&4\\ \end{bmatrix}\) and I =\(\begin{bmatrix}1&0\\0&1\\ \end{bmatrix}\)
Then,AI =\(\begin{bmatrix}1&2\\3&4\\ \end{bmatrix}\) \(\begin{bmatrix}1&0\\0&1\\ \end{bmatrix}\) =\(\begin{bmatrix}1+0&0+2\\3+0&0+4\\ \end{bmatrix}\) =\(\begin{bmatrix}1&2\\3&4\\ \end{bmatrix}\)
IA| =\(\begin{bmatrix}1&0\\0&1\\ \end{bmatrix}\) \(\begin{bmatrix}1&2\\3&4\\ \end{bmatrix}\) =\(\begin{bmatrix}1+0&2+0\\0+3&0+4\\ \end{bmatrix}\) =\(\begin{bmatrix}1&2\\3&4\\ \end{bmatrix}\)
∴ AI = IA = A
Here, I is a multiplicative identity.
Note that results which are different from the result obtained in the case of numbers are given by multiplication of matrices.Examples of these result are given below:
(i) If AB is a null matrix, it does not imply that at least one of the matrices A and B must be a zero matrix.
Let A =\(\begin{bmatrix}1&1\\1&1\\ \end{bmatrix}\) and B =\(\begin{bmatrix}1&0\\−1&0\\ \end{bmatrix}\)
Then AB =\(\begin{bmatrix}1&1\\1&1\\ \end{bmatrix}\) \(\begin{bmatrix}1&0\\−1&0\\ \end{bmatrix}\) =\(\begin{bmatrix}1−1&0+0\\1−1&0+0\\ \end{bmatrix}\) = \(\begin{bmatrix}0&0\\0&0\\ \end{bmatrix}\)
Thus AB is a zero matrix though neither A or B is zero matrix.
(ii) Cancellation law for the multiplication of the matrices may not hold.
Let A =\(\begin{bmatrix}1&−1\\2&−2\\ \end{bmatrix}\) , B =\(\begin{bmatrix}4&5\\3&3\\ \end{bmatrix}\) and C =\(\begin{bmatrix}2&7\\1&5\\ \end{bmatrix}\).
Then AB =\(\begin{bmatrix}1&−1\\2&−2\\ \end{bmatrix}\) \(\begin{bmatrix}4&5\\3&3\\ \end{bmatrix}\) =\(\begin{bmatrix}4−3&5−3\\8−6&10−6\\ \end{bmatrix}\) =\(\begin{bmatrix}1&2\\2&4\\ \end{bmatrix}\).........(i)
AC =\(\begin{bmatrix}1&−1\\2&−2\\ \end{bmatrix}\) \(\begin{bmatrix}2&7\\1&5\\ \end{bmatrix}\) =\(\begin{bmatrix}2−1&7−5\\4−2&14−10\\ \end{bmatrix}\) =\(\begin{bmatrix}1&2\\2&4\\ \end{bmatrix}\) ..........(ii)
From (i) and (ii) AB = AC.
It follows that AB = AC does not necessarily imply that B = C. Thus cancellation law for the multiplication of matrices may not hold.
Determinants
We can call determinants of order 2 if we arrange 4 numbers in 2 rows and 2 columns between two vertical lines.
It is written as \(\begin{vmatrix}a&b\\c&d\\ \end{vmatrix}\).
and its value is defined as ad - bc. To get this value, we take the product of diagonal elements a and d and subtract from it te product of the diagonal elements can add.
We denote the determinant by Δ, read as delta.
∴ Δ =\(\begin{vmatrix}a&b\\c&d\\ \end{vmatrix}\) = ad - bc
If we arrange 9 numbers in 3 rows and 3 columns between two verticles lines, we get a determinant of order 3.
It is written as \(\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\\ \end{vmatrix}\) and its value is defined as
Δ = a\(\begin{vmatrix}e&f\\h&i\\ \end{vmatrix}\) -b\(\begin{vmatrix}d&f\\g&i\\ \end{vmatrix}\) +c\(\begin{vmatrix}d&e\\g&h\\ \end{vmatrix}\)
= a(ei - hf) -b(di - gf) +c(dh - ge)
= aei - ahf - bdi + bgf + cdh - cge
This is called the expansion of the determinant along its first row. To get this value, we start with the element a in the top left-hand corner. We delete the other element of row and column in which a occurs and multiply a by the second order determinant that remains. We proceed in the same way to get the determinants to be multiplied by B and. Thea, band c are taken to be alternatively positive and negative.
The determinant can be similarly expanded by the elements of the first column as
Δ =a\(\begin{vmatrix}e&f\\h&i\\ \end{vmatrix}\) -d\(\begin{vmatrix}b&c\\h&i\\ \end{vmatrix}\) +c\(\begin{vmatrix}b&c\\&f\\ \end{vmatrix}\)
= a(ei - hf) -d(bi - hc) +g(bf - ec)
= aei - ahf - dbi + dhc + gbf - gec, which gives the same value.
By this, we can assume the same values of the determinant by expanding it along row or column.
Matrix and its determinant
Determinant of the matrix is defined as the determinant which has the same elements in the same position as the matrix.
For example :
Let A = \(\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\\ \end{bmatrix}\) be the 2x2 square matrix.
Then the determinant of A is defined as,
|A| =\(\begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\\ \end{vmatrix}\)
= a11a22 - a21a12
Singular and non-singular matrices
A square matrix A is to be singular if its determinant |A| =0 and non-singular if |A|≠ 0.
If A =\(\begin{bmatrix}9&6\\8&6\\ \end{bmatrix}\) and B =\(\begin{bmatrix}6&2\\9&3\\ \end{bmatrix}\),
Then |A| =\(\begin{bmatrix}9&6\\8&6\\ \end{bmatrix}\) = 9.6 - 8.6= 54 - 48 = 6.
|B| =\(\begin{bmatrix}6&2\\9&3\\ \end{bmatrix}\) =6.3 - 9.2 =18 - 18 = 0
∴ Ais a non-singular matrix and B is a singular matrix.
The inverse of a Matrix
Let A be the square matrix. If there exists a matrix B such that AB =BA = I, then B is called the inverse of A. The inverse of A is denoted by A-1. Therefore, B = A-1
Thus, AA-1 = A-1A = I
Note that B is the inverse of A, then A is the inverse of B.
Requirements that are necessary for the existence of the inverse of a matrix:
(a) The matrix must be a square matrix.
In order that both the products AB and BA may be defined, either (i) A and B must be a square matrix of the same order or (ii) A and B must be of order m x n and n x m respectively.
In A and B are orders m x n and n x m respectively, then the order of AB will be m x m and the order of BA will be n x n. And hence AB ≠ BA.
So, for AB = BA, A and B must be a square matrix of the same order.
(b) The equation AB = BA = I must be satisfied.
(c) The matrix must be nonsingular.
The process of finding the inverse of a matrix.
For example :
|A| =\(\begin{vmatrix}2&1\\5&3\\ \end{vmatrix}\) = 2.3-5.1 =6-5 =1
Then, A-1 = \(\frac {1}{|A|}\)\(\begin{bmatrix}3&-1\\-5&2\\ \end{bmatrix}\)
=\(\frac{1}{1}\)\(\begin{bmatrix}3&-1\\-5&-2\\ \end{bmatrix}\)
=\(\begin{bmatrix}3&-1\\-5&2\\ \end{bmatrix}\)
Thus the inverse of a non-singular matrix A =\(\begin{bmatrix}a&c\\b&d\\ \end{bmatrix}\) can be found by the following steps:
(i) Find |A| = ad - bc. Note that if |A| = 0, we cannot find A-1
(ii) Exchange the elements in the leading diagonal and obtain \(\begin{bmatrix}d&.....\\.....&a\\ \end{bmatrix}\)
(iii) change the sign of the elements in the order diagonal and obtain \(\begin{bmatrix}d&-c\\-b&a\\ \end{bmatrix}\)
(iv) Find A-1 by using the following formula: A-1 =\(\frac{1}{|A|}\) \(\begin{bmatrix}d&-c\\-b&a\\ \end{bmatrix}\)
Some properties of inverses:
(a) The inverse of the product of two non -singular matrices is equal to the product of the inverse taken in the reverse order.
i.e. If A and B are non-singular square matrices of the same order, then AB is also non-singular and (AB)-1 = B-1A-1.
(b) The operations of transposing and inverting a non-singular matrix is commutative .i.e. (A')-1 = (A-1)'.
Let A =\(\begin{bmatrix}4&1\\7&2\\ \end{bmatrix}\)
Then A' = \(\begin{bmatrix}4&7\\1&2\\ \end{bmatrix}\)
Now, |A| =\(\begin{bmatrix}4&1\\7&2\\ \end{bmatrix}\) = 8 - 7 = 1 and
|A'| =\(\frac{1}{|A|}\) \(\begin{bmatrix}2&-1\\-7&4\\ \end{bmatrix}\) =\(\begin{bmatrix}2&-1\\-7&4\\ \end{bmatrix}\) and
(A-1)' = \(\begin{bmatrix}2&-7\\-1&4\\ \end{bmatrix}\) ........(i)
(A')-1 = \(\frac{1}{|A|}\) \(\begin{bmatrix}2&-7\\-1&4\\ \end{bmatrix}\) =\(\begin{bmatrix}2&-7\\-1&4\\ \end{bmatrix}\) .........(ii)
From (i) and (ii),(A')-1 = (A-1)'.
A solution of a system of linear equations by using inverse.
Suppose that we are required to solve the following 2 linear equtions in 2 unknown x and y.:
a11x + a12y =b1
a21 + a22 = b2
In matrix form the equations can be written as
\(\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\\ \end{bmatrix}\) \(\begin{bmatrix}x\\y\\ \end{bmatrix}\) =\(\begin{bmatrix}b_{1}\\b_{2}\\ \end{bmatrix}\)
Which can again be written as a single matrix equation
AX =B .............(i)
Where A =\(\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\\ \end{bmatrix}\), X =\(\begin{bmatrix}x\\y\\ \end{bmatrix}\) and B=\(\begin{bmatrix}b_{1}\\b_{2}\\ \end{bmatrix}\)
If A is non-singular, A-1exists.
∴ Pre-multiplying (i) by A-1, we have
A-1(AX) = A-1B
or (A-1A)X = A-1B
or IX = A-1B
or X = A-1B, which gives the solution of the equation.
Note: Since A-1 is a unique the above solution is also unique. If, however, the matrixAis singular,
i.e. |A| = 0, then A-1 does not exist and hence the above method fails to give any unique solution.
Things to remember
The matrix obtained by interchanging rows and columns is called transpose of the matrix. The transpose of matrix A is denoted by A'.
If the determinant of the matrix is zero, then the matrix is called the singular matrix.
If the determinant of the matrix is not zero, then the matrix is called the non-singular matrix.
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Videos for Matrix
Determinant of 3x3 matrix
Introduction to matrices
Inverse Matrix (part 1)
Inverse of 2x2 matrix
Inverse of 3x3 matrix
Inverting matrices (part 2)
Inverting Matrices (part 3)
Matrices and Gaussian Elimination
Matrix multiplication (part 1)
Matrix multiplication (part 2)
Questions and Answers
What do you mean by transpose of a matrix? Illustrate with example.
Transpose of matrix: A matrix obtained by interchanging rows and columns of the given matrix is called transpose of the matrix.
A = \(\begin{pmatrix} a & b& c\\ d& e& f\\ \end{pmatrix}\)
AT or AI =\(\begin{pmatrix} a & d \\ b & e \\ c & f \\ \end{pmatrix}\)
If \(\begin {pmatrix} 4 & 1 \\ 7 & -3\\ \end {pmatrix}\) \(\begin {pmatrix} 2 & -1 \\ 1 & 3\\ \end {pmatrix}\)= \(\begin {pmatrix} x & -1 \\ 11 & y\\ \end {pmatrix}\), find the value of x and y.
\(\begin {pmatrix} 4 & 1 \\ 7 & -3\\ \end {pmatrix}\) \(\begin {pmatrix} 2 & -1 \\ 1 & 3\\ \end {pmatrix}\)= \(\begin {pmatrix} x & -1 \\ 11 & y\\ \end {pmatrix}\)
or, \(\begin {pmatrix} 4 × 2 + 1× 1 & 4× -1 + 1× 3 \\ 7× 2 + 1× -3 & 7× -1 + 3 × -3 \\ \end {pmatrix}\) = \(\begin {pmatrix} x & -1 \\ 11 & y\\ \end {pmatrix}\)
or, \(\begin {pmatrix} 8 + 1 & -4 + 3 \\ 14 - 3 & -7 -9 \\ \end {pmatrix}\) = \(\begin {pmatrix} x & -1 \\ 11 & y\\ \end {pmatrix}\)
or, \(\begin {pmatrix} 9 & -1 \\ 11 & -16\\ \end{pmatrix}\) = \(\begin {pmatrix} x & -1 \\ 11 & y\\ \end {pmatrix}\)
∴ x = 9 and y = -16 Ans
Give A = \(\begin {bmatrix} -2 & 1 \\ -1 & 8 \\ \end {bmatrix}\), find:
(i) The transpose of A, (ii) The value of determinant A.
A = \(\begin {bmatrix} -2 & 1 \\ -1 & 8 \\ \end {bmatrix}\)
Transpose of A (AT) or AI= \(\begin {bmatrix} -2 & -1 \\ 1 & 8 \\ \end {bmatrix}\)
\(\begin {vmatrix} A \end {vmatrix}\) = \(\begin {vmatrix} -2 & 1 \\ -1 & 8 \\ \end {vmatrix}\) = -16 -(-1)= -16 + 1 = - 15 Ans
If M = \(\begin {bmatrix} 3 & 2 \\ 7 & -3 \\ \end {bmatrix}\) and N = \(\begin {bmatrix} 2 & -3 \\ -5 & 0 \\ \end {bmatrix}\) calculate the value of 2M + 3N.
Here,
M = \(\begin {bmatrix} 3 & 2 \\ 7 & -3 \\ \end {bmatrix}\) and N = \(\begin {bmatrix} 2 & -3 \\ -5 & 0 \\ \end {bmatrix}\)
2M = 2\(\begin {bmatrix} 3 & 2 \\ 7 & -3 \\ \end {bmatrix}\) =\(\begin {bmatrix} 6 & 4 \\ 14 & -6 \\ \end {bmatrix}\)
3N = 3\(\begin {bmatrix} 2& -3 \\ -5 & 0 \\ \end {bmatrix}\) =\(\begin {bmatrix} 6 & -9\\ -15 & 0\\ \end {bmatrix}\)
2M + 3N =\(\begin {bmatrix} 6 & 4 \\ 14 & -6 \\ \end {bmatrix}\) +\(\begin {bmatrix} 6 & -9\\ -15 & 0\\ \end {bmatrix}\) = \(\begin {bmatrix} 6+6 & 4-9 \\ 14-15 & -6+0\\ \end {bmatrix}\) =\(\begin {bmatrix} 12 & -5 \\ -1 & -6 \\ \end {bmatrix}\) Ans
If A = \(\begin {pmatrix} 2 & 3 \\ 5 & 0 \\ \end {pmatrix}\) and B = \(\begin {pmatrix} -3 & 4 \\ 2 & -5 \\ \end {pmatrix}\) then find the value of 2A - 3B.
Here,
A = \(\begin {pmatrix} 2 & 3 \\ 5 & 0 \\ \end {pmatrix}\) and B = \(\begin {pmatrix} -3 & 4 \\ 2 & -5 \\ \end {pmatrix}\)
2A - 3B = 2 \(\begin {pmatrix} 2 & 3 \\ 5 & 0 \\ \end {pmatrix}\) - 3 \(\begin {pmatrix} -3 & 4 \\ 2 & -5 \\ \end {pmatrix}\)
=\(\begin {pmatrix} 4 & 6 \\ 10 & 0 \\ \end {pmatrix}\) -\(\begin {pmatrix} -9 & 12 \\ 6 & -15 \\ \end {pmatrix}\)
=\(\begin {pmatrix} 4 + 9 & 6 - 12 \\ 10 - 6 & 0 + 15 \\ \end {pmatrix}\)
=\(\begin {pmatrix} 13 & -6\\ 4 & 15 \\ \end {pmatrix}\) Ans
If A = \(\begin {pmatrix}1 & -3 \\ -2 & 4\\ \end {pmatrix}\) and B = \(\begin {pmatrix}4 & 2 \\ 3 & -5\\ \end {pmatrix}\), write the transpose of A + B.
Here,
A = \(\begin {pmatrix}1 & -3 \\ -2 & 4\\ \end {pmatrix}\) and B = \(\begin {pmatrix}4 & 2 \\ 3 & -5\\ \end {pmatrix}\)
A + B =\(\begin {pmatrix}1 & -3 \\ -2 & 4\\ \end {pmatrix}\) +\(\begin {pmatrix}4 & 2 \\ 3 & -5\\ \end {pmatrix}\)
=\(\begin {pmatrix}1+4 & -3+2 \\ -2+3 & 4-5\\ \end {pmatrix}\)
=\(\begin {pmatrix}5 & -1 \\ 1 & -1\\ \end {pmatrix}\)
(A + B)T =\(\begin {pmatrix}5 & 1 \\ -1 & -1\\ \end {pmatrix}\) Ans
A2×3 amd B3×3 are two matrices. Are AB and BA defined? Write with reasons.
A2×3 and B3×3
AB is defined because the number of column of A is equal to the number of rows of B.
A2×3 and B3×3= AB2×3
BA is not defined because the number of column of B is not equal to the number of rows of A.
B3×3 and A2×3 , BA does not exist.
Prove that matrix A = \(\begin {pmatrix} -3 & 2 \\ -9 & 6 \\ \end {pmatrix}\) has no inverse.
A = \(\begin {pmatrix} -3 & 2 \\ -9 & 6 \\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) = \(\begin {vmatrix} -3 & 2 \\ -9 & 6 \\ \end {vmatrix}\) = -18 + 18 = 0
Hence, A-1 does not exist.
∴ Given matrix has no inverse matrix. Proved
Construct a matrix of order 2×2 whose (ij)th element is given by aij = 2\(\vec i\) - 3\(\vec j\).
a11 = 2(1) - 3(1) = 2 - 3 = -1
a12= 2(1) - 3(2) = 2 - 6 = -4
a21= 2(2) - 3(1) = 4 - 3 = 1
a22= 2(2) - 3(2) = 4 - 6 = -2
∴ Required matrix = \(\begin {pmatrix}-1 & -4 \\ 1 & -2 \\ \end{pmatrix}\)2×2 Ans
Find the matrix x, y = \(\begin {pmatrix} 3 & 2 \\ 2 & 4 \\ \end {pmatrix}\) and 2x + y = \(\begin {pmatrix} 1 & 0 \\ -3 & 2 \\ \end {pmatrix}\).
2x + y =\(\begin {pmatrix} 1 & 0 \\ -3 & 2 \\ \end {pmatrix}\)
or, 2x + \(\begin {pmatrix} 3 & 2 \\ 2 & 4 \\ \end {pmatrix}\) = \(\begin {pmatrix} 1 & 0 \\ -3 & 2 \\ \end {pmatrix}\)
or, 2x =\(\begin {pmatrix} 1 & 0 \\ -3 & 2 \\ \end {pmatrix}\) -\(\begin {pmatrix} 3 & 2 \\ 2 & 4 \\ \end {pmatrix}\)
or, 2x =\(\begin {pmatrix} 1-3 & 0-2 \\ -3-2 & 2-4 \\ \end {pmatrix}\)
or, 2x =\(\begin {pmatrix} -2 & -2 \\ -5 & -2\\ \end {pmatrix}\)
or, x = \(\begin {pmatrix} \frac {-2}{2} & \frac {-2}{2} \\ \frac {-5}{2} & \frac {-2}{2} \\ \end {pmatrix}\)
∴ x =\(\begin {pmatrix} -1 & -1 \\ \frac {-5}{2} & -1\\ \end {pmatrix}\) Ans
Construct a 3×3 matrix A whose elements are given by aij = 3i + 2j.
Here,
aij = 3i + 2j
a11 = 3× 1 + 2 × 1 = 3 + 2 = 5
a12= 3× 1 + 2 × 2= 3 + 4 = 8
a13= 3× 1 + 2 × 3 = 3 + 6 = 9
a21 = 3× 2 + 2 × 1 = 6 + 2 = 8
a22= 3× 2 + 2 × 2 = 6 + 4 = 10
a23= 3× 2 + 2 × 3 = 6 + 6 = 12
a31 = 3× 3 + 2 × 1 = 9 + 2 = 11
a32= 3× 3 + 2 × 2 = 9 + 4 = 13
a33= 3× 3 + 2 × 3 = 9 + 6 = 15
∴ The required matrix is A = \(\begin {bmatrix}5 & 7 &9 \\ 8 & 10 & 12 \\ 11 & 13 & 15 \\ \end {bmatrix}\) Ans
If A + B = \(\begin {bmatrix} 7 & 0 \\ 2 & 5 \\ \end {bmatrix}\) and A - B = \(\begin {bmatrix} 3 & 0 \\ 0 & 3 \\ \end {bmatrix}\).
If:
A + B = \(\begin {bmatrix} 7 & 0 \\ 2 & 5 \\ \end {bmatrix}\) ..................(1)
A - B = \(\begin {bmatrix} 3& 0 \\ 0 & 3 \\ \end {bmatrix}\)....................(2)
Adding (1) and (2), we get:
A + B + A - B = \(\begin {bmatrix} 7 & 0 \\ 2 & 5 \\ \end {bmatrix}\) + \(\begin {bmatrix} 3& 0 \\ 0 & 3 \\ \end {bmatrix}\)
or, 2A = \(\begin {bmatrix} 10 & 0 \\ 2 & 8 \\ \end {bmatrix}\)
or, \(\frac 12\) × 2A = \(\frac 12\) \(\begin {bmatrix} 10 & 0 \\ 2 & 8 \\ \end {bmatrix}\)
∴ A = \(\begin {bmatrix} 5 & 0 \\ 1 & 4 \\ \end {bmatrix}\)
Now,
From (1)
A + B =\(\begin {bmatrix} 7 & 0 \\ 2 & 5 \\ \end {bmatrix}\)
or,\(\begin {bmatrix} 5 & 0 \\ 1 & 4 \\ \end {bmatrix}\) + B=\(\begin {bmatrix} 7 & 0 \\ 2 & 5 \\ \end {bmatrix}\)
or, B =\(\begin {bmatrix} 7 & 0 \\ 2 & 5 \\ \end {bmatrix}\) -\(\begin {bmatrix} 5 & 0 \\ 1 & 4 \\ \end {bmatrix}\)
or, B =\(\begin {bmatrix} 7-5 & 0-0 \\ 2-1 & 5-4 \\ \end {bmatrix}\)
∴ B =\(\begin {bmatrix} 2 & 0 \\ 1 & 1 \\ \end {bmatrix}\)
If A = \(\begin {bmatrix} 3 & -5 \\ 2 & 0 \\ \end {bmatrix}\) and B = \(\begin {bmatrix} -1 & 4 \\ 0 & -3 \\ \end {bmatrix}\), find the determinant of 2AT - 3BT.
Here,
A = \(\begin {bmatrix} 3 & -5 \\ 2 & 0 \\ \end {bmatrix}\) and B = \(\begin {bmatrix} -1 & 4 \\ 0 & -3 \\ \end {bmatrix}\)
AT =\(\begin {bmatrix} 3 & 2\\ -5 & 0 \\ \end {bmatrix}\) and B =\(\begin {bmatrix} -1 & 0 \\ 4 & -3 \\ \end {bmatrix}\)
Now,
2AT - 3BT = 2 \(\begin {bmatrix} 3 & 2\\ -5 & 0 \\ \end {bmatrix}\) - 3\(\begin {bmatrix} -1 & 0 \\ 4 & -3 \\ \end {bmatrix}\)
=\(\begin {bmatrix} 6 & 4 \\ -10 & 0 \\ \end {bmatrix}\) -\(\begin {bmatrix} -3 & 0 \\ 12 & -9 \\ \end {bmatrix}\)
=\(\begin {bmatrix} 6+3 & 4-0 \\-10-12 & 0+9 \\ \end {bmatrix}\)
=\(\begin {bmatrix} 9 & 4 \\ -22 & 9 \\ \end {bmatrix}\)
Again,
\(\begin {vmatrix} 2A^T- 3B^T \\ \end {vmatrix}\) =\(\begin {vmatrix} 9 & 4 \\ -22 & 9 \\ \end {vmatrix}\) = 81 + 88 = 169 Ans
Define inverse of a matrix with example.
Two matrices A and B are said to be inverse to each other, if AB = I = BA then B is the inverse of A. i.e. A-1 = B. Similarly, A is called the inverse of B i.e. B-1 = A.
If A = \(\begin {pmatrix} 1 & x \\ 0 & 1 \\ \end {pmatrix}\) , B = \(\begin {pmatrix} 3 & 2 \\ -5 & 4 \\ \end {pmatrix}\) and determinant of A2 - BT = 8, find the value of x.
A = \(\begin {pmatrix} 1 & x \\ 0 & 1 \\ \end {pmatrix}\) , B =\(\begin {pmatrix} 3 & 2 \\ -5 & 4 \\ \end {pmatrix}\)
A2 - BT
=\(\begin {pmatrix} 1 & x \\ 0 & 1 \\ \end {pmatrix}\) .\(\begin {pmatrix} 1 & x \\ 0 & 1 \\ \end {pmatrix}\) -\(\begin {pmatrix} 3 & -5 \\ 2 & 4 \\ \end {pmatrix}\)
=\(\begin {pmatrix} 1 + 0 & x + x \\ 0 + 0 & 0 + 1 \\ \end {pmatrix}\) -\(\begin {pmatrix} 3 &-5 \\ 2 & 4 \\ \end {pmatrix}\)
=\(\begin {pmatrix} 1 & 2x \\ 0 & 1 \\ \end {pmatrix}\) -\(\begin {pmatrix} 3 & -5 \\ 2 & 4 \\ \end {pmatrix}\)
=\(\begin {pmatrix} -2 & 2x + 5 \\ -2 & -3 \\ \end {pmatrix}\)
Now,
\(\begin {vmatrix} -2 & 2x + 5 \\ -2 & -3 \\ \end {vmatrix}\) = 8
or, 6 + 4x + 10 = 8
or, 4x = 8 - 16
or, x = \(\frac {-8}{4}\)
∴ x = -2 Ans
If \(\begin {pmatrix} 2x + 3 \\ y - 2 \\ 3z - y \\ \end {pmatrix}\) = \(\begin {pmatrix} 5 & -8 \\ 12 & -40 \\ -3 & 4 \\ \end {pmatrix}\) \(\begin {pmatrix} 7 \\ 2 \\ \end {pmatrix}\), find the value of x, y and z.
\(\begin {pmatrix} 2x + 3 \\ y - 2 \\ 3z - y \\ \end {pmatrix}\) = \(\begin {pmatrix} 5 & -8 \\ 12 & -40 \\ -3 & 4 \\ \end {pmatrix}\)\(\begin {pmatrix} 7 \\ 2 \\ \end {pmatrix}\)
or,\(\begin {pmatrix} 2x + 3 \\ y - 2 \\ 3z - y \\ \end {pmatrix}\) = \(\begin {pmatrix} 35 - 16 \\ 84 - 80 \\ -21 + 8 \\ \end {pmatrix}\) = \(\begin {pmatrix} 19 \\ 4 \\ -13\\ \end {pmatrix}\)
Taking the corresponding elements of the equal matrix
2x + 3 = 19
or, 2x = 19 - 3
or, x = \(\frac {16}{2}\)
∴ x = 8
y - 2 = 4
or, y = 4 + 2
∴ y = 6
3z - 1 = - 13
or, 3z = - 13 + 1
or, z = \(\frac {-12}{3}\)
∴ z = -4
∴ x = 8, y = 6 and z = -4 Ans
Define the transpose of a matrix.If A = \(\begin {pmatrix} 1 & 3 & 5 \\ -2 & 4 & 7 \\ \end {pmatrix}\), find AT.
A matrix obtained by interchanging rows and columns of the given matrix is called transpose of the matrix.
A = \(\begin {pmatrix} 1 & 3 & 5 \\ -2 & 4 & 7 \\ \end {pmatrix}\)
AT =A = \(\begin {pmatrix} 1 & -2 \\ 3 & 4 \\ 5 & 7\\ \end {pmatrix}\) Ans
If \(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) = \(\begin {pmatrix} 3 & 5\\ 4 & 6\\ \end {pmatrix}\) \(\begin {pmatrix} 1\\ 2\\ \end {pmatrix}\), find the value of x and y.
\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) = \(\begin {pmatrix} 3 & 5\\ 4 & 6\\ \end {pmatrix}\)\(\begin {pmatrix} 1\\ 2\\ \end {pmatrix}\)
or,\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} 3 + 10 \\ 4 + 12 \\ \end {pmatrix}\)
or,\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} 13\\ 16\\ \end {pmatrix}\)
∴ x = 13 and y = 16 Ans
Find the value of \(\begin {pmatrix} x\\ y\\ \end {pmatrix}\). If \(\begin {pmatrix} -1 & 0\\ 0 & -2\\ \end {pmatrix}\) \(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) = \(\begin {pmatrix} -2\\ 4\\ \end {pmatrix}\).
Here,
\(\begin {pmatrix} -1 & 0\\ 0 & -2\\ \end {pmatrix}\)\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} -2\\ 4\\ \end {pmatrix}\)
or,\(\begin {pmatrix} -x + 0\\ 0 - 2\\ \end {pmatrix}\) =\(\begin {pmatrix} -2\\ 4\\ \end {pmatrix}\)
or,\(\begin {pmatrix} -x\\ -2y\\ \end {pmatrix}\) =\(\begin {pmatrix} -2\\ 4\\ \end {pmatrix}\)
Taking corresponding elements of the equal matrix
-x = -2
∴ x = 2
-2y = 4
or, y = \(\frac {4}{-2}\)
∴ y = -2
∴ x = 2 & y = -2 Ans
Find the value of x and y: \(\begin {pmatrix} 1 & 1\\ 3 & y\\ \end {pmatrix}\) \(\begin {pmatrix} x\\ 1\\ \end {pmatrix}\) = \(\begin {pmatrix} 4\\ 1\\ \end {pmatrix}\).
\(\begin {pmatrix} 1 & 1\\ 3 & y\\ \end {pmatrix}\) \(\begin {pmatrix} x\\ 1\\ \end {pmatrix}\) =\(\begin {pmatrix} 4\\ 1\\ \end {pmatrix}\)
or, \(\begin {pmatrix} x + 1\\ 3x + y\\ \end {pmatrix}\) =\(\begin {pmatrix} 4\\ 1\\ \end {pmatrix}\)
Taking corresponding elements of the equal matrix:
x + 1 = 4
or, x = 4 -1
∴ x = 3
3x + y = 1
or, 3 × 3 + y = 1
or, 9 + y = 1
or, y = 1 - 9
∴ y = -8
∴ x = 2 and y = -8 Ans
If A = \(\begin {pmatrix} 3 & 4\\ -2 & 5\\ \end {pmatrix}\) and B = \(\begin {pmatrix} 2 & 1\\ -1 & 3\\ \end {pmatrix}\), find the determinant of 3A - 4B.
Given matrix A = \(\begin {pmatrix} 3 & 4\\ -2 & 5\\ \end {pmatrix}\) and B =\(\begin {pmatrix} 2 & 1\\ -1 & 3\\ \end {pmatrix}\)
3A - 4B = 3\(\begin {pmatrix} 3 & 4\\ -2 & 5\\ \end {pmatrix}\) - 4\(\begin {pmatrix} 3 & 4\\ -2 & 5\\ \end {pmatrix}\)
=\(\begin {pmatrix} 9 & 12\\ -6 & 15\\ \end {pmatrix}\) +\(\begin {pmatrix} -8 & -4\\ 4 & -12\\ \end {pmatrix}\)
=\(\begin {pmatrix} 9 - 8 & 12 - 4\\ -6 + 4 & 15 - 12\\ \end {pmatrix}\)
=\(\begin {pmatrix} 1 & 8\\ -2 & 3\\ \end {pmatrix}\)
\(\begin {vmatrix} 3A - 4B\\ \end {vmatrix}\)
=\(\begin {vmatrix} 1 & 8\\ -2 & 3\\ \end {vmatrix}\)
= 1× 3 - 8 (-2)
= 3 + 16
= 19 Ans
If P = \(\begin {pmatrix} 1 & -1\\ -1 & 1\\ \end {pmatrix}\) and Q = \(\begin {pmatrix} 2 & 3\\ 1 & 4\\ \end {pmatrix}\); find the determinant of PQ.
Here,
P = \(\begin {pmatrix} 1 & -1\\ -1 & 1\\ \end {pmatrix}\) and Q =\(\begin {pmatrix} 2 & 3\\ 1 & 4\\ \end {pmatrix}\)
PQ = \(\begin {pmatrix} 1 × 2 + 1 × -1 & 1 × 3 + 4 × -1\\ -1 × 2 + 1 × 1& -1 × 3 + 1 × 4\ \ \end {pmatrix}\)
=\(\begin {pmatrix} 2-1 & 3-4\\ -2+1 & -3+4\\ \end {pmatrix}\)
=\(\begin {pmatrix} 1 & -1\\ -1 & 1\\ \end {pmatrix}\)
\(\begin {vmatrix} PQ\\ \end {vmatrix}\) =\(\begin {vmatrix} 1 & -1\\ -1 & 1\\ \end {vmatrix}\)
= -1× 1 - (-1)× (-1)
= 1 -1
= 0 Ans
If P = \(\begin {pmatrix} 1 & 3\\ 2 & -1\\ \end {pmatrix}\) and Q = \(\begin {pmatrix} 1 & 2\\ 3 & -4\\ \end {pmatrix}\), find the determinant of 2P - 3Q.
2P - 3Q = 2 \(\begin {pmatrix} 1 & 3\\ 2 & -1\\ \end {pmatrix}\) - 3\(\begin {pmatrix} 1 & 2\\ 3 & -4\\ \end {pmatrix}\)
=\(\begin {pmatrix} 2 & 6\\ 4 & -2\\ \end {pmatrix}\) -\(\begin {pmatrix} 3 & 6\\ 9 & -12\\ \end {pmatrix}\)
=\(\begin {pmatrix} 2-3 & 6-6\\ 4-9 & -2+12\\ \end {pmatrix}\)
=\(\begin {pmatrix} -1 & 0\\ -5 & 10\\ \end {pmatrix}\)
\(\begin {vmatrix} 2P - 3Q\\ \end {vmatrix}\) =\(\begin {vmatrix} -1 & 0\\ -5 & 10\\ \end {vmatrix}\) = -1× 10 - (-5)× 0 = -10 + 0 = -10 Ans
If P = \(\begin {pmatrix} 2 & 0\\ -1 & 3\\ \end {pmatrix} \) and Q = \(\begin {pmatrix} 2 & 4\\ -6 & 2\\ \end {pmatrix}\), find the determinant of 5P - \(\frac 12\)Q + 2I.
Here,
P = \(\begin {pmatrix} 2 & 0\\ -1 & 3\\ \end {pmatrix} \), Q = \(\begin {pmatrix} 2 & 4\\ -6 & 2\\ \end {pmatrix}\) and I = \(\begin {pmatrix} 1 & 0\\ 0 & 1\\ \end {pmatrix} \)
5P - \(\frac 12\)Q + 2I = 5 \(\begin {pmatrix} 2 & 0\\ -1 & 3\\ \end {pmatrix} \) - \(\frac 12\)\(\begin {pmatrix} 2 & 4\\ -6 & 2\\ \end {pmatrix}\) +2 \(\begin {pmatrix} 1 & 0\\ 0 & 1\\ \end {pmatrix} \) = \(\begin {pmatrix} 10 & 0\\ -5 & 15\\ \end {pmatrix} \) - \(\begin {pmatrix} 1 & 2\\ -3 & 1\\ \end {pmatrix} \) + \(\begin {pmatrix} 2 & 0\\ 0 & 2\\ \end {pmatrix} \) = \(\begin {pmatrix} 12 & 0\\ -5 & 17\\ \end {pmatrix} \) - \(\begin {pmatrix} 1 & 2\\ -3 & 1\\ \end {pmatrix} \) = \(\begin {pmatrix} 11 & -2\\ -2 & 16\\ \end {pmatrix} \)
\(\begin {vmatrix} 5P - (\frac 12)Q + 2I\\ \end {vmatrix} \) =\(\begin {vmatrix} 11 & -2\\ -2 & 16\\ \end {vmatrix} \) = 16× 11 - (-2)× (-2) = 176 - 4 = 172 Ans
If A = \(\begin {pmatrix} 3 & 5\\ -6 & 0\\ \end {pmatrix}\) and B = \(\begin {pmatrix} 18 & -6\\ 12 & 3\\ \end {pmatrix}\) find the determinant of 3A - \(\frac 13\)B.
Here,
A = \(\begin {pmatrix} 3 & 5\\ -6 & 0\\ \end {pmatrix}\) and B =\(\begin {pmatrix} 18 & -6\\ 12 & 3\\ \end {pmatrix}\)
3A - \(\frac 13\)B
= 3\(\begin {pmatrix} 3 & 5\\ -6 & 0\\ \end {pmatrix}\) - \(\frac 13\)\(\begin {pmatrix} 18 & -6\\ 12 & 3\\ \end {pmatrix}\)
=\(\begin {pmatrix} 9 & 15\\ -18 & 0\\ \end {pmatrix}\) -\(\begin {pmatrix} 6 & -2\\ 4 & 1\\ \end {pmatrix}\)
= \(\begin {pmatrix} 9-6 & 15+2\\ -18-4 & 0-1\\ \end {pmatrix}\)
=\(\begin {pmatrix} 3 & 17\\ -22 & -1\\ \end {pmatrix}\)
\(\begin {vmatrix} 3A - (\frac 13) B\\ \end {vmatrix}\) =\(\begin {pmatrix} 3 & 17\\ -22 & -1\\ \end {pmatrix}\) = -3 +374 = 371 Ans
Which matrix multiplies to the matrix\(\begin{pmatrix} 1 & 1\\ 3 & 4\\ \end {pmatrix}\) to get a matrix\(\begin{pmatrix} 4 & 5\\ 6 & 2\\ \end {pmatrix}\)?
Let: M =\(\begin{pmatrix} a & c\\ b & d\\ \end {pmatrix}\)
\(\begin{pmatrix} a & c\\ b & d\\ \end {pmatrix}\)\(\begin{pmatrix} 1 & 1\\ 3 & 4\\ \end {pmatrix}\) =\(\begin{pmatrix} 4 & 5\\ 6 & 2\\ \end {pmatrix}\)
or,\(\begin{pmatrix} a×1+c×3 & a×1+c×4\\ b×1+d×3 & b×1+d×4\\ \end {pmatrix}\) =\(\begin{pmatrix} 4 & 5\\ 6 & 2\\ \end {pmatrix}\)
or,\(\begin{pmatrix} a+3c & a+4c\\ b+3d & b+4d\\ \end {pmatrix}\) =\(\begin{pmatrix} 4 & 5\\ 6 & 2\\ \end {pmatrix}\)
Equating both sides,
a + 3c = 4
or, a = 4 - 3c .........................(1)
a + 4c = 5................................(2)
Putting the value of a in equation (2)
4 - 3c + 4c = 5
or, 4 + c = 5
or, c = 5 - 4
∴ c = 1
Putting the value of c in equation (1)
∴ a = 4 - 3c = 4 - 3× 1 = 4 - 3 = 1
b + 3d = 6
b = 6 - 3d..........................(3)
b + 4d = 2.........................(4)
Putting the value of b in equation (4)
6 - 3d + 4d = 2
or, 6 + d = 2
or, d = 2 - 6
∴ d = -4
Puting the value of d in equation (3)
∴ b = 6 - 3 × (-4) = 6 + 12 = 18
∴ \(\begin{pmatrix} a & b\\ c & d\\ \end {pmatrix}\) = \(\begin{pmatrix} 1 & 1\\ 18 & -4\\ \end {pmatrix}\) Ans
If A = \(\begin {pmatrix} -1 & 4\\ 0 & -3\\ \end {pmatrix}\) and B = \(\begin {pmatrix} 3 & -5\\ 2 & 0\\ \end {pmatrix}\) find the determinant of 2AT - 3BT.
Here,
A = \(\begin {pmatrix} -1 & 4\\ 0 & -3\\ \end {pmatrix}\) and B =\(\begin {pmatrix} 3 & -5\\ 2 & 0\\ \end {pmatrix}\)
AT =\(\begin {pmatrix} -1 & 0\\ 4 & -3\\ \end {pmatrix}\) and BT =\(\begin {pmatrix} 3 & 2\\ -5 & 0\\ \end {pmatrix}\)
2AT - 3BT
= 2\(\begin {pmatrix} -1 & 0\\ 4 & -3\\ \end {pmatrix}\) - 3\(\begin {pmatrix} 3 & 2\\ -5 & 0\\ \end {pmatrix}\)
=\(\begin {pmatrix} -2 & 0\\ 8 & -6\\ \end {pmatrix}\) - \(\begin {pmatrix} 9 & 6\\ 15 & 0\\ \end {pmatrix}\)
=\(\begin {pmatrix} -2-9 & 0-6\\ 8-15 & -6-0\\ \end {pmatrix}\)
=\(\begin {pmatrix} -11 & -6\\ -7 & -6\\ \end {pmatrix}\)
\(\begin {vmatrix} 2A^T - 3B^T\\ \end {vmatrix}\) =\(\begin {vmatrix} -11 & -6\\ -7 & -6\\ \end {vmatrix}\) = -11 ×(-6) - (-7) ×(-6) = 66 - 42 = 24 Ans
If P = \(\begin {pmatrix} -3 & 0\\ 1 & -2\\ \end {pmatrix}\) and Q = \(\begin {pmatrix} 1 & -2\\ 3 & 4\\ \end {pmatrix}\), find the determinant of 5P - 2Q - 3I.
Here,
P = \(\begin {pmatrix} -3 & 0\\ 1 & -2\\ \end {pmatrix}\) and Q = \(\begin {pmatrix} 1 & -2\\ 3 & 4\\ \end {pmatrix}\) and I =\(\begin {pmatrix} 1& 0\\ 0 & 1\\ \end {pmatrix}\)
5P - 2Q - 3I
= 5 \(\begin {pmatrix} -3 & 0\\ 1 & -2\\ \end {pmatrix}\) - 2\(\begin {pmatrix} 1 & -2\\ 3 & 4\\ \end {pmatrix}\) -3\(\begin {pmatrix} 1& 0\\ 0 & 1\\ \end {pmatrix}\)
=\(\begin {pmatrix}-15 & 0\\ 5 & -10\\ \end {pmatrix}\) -\(\begin {pmatrix} 2 & -4\\ 6 & 8\\ \end {pmatrix}\) -\(\begin {pmatrix} 3 & 0\\ 0 & 3\\ \end {pmatrix}\)
=\(\begin {pmatrix} -15 - 2 - 3 & 0 - (-4) - 0\\ 5 - 6 -0 & -10 - 8 - 3\\ \end {pmatrix}\)
=\(\begin {pmatrix} -20 & 4\\ -1 & -21\\ \end {pmatrix}\)
\(\begin {vmatrix} 5P - 2Q - 3I\\ \end {vmatrix}\) =\(\begin {vmatrix} -20 & 4\\ -1 & -21\\ \end {vmatrix}\) = -20 × (-21) - 4 × (-1) = 420 + 4 = 424 Ans
If M = \(\begin {bmatrix} 3 & -2\\ 1 & 2\\ \end {bmatrix}\) and N = \(\begin {bmatrix} 1 & 3\\ 4 & -1\\ \end {bmatrix}\), find the determinant of 2M + 3N.
Here,
M = \(\begin {bmatrix} 3& -2\\ 1 & 2\\ \end {bmatrix}\) and N = \(\begin {bmatrix} 1 & 3\\ 4 & -1\\ \end {bmatrix}\)
2M + 3N
= 2\(\begin {bmatrix} 3& -2\\ 1 & 2\\ \end {bmatrix}\) + 3\(\begin {bmatrix} 1 & 3\\ 4 & -1\\ \end {bmatrix}\)
=\(\begin {bmatrix} 6 & -4\\ 2 & 4\\ \end {bmatrix}\) +\(\begin {bmatrix} 3 & 9\\ 12 & -3\\ \end {bmatrix}\)
=\(\begin {bmatrix} 6 + 3 & -4 + 9\\ 2 + 12 & 4 - 3\\ \end {bmatrix}\)
=\(\begin {bmatrix} 9 & 5\\ 14 & 1\\ \end {bmatrix}\)
\(\begin {vmatrix} 2M + 3N\\ \end {vmatrix}\) =\(\begin {bmatrix} 9 & 5\\ 14 & 1\\ \end {bmatrix}\) = 9× 1 - 5× 14 = 9 - 70 = -61 Ans
If P = \(\begin {pmatrix} 1 & -2\\ 3 & -1\\ \end {pmatrix}\) and Q = \(\begin {pmatrix} 3 & -4\\ 2 & 1\\ \end {pmatrix}\), find the determinant of 3P - 2Q.
Here,
P = \(\begin {pmatrix} 1& -2\\ 3 & -1\\ \end {pmatrix}\) and Q =\(\begin {pmatrix} 3 & -4\\ 2 & 1\\ \end {pmatrix}\)
3P - 2Q
= 3 \(\begin {pmatrix} 1& -2\\ 3 & -1\\ \end {pmatrix}\) - 2 \(\begin {pmatrix} 3 & -4\\ 2 & 1\\ \end {pmatrix}\)
= \(\begin {pmatrix} 3 & -6\ 9 & -3\\ \end {pmatrix}\) -\(\begin {pmatrix} 6 & -8\\ 4 & 2\\ \end {pmatrix}\)
=\(\begin {pmatrix} 3-6 & -6+8\\ 9-4 & -3-2\\ \end {pmatrix}\)
=\(\begin {pmatrix} -3 & 2\\ 5 & -5\\ \end {pmatrix}\)
\(\begin {vmatrix} 3P - 2Q\\ \end {vmatrix}\) =\(\begin {pmatrix} -3 & 2\\ 5 & -5\\ \end {pmatrix}\) = -3× (-5) - 5× 2 = 15 - 10 = 5 Ans
If A = \(\begin {pmatrix} 4 & 6\\ x & 3\\ \end {pmatrix}\), B = \(\begin {pmatrix} -4 & 5\\ 7 & 8\\ \end {pmatrix}\) and the determinant of A - B - 5I is 14, calculate the value of x.
Here,
A = \(\begin {pmatrix} 4 & 6\\ x & 3\\ \end {pmatrix}\) and B = \(\begin {pmatrix} -4 & 5\\ 7 & 8\\ \end {pmatrix}\) and I =\(\begin {pmatrix} 1 & 0\\ 0 & 1\\ \end {pmatrix}\)
Now,
A - B - 5I
=\(\begin {pmatrix} 4 & 6\\ x & 3\\ \end {pmatrix}\) -\(\begin {pmatrix} -4 & 5\\ 7 & 8\\ \end {pmatrix}\) - 5\(\begin {pmatrix} 1 & 0\\ 0 & 1\\ \end {pmatrix}\)
=\(\begin {pmatrix} 4-(-4) & 6-5\\ x-7 & 3-8\\ \end {pmatrix}\) -\(\begin {pmatrix} 5 & 0\\ 0 & 5\\ \end {pmatrix}\)
=\(\begin {pmatrix} 8 & 1\\ x-7 & -5\\ \end {pmatrix}\) -\(\begin {pmatrix} 5 & 0\\ 0 & 5\\ \end {pmatrix}\)
=\(\begin {pmatrix} 8-5 & 1-0\\ x-7-0 & -5-5\\ \end {pmatrix}\)
=\(\begin {pmatrix} 3 & 1\\ x - 7 & -10\\ \end {pmatrix}\)
We have,
\(\begin {vmatrix} A - B - 2I\\ \end {vmatrix}\) = 14
or,\(\begin {vmatrix} 3 & 1\\ x - 7 & -10\\ \end {vmatrix}\) = 14
or, 3× -10 - 1× (x-7) = 14
or, -30 - x + 7 = 14
or, -x = 14 - 7 + 30
or, -x = 37
∴ x = -37 Ans
If the inverse of matrix A = \(\begin {pmatrix} 1 & -2\\ 0 & x\\ \end {pmatrix}\) and B = \(\begin {pmatrix} 1 & 4\\ y & 2\\ \end {pmatrix}\), determine the values of x and y.
Here,
A = \(\begin {pmatrix} 1 & -2\\ 0 & x\\ \end {pmatrix}\) and B =\(\begin {pmatrix} 1 & 4\\ y & 2\\ \end {pmatrix}\)
If A and B are inverse matrix, then
AB = I, where I =\(\begin {pmatrix} 1 & 0\\ 0 & 1\\ \end {pmatrix}\)
or,\(\begin {pmatrix} 1 & -2\\ 0 & x\\ \end {pmatrix}\)\(\begin {pmatrix} 1 & 4\\ y & 2\\ \end {pmatrix}\) =\(\begin {pmatrix} 1 & 0\\ 0 & 1\\ \end {pmatrix}\)
or,\(\begin {pmatrix} 1 × 1 + y × (-2) & 1 × 4 + 2 × (-2) \\ 0 × 1 + y × x & 0 × 4 + 2 × x \\ \end {pmatrix}\) =\(\begin {pmatrix} 1 & 0\\ 0 & 1\\ \end {pmatrix}\)
or,\(\begin {pmatrix} -2y + 1 & 0\\ xy & 2x\\ \end {pmatrix}\) =\(\begin {pmatrix} 1 & 0\\ 0 & 1\\ \end {pmatrix}\)
Taking corresponding elements:
-2y + 1 = 1
or, -2y = 1 - 1
or, y = \(\frac 02\)
∴ y = 0
Similarly,
2x = 1
∴ x= \(\frac 12\)
∴ x = \(\frac 12\) and y = 0 Ans
If A = \(\begin {pmatrix} 7 & 4\\ 3 & 2\\ \end {pmatrix}\) then find A-1.
Given,
A = \(\begin {pmatrix} 7 & 4\\ 3 & 2\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) = \(\begin {vmatrix} 7 & 4\\ 3 & 2\\ \end {vmatrix}\) = 7× 2 - 4× 3 = 14 - 12 = 2≠ 0.
It is possible to find A-1.
A-1 = \(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj (A)
A = \(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) \(\begin {pmatrix} 2 & -4\\ -3 & 7\\ \end {pmatrix}\) = \(\frac 12\)\(\begin {pmatrix} 2 & -4\\ -3 & 7\\ \end {pmatrix}\) = \(\begin {pmatrix} (\frac 22) & (\frac {-4}{2})\\ (\frac {-3}{2}) & (\frac 72)\\ \end {pmatrix}\) =\(\begin {pmatrix} 1 & -2\\ (\frac {-3}{2} & (\frac 72)\\ \end {pmatrix}\)
∴ A-1 =\(\begin {pmatrix} 1 & -2\\ (\frac {-3}{2} & (\frac 72)\\ \end {pmatrix}\) Ans
If the inverse of the matrix \(\begin {pmatrix} 2m & 7\\ 5 & 9\\ \end {pmatrix}\) is the matrix \(\begin {pmatrix} 9 & n\\ -5 & 4\\ \end {pmatrix}\). Calculate the value of m and n.
Let:
A =\(\begin {pmatrix} 2m & 7\\ 5 & 9\\ \end {pmatrix}\) and A-1=\(\begin {pmatrix} 9 & n\\ -5 & 4\\ \end {pmatrix}\)
We know that,
AA-1 = I, where I =\(\begin {pmatrix} 1 & 0\\ 0 & 1\\ \end {pmatrix}\)
or,\(\begin {pmatrix} 2m & 7\\ 5 & 9\\ \end {pmatrix}\) \(\begin {pmatrix} 9 & n\\ -5 & 4\\ \end {pmatrix}\) =\(\begin {pmatrix} 1 & 0\\ 0 & 1\\ \end {pmatrix}\)
or,\(\begin {pmatrix} 18m - 35 & 2mn + 28\\ 45 - 45 & 5n + 36\\ \end {pmatrix}\) =\(\begin {pmatrix} 1 & 0\\ 0 & 1\\ \end {pmatrix}\)
or,\(\begin {pmatrix} 18m - 35 & 2mn + 28\\ 0 & 5n + 36\\ \end {pmatrix}\) =\(\begin {pmatrix} 1 & 0\\ 0 & 1\\ \end {pmatrix}\)
Taking the corresponding elements both sides
18m - 35 = 1
or, 18 m = 1 + 35
or, m = \(\frac {36}{18}\)
∴ m = 2
5n + 36 = 1
or, 5n = 1 - 36
or, n = \(\frac {-35}{5}\)
∴ n = -7
∴ m = 2 and n = -7 Ans
If \(\begin {vmatrix} x-1 & x-2\\ x & x-3\\ \end {vmatrix}\) = 0, find the value of x.
Here,
\(\begin {vmatrix} x-1 & x-2\\ x & x-3\\ \end {vmatrix}\) = 0
or, (x - 1) (x - 3) - x(x - 2) = 0
or, x2 - 4x + 3 - x2 + 2x = 0
or, - 2x + 3 = 0
or, 2x = 3
∴ x = \(\frac 32\) Ans
If A = \(\begin {bmatrix} 3 & -5\\ -4 & 2\\ \end {bmatrix}\) show that A2 - 5A = 14I, where I is an identity matrix.
Here,
A = \(\begin {bmatrix} 3 & -5\\ -4 & 2\\ \end {bmatrix}\)
L.H.S. = A2 - 5A = A× A - 5× A
= \(\begin {bmatrix} 3 & -5\\ -4 & 2\\ \end {bmatrix}\)× \(\begin {bmatrix} 3 & -5\\ -4 & 2\\ \end {bmatrix}\) - 5 \(\begin {bmatrix} 3 & -5\\ -4 & 2\\ \end {bmatrix}\)
= \(\begin {bmatrix} 3×3 + (-5)×(-4) & 3×(-5) + (-5)×2 \\ -4×3 + 2×(-4) & -4×(-5) + 2×2\\ \end {bmatrix}\) + \(\begin {bmatrix} -5×3 & -5×(-5)\\ -5×(-4) & -5×2\\ \end {bmatrix}\)
= \(\begin {bmatrix} 9+20 & -15-10\\ -12-8 & 20+4\\ \end {bmatrix}\) + \(\begin {bmatrix} -15 & 25\\ 20 & -10\\ \end {bmatrix}\)
= \(\begin {bmatrix} 29 & -25\\ -20 & 24\\ \end {bmatrix}\) + \(\begin {bmatrix} -15 & 25\\ 20 & -10\\ \end {bmatrix}\)
= \(\begin {bmatrix} 29-15 & -25+25\\ -20+20 & 24-10\\ \end {bmatrix}\)
= \(\begin {bmatrix} 14 & 0\\ 0 & 14\\ \end {bmatrix}\)
= 14 \(\begin {bmatrix} 1 & 0\\ 0 & 1\\ \end {bmatrix}\)
= 14I Ans
If the inverse of the matrix \(\begin {pmatrix} x & 2x - 9\\ -y & 3\\ \end {pmatrix}\) is the matrix \(\begin {pmatrix} 3 & 5\\ y & x\\ \end {pmatrix}\). Find the value of x and y.
We know that,
\(\begin {pmatrix} x & 2x - 9\\ -y & 3\\ \end {pmatrix}\)\(\begin {pmatrix} 3& 5\\ y & x\\ \end {pmatrix}\) = I where I =\(\begin {pmatrix} 1 & 0\\ 0 & 1\\ \end {pmatrix}\)
or,\(\begin {pmatrix} 3x + y(2x - 9) & 5x + x(2x - 9)\\ -3y + 3y & -5y + 3x\\ \end {pmatrix}\) =\(\begin {pmatrix} 1 & 0\\ 0 & 1\\ \end {pmatrix}\)
or,\(\begin {pmatrix} 3x + 2xy - 9y & 5x + 2x^2 - 9x\\ 0 & -5y + 3x\\ \end {pmatrix}\) =\(\begin {pmatrix} 1 & 0\\ 0 & 1\\ \end {pmatrix}\)
or,\(\begin {pmatrix} 3x + 2xy - 9y & 2x^2 - 4x\\ 0 & 3x -5y\\ \end {pmatrix}\) =\(\begin {pmatrix} 1 & 0\\ 0 & 1\\ \end {pmatrix}\)
Taking the corresponding element of the equal matrix
2x2 - 4x = 0
2x(x - 2) = 0
Either: 2x = 0 ∴x = 0
Or, x - 2 = 0 ∴x = 2
3x - 5y = 1
or, 3x - 1 = 5y
or, y = \(\frac {3x - 1}{5}\)
If x = 0 then y = \(\frac {3 × 0 - 1}{5}\) = -\(\frac 15\)
If x = 2 then y =\(\frac {3 × 2 - 1}{5}\) = 1
∴ x = 0 or 2
y = -\(\frac 15\) or 1 Ans
if I is the unit matrix of order 2×2 and 3A - 4I = 5\(\begin {pmatrix} 1 & -2\\ 0 & -1\\ \end {pmatrix}\) then find the matrix A.
Here,
3A - 4I = 5\(\begin {pmatrix} 1 & -2\\ 0 & -1\\ \end {pmatrix}\)
or, 3A - 4\(\begin {pmatrix} 1 & 0\\ 0 & 1\\ \end {pmatrix}\) =\(\begin {pmatrix} 5 & -10\\ 0 & -5\\ \end {pmatrix}\)
or, 3A -\(\begin {pmatrix} 4 & 0\\ 0 & 4\\ \end {pmatrix}\) =\(\begin {pmatrix} 5 & -10\\ 0 & -5\\ \end {pmatrix}\)
or, 3A =\(\begin {pmatrix} 5 & -10\\ 0 & -5\\ \end {pmatrix}\) +\(\begin {pmatrix} 4 & 0\\ 0 & 4\\ \end {pmatrix}\)
or, 3A = \(\begin {pmatrix} 5 + 4 & -10 + 0\\ 0 + 0 & -5 + 4\\ \end {pmatrix}\)
or, 3A =\(\begin {pmatrix} 9& -10\\ 0 & -1\\ \end {pmatrix}\)
or, A = \(\begin {pmatrix} (\frac 93) & (\frac {-10}{3})\\ (\frac {0}{3}) & (\frac {-1}{3})\\ \end {pmatrix}\)
∴ A =\(\begin {pmatrix} 3 & (\frac {-10}{3})\\ 0 & (\frac {-1}{3})\\ \end {pmatrix}\) Ans
If A = \(\begin {pmatrix} 3 & 5\\ 1 & 2\\ \end {pmatrix}\), find the determinant of A2 + 5A-1 - 14I, where I is a 2×2 matrix.
Here,
A =\(\begin {pmatrix} 3 & 5\\ 1 & 2\\ \end {pmatrix}\) and I =\(\begin {pmatrix} 1 & 0\\ 0 & 1\\ \end {pmatrix}\)
A2 + 5A-1 - 14I
=\(\begin {pmatrix} 3 & 5\\ 1 & 2\\ \end {pmatrix}\)⋅\(\begin {pmatrix} 3 & 5\\ 1 & 2\\ \end {pmatrix}\) + 5\(\begin {pmatrix} 3 & 5\\ 1 & 2\\ \end {pmatrix}\)-1 -14\(\begin {pmatrix} 1 & 0\\ 0 & 1\\ \end {pmatrix}\)
=\(\begin {pmatrix} 9+5 & 15+10\\ 3+2 & 5+4\\ \end {pmatrix}\) + \(\frac {5}{6-5}\)\(\begin {pmatrix} 2 & -5\\ -1 & 3\\ \end {pmatrix}\) -\(\begin {pmatrix} 14 & 0\\ 0 & 14\\ \end {pmatrix}\)
=\(\begin {pmatrix} 14 & 25\\ 5 & 9\\ \end {pmatrix}\) +\(\begin {pmatrix} 10 & -25\\ -5 & 15\\ \end {pmatrix}\) -\(\begin {pmatrix} 14 & 0\\ 0 & 14\\ \end {pmatrix}\)
=\(\begin {pmatrix} 24 & 0\\ 0 & 24\\ \end {pmatrix}\) -\(\begin {pmatrix} 14 & 0\\ 0 & 14\\ \end {pmatrix}\)
=\(\begin {pmatrix} 10 & 0\\ 0 & 10\\ \end {pmatrix}\)
\(\begin {vmatrix} A^2 + 5A^{-1} - 14A\\ \end {vmatrix}\) =\(\begin {vmatrix} 10 & 0\\ 0 & 10\\ \end {vmatrix}\) = 100 - 0 = 100 Ans
If A = \(\begin {pmatrix} 4 & 2\\ -1 & 1\\ \end {pmatrix}\) prove that A2 - 5A + 6I = 0 where I is an 2×2 unit matrix.
Here,
A = \(\begin {pmatrix} 4 & 2\\ -1 & 1\\ \end {pmatrix}\), I = \(\begin {pmatrix} 1 & 0\\ 0 & 1\\ \end {pmatrix}\)
L.H.S.
= A2 - 5A + 6I
= \(\begin {pmatrix}4 & 2\\ -1 & 1\\ \end {pmatrix}\)⋅ \(\begin {pmatrix}4 & 2\\ -1 & 1\\ \end {pmatrix}\) - 5 \(\begin {pmatrix}4 & 2\\ -1 & 1\\ \end {pmatrix}\) + 6 \(\begin {pmatrix} 1 & 0\\ 0 & 1\\ \end {pmatrix}\)
= \(\begin {pmatrix} 16-2 & 8+2\\ -4-1 & -2+1\\ \end {pmatrix}\) - \(\begin {pmatrix} 20 & 10\\ -5 & 5\\ \end {pmatrix}\) + \(\begin {pmatrix} 6 & 0\\ 0 & 6\\ \end {pmatrix}\)
= \(\begin {pmatrix} 14 & 10\\ -5 & -1\\ \end {pmatrix}\) - \(\begin {pmatrix} 20 & 10\\ -5 & 5\\ \end {pmatrix}\) + \(\begin {pmatrix} 6 & 0\\ 0 & 6\\ \end {pmatrix}\)
= \(\begin {pmatrix} 14 - 20 + 6 & 10 - 10 + 0\\ 5 + 5 + 0 & -1 - 5 + 6\\ \end {pmatrix}\)
= \(\begin {pmatrix} 0 & 0\\ 0 & 0\\ \end {pmatrix}\)
= 0
= R.H.S Proved
if A-1 = \(\begin {pmatrix} 4 & -13\\ 1 & -3\\ \end {pmatrix}\), find the matrix A.
Here,
A-1 =\(\begin {pmatrix} 4 & -13\\ 1 & -3\\ \end {pmatrix}\)
Adj1 (A) =\(\begin {pmatrix} -3 & -1\\ 13 & 4\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) = -12 + 13 = 1
A = \(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj.(A) = \(\frac 11\)\(\begin {pmatrix} -3 & 13\\ -1 & 4\\ \end {pmatrix}\) =\(\begin {pmatrix} -3 & 13\\ -1 & 4\\ \end {pmatrix}\) Ans
If M = \(\begin {pmatrix} 4 & 0\\ 0 & 5\\ \end {pmatrix}\), find a matrix P such that MP = \(\begin {pmatrix} 1 & 2\\ 2 & 4\\ \end {pmatrix}\).
Let:
P = \(\begin {pmatrix} a & b\\ c & d\\ \end {pmatrix}\)
MP =\(\begin {pmatrix} 1 & 2\\ 2 & 4\\ \end {pmatrix}\)
or,\(\begin {pmatrix} 4 & 0\\ 0 & 5\\ \end {pmatrix}\)⋅\(\begin {pmatrix} a & b\\ c & d\\ \end {pmatrix}\) =\(\begin {pmatrix} 1 & 2\\ 2 & 4\\ \end {pmatrix}\)
or,\(\begin {pmatrix} 4a + 0 & 4b + 0\\ 0 + 5c & 0 + 5d\\ \end {pmatrix}\) =\(\begin {pmatrix} 1 & 2\\ 2 & 4\\ \end {pmatrix}\)
or,\(\begin {pmatrix} 4a & 4b\\ 5c & 5d\\ \end {pmatrix}\) =\(\begin {pmatrix} 1 & 2\\ 2 & 4\\ \end {pmatrix}\)
Taking corresponding elements of the equal matrix:
4a = 1
∴ a = \(\frac 14\)
4b = 2
∴ b = \(\frac 12\)
5c = 2
∴ c = \(\frac 25\)
5d = 4
∴ d = \(\frac 45\)
∴ P =\(\begin {pmatrix} \frac 14 & \frac 12\\ \frac 25 & \frac 45\\ \end {pmatrix}\) Ans
Define inverse of a matrix. Find the inverse A-1 to matrix A if A = \(\begin {pmatrix} 2 & 1\\ 3 & 4\\ \end {pmatrix}\).
Two matrices A and B are said to be inverse to each other, if AB = I = BA then B is the inverse of A. i.e. A-1 = B. Similarly, A is called the inverse of B i.e. B-1 = A.
A =\(\begin {pmatrix} 2 & 1\\ 3 & 4\\ \end {pmatrix}\)
A-1= \(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj.(A)
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 2 & 1\\ 3 & 4\\ \end {vmatrix}\) = 8 - 3 = 5
A-1 = \(\frac 15\)\(\begin {pmatrix} 4 & -1\\ -3 & 2\\ \end {pmatrix}\) =\(\begin {pmatrix} \frac 45 & -\frac 15\\ -\frac35 & \frac 25\\ \end {pmatrix}\) Ans
If A = \(\begin {pmatrix} 1 & 1\\ 5 & 3\\ \end {pmatrix}\) find the A-1.
Here,
A =\(\begin {pmatrix} 1 & 1\\ 5 & 3\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 1 & 1\\ 5 & 3\\ \end {vmatrix}\) = 3 - 5 = -2
A-1=\(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj.(A)
= \(\frac {1}{-2}\)\(\begin {vmatrix} 3 & -1\\ -5 &1\\ \end {vmatrix}\)
=\(\begin {pmatrix} -\frac 32 & \frac 12\\ \frac 52 & -\frac 12\\ \end {pmatrix}\) Ans
Solve by matrix method:
4x - 3y = 11
3x + 7y = -1
Here,
4x - 3y = 11..........................................(1)
3x + 7y = -1..........................................(2)
Given equations in matrix form:
\(\begin {pmatrix} 4 & -3\\ 3 & 7\\ \end {pmatrix}\)\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} 11\\ -1\\ \end {pmatrix}\)
We know,
AX = B
where. A =\(\begin {pmatrix} 4 & -3\\ 3 & 7\\ \end {pmatrix}\), X =\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) and B =\(\begin {pmatrix} 11\\ -1\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 4 & -3\\ 3 & 7\\ \end {vmatrix}\) = 7× 4 - 3× -3 = 28 + 9 = 37≠ 0
Hence, it has unique solution.
X = A-1B where A-1 =\(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj.(A) = \(\frac {1}{37}\)\(\begin {pmatrix} 7 & 3\\ -3 & 4\\ \end {pmatrix}\)
X = A-1B = \(\frac {1}{37} \begin {pmatrix} 7 & 3\\ -3 & 4\\ \end {pmatrix}\)\(\begin {pmatrix} 11\\ -1\\ \end {pmatrix}\)
= \(\frac {1}{37} \begin {pmatrix} 77 & -3\\ -33 & -4\\ \end {pmatrix}\)
= \(\frac {1}{37} \begin {pmatrix} 74\\ 37\\ \end {pmatrix}\)\(\begin {pmatrix} \frac {74}{37}\\ -\frac {37}{37}\\ \end {pmatrix}\)
= \(\begin {pmatrix} 2\\ -1\\ \end {pmatrix}\)
∴ x = 2 and y = -1 Ans
Solve by matrix method:
2x + 3y - 18 = 0
3x - 2y - 1 = 0
Here,
2x + 3y = 18......................(1)
3x - 2y = 1 .........................(2)
Given equation in matrix form:
\(\begin {pmatrix} 2 & 3\\ 3 & -2\\ \end {pmatrix}\)\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) = \(\begin {pmatrix} 18\\ 1\\ \end {pmatrix}\), AX = B
where, A =\(\begin {pmatrix} 2 & 3\\ 3 & -2\\ \end {pmatrix}\), X =\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) and B =\(\begin {pmatrix} 18\\ 1\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 2 & 3\\ 3 & -2\\ \end {vmatrix}\) = 2× (-2) - 3× 3 = -4 - 9 = -13≠ 0
It has a unique solution.
A-1=\(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj.A =\(\begin {pmatrix} -2 & -3\\ -3 & 2\\ \end {pmatrix}\)
X = A-1B
or, X = \(\frac {1}{-13}\)\(\begin {pmatrix} -2 & -3\\ -3 & 2\\ \end {pmatrix}\)\(\begin {pmatrix} 18\\ 1\\ \end {pmatrix}\)
or, X = \(\frac {1}{-13}\)\(\begin {pmatrix} -2 × 18 + 1 × -3\\ -3 × 18 + 2 × 1\\ \end {pmatrix}\)
or, X = \(\frac {1}{-13}\)\(\begin {pmatrix} -36 - 3\\ -54 + 2\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {-39}{-13}\\ \frac {-52}{-13}\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} 3\\ 4\\ \end {pmatrix}\)
∴ x = 3 and y = -4 Ans
Solve by matrix method:
3x - 2y = 5
x + y = 5
Here,
3x - 2y = 5 ......................(1)
x + y = 5 ...........................(2)
Given equation in matrix form:
\(\begin {pmatrix} 3 & -2\\ 1 & 1\\ \end {pmatrix}\)\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} 5\\ 5\\ \end {pmatrix}\)
We know, AX = B
where, A =\(\begin {pmatrix} 3 & -2\\ 1 & 1\\ \end {pmatrix}\), X =\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) and B =\(\begin {pmatrix} 5\\ 5\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 3 & -2\\ 1 & 1\\ \end {vmatrix}\) = 3× 1 -1× -2 = 5≠ 0
It has a unique solution.
A-1=\(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj. A = \(\frac 15\)\(\begin {pmatrix} 1 & 2\\ -1 & 3\\ \end {pmatrix}\) =\(\begin {pmatrix} \frac 15 & \frac 25\\ \frac {-1}{5} & \frac 35\\ \end {pmatrix}\)
X = A-1B =\(\begin {pmatrix} \frac 15 & \frac 25\\ \frac {-1}{5} & \frac 35\\ \end {pmatrix}\)\(\begin {pmatrix} 5\\ 5\\ \end {pmatrix}\) =\(\begin {pmatrix} 1 + 2\\ -1 + 3\\ \end {pmatrix}\) =\(\begin {pmatrix} 3\\ 2\\ \end {pmatrix}\)
∴ x = 3 and y = 2 Ans
Solve by matrix method:
3x + 5y = 21
2x + 3y = 13
Here,
3x + 5y = 21................................(1)
2x + 3y = 13................................(2)
Given equation in matrix form:
\(\begin {pmatrix} 3 & 5\\ 2 & 3\\ \end {pmatrix}\)\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} 21\\ 13\\ \end {pmatrix}\)
We know, AX = B
where, A =\(\begin {pmatrix} 3 & 5\\ 2 & 3\\ \end {pmatrix}\), X =\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) and B =\(\begin {pmatrix} 21\\ 13\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 3 & 5\\ 2 & 3\\ \end {vmatrix}\) = 3× 3 - 5× 2 = 9 -10 = -1≠ 0
It has a unique solution.
X = A-1B
A-1=\(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj. A = \(\frac {1}{-1}\)\(\begin {pmatrix} 3 & -5\\ -2 & 3\\ \end {pmatrix}\) =\(\begin {pmatrix} -3 & 5\\ 2 & -3\\ \end {pmatrix}\)
X = A-1B
=\(\begin {pmatrix} -3 & 5\\ 2 & -3\\ \end {pmatrix}\)\(\begin {pmatrix} 21\\ 13\\ \end {pmatrix}\)
=\(\begin {pmatrix} -63 + 65\\ 42 - 39\\ \end {pmatrix}\)
= \(\begin {pmatrix} 2\\ 3\\ \end {pmatrix}\)
∴ x = 2 and y = 3 Ans
Solve by matrix method:
2x + 3y = 5
5x - 2y = 3
Here,
2x + 3y = 5.........................(1)
5x - 2y = 3..........................(2)
Given equation in matrix form:
\(\begin {pmatrix} 2 & 3\\ 5 & -2\\ \end {pmatrix}\) \(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} 5\\ 3\\ \end {pmatrix}\) or, AX = B
where, A =\(\begin {pmatrix} 2 & 3\\ 5 & -2\\ \end {pmatrix}\), X =\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) and B =\(\begin {pmatrix} 5\\ 3\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 2 & 3\\ 5 & -2\\ \end {vmatrix}\) = -4 - 15 = -19≠ 0
It has a unique solution.
X = A-1B
A-1=\(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj. A = \(\frac {1}{-19}\)\(\begin {pmatrix} -2 & -3\\ -5 & 2\\ \end {pmatrix}\)
X = A-1B
or, X =\(\frac {1}{-19}\)\(\begin {pmatrix} -2 & -3\\ -5 & 2\\ \end {pmatrix}\)\(\begin {pmatrix} 5\\ 3\\ \end {pmatrix}\)
or, X =\(\frac {1}{-19}\)\(\begin {pmatrix} -10 - 9\\ -25 + 6\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {-19}{-19}\\ \frac {-19}{-19}\\ \end {pmatrix}\)
∴ X =\(\begin {pmatrix} 1\\ 1\\ \end {pmatrix}\)
∴ x = 1 and y = 1 Ans
Solve by matrix method:
3x- 5y = 3
4x + 3y = 4
Here,
3x - 5y = 3..........................(1)
4x + 3y = 4.........................(2)
Given equation in matrix form;
\(\begin {pmatrix} 3 & -5\\ 4 & 3\\ \end {pmatrix}\)\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} 3\\ 4\\ \end {pmatrix}\) or, AX = B
where, A = \(\begin {pmatrix} 3 & -5\\ 4 & 3\\ \end {pmatrix}\), X =\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) and B =\(\begin {pmatrix} 3\\ 4\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 3 & -5\\ 4 & 3\\ \end {vmatrix}\) = 9 + 20 = 29≠ 0
It has a unique solution.
X = A-1B
X =\(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\)\(\begin {pmatrix} 3 & 5\\ -4 & 3\\ \end {pmatrix}\)\(\begin {pmatrix} 3\\ 4\\ \end {pmatrix}\)
or, X = \(\frac {1}{29}\) \(\begin {pmatrix} 3 × 3 + 5 × 4\\ -4 × 3 + 3 × 4\\ \end {pmatrix}\)
or, X = \(\frac {1}{29}\) \(\begin {pmatrix} 9 + 20\\ -12 + 12\\ \end {pmatrix}\)
or, X = \(\frac {1}{29}\) \(\begin {pmatrix} 29\\ 0\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {29}{29}\\ \frac {0}{29}\\ \end {pmatrix}\)
∴ X =\(\begin {pmatrix} 1\\ 0\\ \end {pmatrix}\)
∴ x = 1 and y = 0 Ans
Solve by matrix method:
2x + y = 3
3x + 2y = 2
Here,
2x + y = 3..................(1)
3x + 2y = 2...............(2)
Given equation in matrix form:
\(\begin {pmatrix} 2 & 1\\ 3 & 2\\ \end {pmatrix}\)\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} 3\\ 2\\ \end {pmatrix}\), AX = B
where, A =\(\begin {pmatrix} 2 & 1\\ 3 & 2\\ \end {pmatrix}\), X =\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) and B =\(\begin {pmatrix} 3\\ 2\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 2 & 1\\ 3 & 2\\ \end {vmatrix}\) = 2 × 2 - 3 × 1 = 4 - 3 = 1 ≠ 0
It has a unique solution.
A-1=\(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj. A = \(\frac 11\) \(\begin {pmatrix} 2 & -1\\ -3 & 2\\ \end {pmatrix}\)
=\(\begin {pmatrix} 2 & -1\\ -3 & 2\\ \end {pmatrix}\)
X = A-1B
or, X =\(\begin {pmatrix} 2 & -1\\ -3 & 2\\ \end {pmatrix}\)\(\begin {pmatrix} 3\\ 2\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} 2 × 3 -1 × 2\\ -3 × 3 + 2 × 2\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} 6 - 2\\ -9 + 4\\ \end {pmatrix}\)
∴ X = \(\begin {pmatrix} 4\\ -5\\ \end {pmatrix}\)
∴ x = 4 and y = -5 Ans
Given: A = \(\begin {bmatrix} a & b\\ 0 & c\\ \end {bmatrix}\), B = \(\begin {bmatrix} 2 & 0\\ 0 & 3\\ \end {bmatrix}\) and AB = A + B, find the values of a, b and c.
Here,
A = \(\begin {bmatrix} a & b\\ 0 & c\\ \end {bmatrix}\) and B =\(\begin {bmatrix} 2 & 0\\ 0 & 3\\ \end {bmatrix}\)
AB =\(\begin {bmatrix} a & b\\ 0 & c\\ \end {bmatrix}\)\(\begin {bmatrix} 2 & 0\\ 0 & 3\\ \end {bmatrix}\) =\(\begin {bmatrix} a × 2 + b × 0 & a × 0 + b × 3\\ 0 × 2 + c × 0& 0 × 0 + c × 3\\ \end {bmatrix}\) =\(\begin {bmatrix} 2a & 3b\\ 0 & 3c\\ \end {bmatrix}\)
A + B =\(\begin {bmatrix} a & b\\ 0 & c\\ \end {bmatrix}\) +\(\begin {bmatrix} 2 & 0\\ 0 & 3\\ \end {bmatrix}\) =\(\begin {bmatrix} a +2 & b\\ 0 & c + 3\\ \end {bmatrix}\)
From Question,
\(\begin {bmatrix} 2a & 3b\\ 0 & 3c\\ \end {bmatrix}\) =\(\begin {bmatrix} a +2 & b\\ 0 & c + 3\\ \end {bmatrix}\)
Taking the corresponding elements of the equal matrix,
a + 2 = 2a
or, 2a - a = -2
∴ a = -2
3b = b
or, 3b - b = 0
or, 2b = 0
∴ b = 0
3c = c + 3
or, 3c - c = 3
or, 2c = 3
∴ c = \(\frac 32\)
∴ a = 2, b = 0 and c = \(\frac 32\) Ans
Solve by matrix method:
2x + 3y + 4 = 0
-5x + 4y + 13 = 0
Here,
2x + 3y = -4....................................(1)
-5x + 4y = -13...............................(2)
Given equation in matrix form:
\(\begin {pmatrix} 2 & 3\\ -5 & 4\\ \end {pmatrix}\)\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} -4\\ -13\\ \end {pmatrix}\), AX = B
where, A =\(\begin {pmatrix} 2 & 3\\ -5 & 4\\ \end {pmatrix}\), X =\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) and B =\(\begin {pmatrix} -4\\ -13\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 2 & 3\\ -5 & 4\\ \end {vmatrix}\) = 2 × 4 - (-5) × 3 = 8 + 15 = 23 ≠ 0
It has a unique solution.
A-1= \(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj. A = \(\frac {1}{23}\) \(\begin {pmatrix} 4 & -3\\ 5 & 2\\ \end {pmatrix}\) =\(\begin {pmatrix} \frac {4}{23} & \frac {-3}{23}\\ \frac {5}{23} & \frac {2}{23}\\ \end {pmatrix}\)
We know that,
AX = B
X = A-1B
or, X =\(\begin {pmatrix} \frac {4}{23} & \frac {-3}{23}\\ \frac {5}{23} & \frac {2}{23}\\ \end {pmatrix}\)\(\begin {pmatrix} -4\\ -13\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {-16}{23} + \frac {39}{23}\\ \frac {-20}{23} - \frac {26}{23}\\ \end {pmatrix}\)
or, X = \(\begin {pmatrix} \frac {-16 + 39}{23}\\ \frac {-20 - 26}{23}\\ \end {pmatrix}\)
or, X = \(\begin {pmatrix} \frac {23}{23}\\ \frac {-46}{23}\\ \end {pmatrix}\)
∴ X =\(\begin {pmatrix} 1\\ -2\ \end {pmatrix}\)
∴ x = 1 and y = -2Ans
Solve the following given equations by matrix method:
4x + 3y = 5
y - 3x = -7
Here,
4x + 3y = 5.........................................(1)
y - 3x = -7...........................................(2)
Given equation in matrix form:
\(\begin {pmatrix} 4 &3\\ -3 & 1\\ \end {pmatrix}\)\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} 5\\ -7\\ \end {pmatrix}\), AX =B
where, A =\(\begin {pmatrix} 4 & 3\\ -3 & 1\\ \end {pmatrix}\), X =\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) and B =\(\begin {pmatrix} 5\\ -7\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 4 & 3\\ -3 & 1\\ \end {vmatrix}\) = 4 × 1 - 3 × -3 = 4 + 9 = 13 ≠ 0
It has unique solution.
A-1=\(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj.A = \(\frac {1}{13}\)\(\begin {pmatrix} 1 & -3\\ 3 & 4\\ \end {pmatrix}\) =\(\begin {pmatrix} \frac {1}{13} & \frac {-3}{13}\\ \frac {3}{13} & \frac {4}{13}\\ \end {pmatrix}\)
We know that,
AX = B
X = A-1B
or, X =\(\begin {pmatrix} \frac {1}{13} & \frac {-3}{13}\\ \frac {3}{13} & \frac {4}{13}\\ \end {pmatrix}\)\(\begin {pmatrix} 5\\ -7\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {5}{13} + \frac {21}{13}\\ \frac {15}{13} - \frac {28}{13}\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {5 + 21}{13}\\ \frac {15 - 28}{13}\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {26}{13}\\ \frac {-13}{13}\\ \end {pmatrix}\)
∴ X =\(\begin {pmatrix} 2\\ -1\\ \end {pmatrix}\)
∴ x = 2 and y = -1 Ans
Solve the given equation by matrix method:
2x - y = 1
2y + x = 3
Here,
2x - y = 1...............................(1)
x + 2y = 3..............................(2)
Given equation in matrix form:
\(\begin {pmatrix} 2 & -1\\ 1 & 2\\ \end {pmatrix}\)\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} 1\\ 3\\ \end {pmatrix}\), AX = B
where, A =\(\begin {pmatrix} 2 & -1\\ 1 & 2\\ \end {pmatrix}\), X = \(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) and B =\(\begin {pmatrix} 1\\ 3\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 2 & -1\\ 1 & 2\\ \end {vmatrix}\) = 2 × 2 - 1 × -1 = 4 + 1 = 5 ≠ 0
It has a unique solution.
A-1= \(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj.A = \(\frac {1}{5}\)\(\begin {pmatrix} 2 & 1\\ -1 & 2\\ \end {pmatrix}\) =\(\begin {pmatrix} \frac {-2}{5} & \frac {1}{5}\\ \frac {-1}{5} & \frac {2}{5}\\ \end {pmatrix}\)
We know that,
AX = B
X = A-1B
or, X =\(\begin {pmatrix} \frac {-2}{5} & \frac {1}{5}\\ \frac {-1}{5} & \frac {2}{5}\\ \end {pmatrix}\)\(\begin {pmatrix} 1\\ 3\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {2}{5} + \frac {3}{5}\\ \frac {-1}{5} + \frac {6}{5}\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {2 + 3}{5}\\ \frac {-1 + 6}{5}\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {5}{5}\\ \frac {5}{5}\\ \end {pmatrix}\)
∴ X =\(\begin {pmatrix} 1\\ 1\\ \end {pmatrix}\)
∴ x = 1 and y = 1 Ans
Solve by matrix method:
x + 2y = 8
2x + 3y = 11
Here,
x + 2y = 8........................................(1)
2x + 3y = 11..................................(2)
Given equation in matrix form:
\(\begin {pmatrix} 1 & 2\\ 2 & 3\\ \end {pmatrix}\)\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} 8\\ 11\\ \end {pmatrix}\) , AX = B
where, A =\(\begin {pmatrix} 1 & 2\\ 2 & 3\\ \end {pmatrix}\), X =\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) and B =\(\begin {pmatrix} 8\\ 11\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 1 & 2\\ 2 & 3\\ \end {vmatrix}\) = 1 × 3 - 2 × 2 = 3 - 4 = -1 ≠ 0
It has a unique solution.
A-1= \(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj.A = \(\frac {1}{-1}\)\(\begin {pmatrix} 3 & -2\\ -2 & 1\\ \end {pmatrix}\) = \(\begin {pmatrix} -3 & 2\\ 2 & -1\\ \end {pmatrix}\)
We know that,
AX = B
X = A-1B
or, X =\(\begin {pmatrix} -3 & 2\\ 2 & -1\\ \end {pmatrix}\)\(\begin {pmatrix} 8\\ 11\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} -24 + 22\\ 16 - 11\\ \end {pmatrix}\)
∴ X =\(\begin {pmatrix} -2\\ 5\\ \end {pmatrix}\)
∴ x = -2 and y = 5 Ans
Solve by matrix method:
2x + 3y = 5
2y - 3x = - 1
Here,
2x + 3y = 5...........................(1)
-3x + 2y = -1.......................(2)
Given equation in matrix form:
\(\begin {pmatrix} 2 & 3\\ -3 & 2\\ \end {pmatrix}\)\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} 5\\ -1\\ \end {pmatrix}\), AX = B
where, A =\(\begin {pmatrix} 2 & 3\\ -3 & 2\\ \end {pmatrix}\), X =\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) and B =\(\begin {pmatrix} 5\\ -1\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 2 & 3\\ -3 & 2\\ \end {vmatrix}\) = 2 × 2 - (-3) × 3 = 4 + 9 = 13 ≠ 0
It has a unique solution.
A-1= \(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj.A = \(\frac {1}{13}\)\(\begin {pmatrix} 2 & -3\\ 3 & 2\\ \end {pmatrix}\)
We know that,
AX = B
X = A-1B
or, X =\(\frac {1}{13}\)\(\begin {pmatrix} 2 & -3\\ 3 & 2\\ \end {pmatrix}\)\(\begin {pmatrix} 5\\ -1\\ \end {pmatrix}\)
or, X = \(\frac {1}{13}\)\(\begin {pmatrix} 10 + 3\\ 15 - 2\\ \end {pmatrix}\)
or, X = \(\frac {1}{13}\)\(\begin {pmatrix} 13\\ 13\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {13}{13}\\ \frac {13}{13}\\ \end {pmatrix}\)
∴ X = \(\begin {pmatrix} 1\\ 1\\ \end {pmatrix}\)
∴ x = 1 and y = 1 Ans
Solve by matrix method:
3x + y = 51
4x - 3y = 3
Given equation:
3x + y = 51...........................(1)
4x - 3y = 3............................(2)
Given equation in matrix form:
\(\begin {pmatrix} 3 & 1\\ 4 & -3\\ \end {pmatrix}\)\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} 51\\ 3\\ \end {pmatrix}\), AX = B
where, A =\(\begin {pmatrix} 3 & 1\\ 4 & -3\\ \end {pmatrix}\), X = \(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) and B =\(\begin {pmatrix} 51\\ 3\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 3 & 1\\ 4 & -3\\ \end {vmatrix}\) = 3 × (-3) - 4 × 1 = -9 - 4 = -13 ≠ 0
It has a unique solution.
A-1= \(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj.A = \(\frac {1}{-13}\)\(\begin {pmatrix} -3 & -1\\ -4 & 3\\ \end {pmatrix}\) =\(\begin {pmatrix} \frac {3}{13} & \frac {1}{13}\\ \frac {4}{13} & \frac {-3}{13}\\ \end {pmatrix}\)
We know that,
AX = B
X = A-1B
or, X =\(\begin {pmatrix} \frac {3}{13} & \frac {1}{13}\\ \frac {4}{13} & \frac {-3}{13}\\ \end {pmatrix}\)\(\begin {pmatrix} 51\\ 3\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {153}{13} & \frac {3}{13}\\ \frac {204}{13} & \frac {-9}{13}\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {153 + 3}{13}\\ \frac {204 - 9}{13}\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {156}{13}\\ \frac {195}{13}\\ \end {pmatrix}\)
∴ X =\(\begin {pmatrix} 12\\ 5\\ \end {pmatrix}\)
∴ x = 12 and y = 5 Ans
Solve by matrix method:
2x - 5y = 1
7x + 3y = 24
Here,
2x - 5y = 1.................................(1)
7x + 3y = 24.............................(2)
Given equation in matrix form:
\(\begin {pmatrix} 2 & -5\\ 7 & 3\\ \end {pmatrix}\)\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} 1\\ 24\\ \end {pmatrix}\), AX = B
where, A =\(\begin {pmatrix} 2 & -5\\ 7 & 3\\ \end {pmatrix}\), X =\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) and B =\(\begin {pmatrix} 1\\ 24\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 2 & -5\\ 7 & 3\\ \end {vmatrix}\) = 2 × 3 - 7 × (-5) = 6 + 35 = 41 ≠ 0
It has a unique solution.
A-1= \(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj.A = \(\frac {1}{41}\)\(\begin {pmatrix} 3 & 5\\ -7 & 2\\ \end {pmatrix}\) =\(\begin {pmatrix} \frac {3}{41} & \frac {5}{41}\\ \frac {-7}{41} & \frac {2}{41}\\ \end {pmatrix}\)
We know that,
AX = B
X = A-1B
or, X =\(\begin {pmatrix} \frac {3}{41} & \frac {5}{41}\\ \frac {-7}{41} & \frac {2}{41}\\ \end {pmatrix}\)\(\begin {pmatrix} 1\\ 24\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {3 × 1}{41} & \frac {5 × 24}{41}\\ \frac {-7 × 1}{41} & \frac {2 × 24}{41}\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {3}{41} & \frac {120}{41}\\ \frac {-7}{41} & \frac {48}{41}\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {3 + 120}{41}\\ \frac {-7 + 48}{41}\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {123}{41}\\ \frac {41}{41}\\ \end {pmatrix}\)
∴ X =\(\begin {vmatrix} 3\\ 1\\ \end {vmatrix}\)
∴ x = 3 and y = 1 Ans
Solve by matrix method;
x - 8y = 39
2x - 3y - 13 = 0
Here,
x - 8y = 39..........................................(1)
2x - 3y = 13.......................................(2)
Given equation in matrix form:
\(\begin {pmatrix} 1 & -8\\ 2 & -3\\ \end {pmatrix}\)\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} 39\\ 13\\ \end {pmatrix}\), AX = B
where, A =\(\begin {pmatrix} 1 & -8\\ 2 & -3\\ \end {pmatrix}\), X =\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) and B =\(\begin {pmatrix} 39\\ 13\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 1 & -8\\ 2 & -3\\ \end {vmatrix}\) = 1 × (-3) - 2 × (-8) = -3 + 16 = 13 ≠ 0
It has a unique solution.
A-1= \(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj.A = \(\frac {1}{13}\)\(\begin {pmatrix} -3 & -8\\ -2 & 1\\ \end {pmatrix}\)
We know that,
AX =B
X = A-1B
or, X =\(\frac {1}{13}\)\(\begin {pmatrix} -3 & -8\\ -2 & 1\\ \end {pmatrix}\)\(\begin {pmatrix} 39\\ 13\\ \end {pmatrix}\)
or, X = \(\frac {1}{13}\)\(\begin {pmatrix} -117 + 104\\ -78 + 13\\ \end {pmatrix}\)
or, X = \(\frac {1}{13}\)\(\begin {pmatrix} -13\\ -65\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {-13}{13}\\ \frac {-65}{13}\\ \end {pmatrix}\)
∴ X = \(\begin {pmatrix} -1\\ -5\\ \end {pmatrix}\)
∴ x = -1 and y = -5 Ans
Solve by matrix method:
\(\frac 2x\) + \(\frac 3y\) = 2
\(\frac 4x\) - \(\frac 9y\) = -1
Here,
\(\frac 2x\) + \(\frac 3y\) = 2..............................................(1)
\(\frac 4x\) - \(\frac 9y\) = -1.............................................(2)
Given equation in matrix form:
\(\begin {pmatrix} 2 & 3\\ 4 & -9\\ \end {pmatrix}\)\(\begin {pmatrix} \frac 1x\\ \frac1y\\ \end {pmatrix}\) =\(\begin {pmatrix} 2\\ -1\\ \end {pmatrix}\), AX = B
where, A =\(\begin {pmatrix} 2 & 3\\ 4 & -9\\ \end {pmatrix}\), X =\(\begin {pmatrix} \frac 1x\\ \frac1y\\ \end {pmatrix}\) and B = \(\begin {pmatrix} 2\\ -1\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 2 & 3\\ 4 & -9\\ \end {vmatrix}\) = 2 × (-9) - 4 × 3 = -18 - 12 = -30 ≠ 0
It has a unique solution.
A-1= \(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj.A = \(\frac {1}{-30}\)\(\begin {pmatrix} -9 & -3\\ -4 & 2\\ \end {pmatrix}\) =\(\begin {pmatrix} \frac {9}{30} & \frac {3}{30}\\ \frac {4}{30} & \frac {-2}{30}\\ \end {pmatrix}\)
We know that,
AX = B
X = A-1B
or, X =\(\begin {pmatrix} \frac {9}{30} & \frac {3}{30}\\ \frac {4}{30} & \frac {-2}{30}\\ \end {pmatrix}\)\(\begin {pmatrix} 2\\ -1\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {9×2}{30} - \frac {3×1}{30}\\ \frac {4×2}{30} + \frac {2×1}{30}\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {18-3}{30}\\ \frac {8+2}{30}\\ \end {pmatrix}\)
or, X = \(\begin {pmatrix} \frac {15}{30}\\ \frac {10}{30}\\ \end {pmatrix}\)
∴ X = \(\begin {pmatrix} \frac 12\\ \frac 13\\ \end {pmatrix}\)
i.e.\(\begin {pmatrix} \frac 1x\\ \frac1y\\ \end {pmatrix}\) =\(\begin {pmatrix} \frac 12\\ \frac 13\\ \end {pmatrix}\)
Taking the corresponding part of the equal matrix:
\(\frac 1x\) = \(\frac 12\)
∴ x = 2
\(\frac 1y\) = \(\frac 13\)
∴ y = 3
∴ x = 2 and y = 3 Ans
Solve by matrix method:
\(\frac {3x + 5y}{4}\) = \(\frac {7x + 3y}{5}\) = 4
Here,
\(\frac {3x + 5y}{4}\) =4, \(\frac {7x + 3y}{5}\) = 4
3x + 5y = 16..............................(1)
7x + 3y = 20..............................(2)
Given equation in matrix form:
\(\begin {pmatrix} 3 & 5\\ 7 & 3\\ \end {pmatrix}\)\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} 16\\ 20\\ \end {pmatrix}\), AX =B
where, A =\(\begin {pmatrix} 3 & 5\\ 7 & 3\\ \end {pmatrix}\), X =\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) and B =\(\begin {pmatrix} 16\\ 20\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 3 & 5\\ 7 & 3\\ \end {vmatrix}\) = 3 × 3 - 7 × 5 = 9 - 35 = -26 ≠ 0
It has a unique solution.
A-1= \(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj.A = \(\frac {1}{-26}\)\(\begin {pmatrix} 3 & -5\\ -7 & 3\\ \end {pmatrix}\)
We know that:
AX = B
X = A-1B
or, X =\(\frac {1}{-26}\)\(\begin {pmatrix} 3 & -5\\ -7 & 3\\ \end {pmatrix}\)\(\begin {pmatrix} 16\\ 20\\ \end {pmatrix}\)
or, X =\(\frac {1}{-26}\)\(\begin {pmatrix} 48 - 100\\ -112 + 60\\ \end {pmatrix}\)
or, X =\(\frac {1}{-26}\)\(\begin {pmatrix} -52\\ -52\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {-52}{-26}\\ \frac {-52}{-26}\\ \end {pmatrix}\)
∴ X =\(\begin {pmatrix} 2\\ 2\\ \end {pmatrix}\)
∴ x = 2 and y = 2 Ans
Solve the equation by matrix method: x = \(\frac 23\)y and 4x - 3y = 1.
Here,
Given equations are:
x = \(\frac 23\)y i.e. 3x - 2y = 0................(1)
4x - 3y = 1............................(2)
Given equation in matrix form;
\(\begin {pmatrix} 3 & -2\\ 4 & -3\\ \end {pmatrix}\)\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} 0\\ 1\\ \end {pmatrix}\), AX = B
where, A =\(\begin {pmatrix} 3 & -2\\ 4 & -3\\ \end {pmatrix}\), X =\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) and B =\(\begin {pmatrix} 0\\ 1\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 3 & -2\\ 4 & -3\\ \end {vmatrix}\) = 3 × -3 - 4 × -2 = -9 + 8 = -1 ≠ 0
It has a unique solution.
A-1= \(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj.A = \(\frac {1}{-1}\)\(\begin {pmatrix} -3 & -2\\ -4 & 3\\ \end {pmatrix}\) = \(\begin {pmatrix} 3 & 2\\ 4 & -3\\ \end {pmatrix}\)
We know that,
AX = B
X = A-1B
or, X =\(\begin {pmatrix} 3 & 2\\ 4 & -3\\ \end {pmatrix}\)\(\begin {pmatrix} 0\\ 1\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} 3 × 0 + 2 × 1\\ 4 × 0 - 3 × 1\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} 0 + 2\\ 0 - 3\\ \end {pmatrix}\)
∴ X =\(\begin {pmatrix} 2\\ -3\\ \end {pmatrix}\)
∴ x = 2 and y = -3 Ans
Solve by matrix method:
\(\frac 4x\) + 3y = 11
3xy - 8 - 5x = 0
Here,
\(\frac 4x\) + 3y = 11.................................(1)
3xy - 8 -5x = 0
or, \(\frac {3xy}{x}\) - \(\frac 8x\) - \(\frac {5x}{x}\) = 0
or, -\(\frac 8x\) + 3y = 5...................................(2)
Given equation in matrix form:
\(\begin {pmatrix} 4 & 3\\ -8 & 3\\ \end {pmatrix}\)\(\begin {pmatrix} \frac 1x\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} 11\\ 5\\ \end {pmatrix}\), AX = B
where, A =\(\begin {pmatrix} 4 & 3\\ -8 & 3\\ \end {pmatrix}\), X =\(\begin {pmatrix} \frac 1x\\ y\\ \end {pmatrix}\) and B =\(\begin {pmatrix} 11\\ 5\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 4 & 3\\ -8 & 3\\ \end {vmatrix}\) = 4 × 3 - (-8) × 3 = 12 + 24 = 36 ≠ 0
It has a unique solution.
A-1= \(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj.A = \(\frac {1}{36}\)\(\begin {pmatrix} 3 & -3\\ 8 & 4\\ \end {pmatrix}\) =\(\begin {pmatrix} \frac {3}{36} & \frac {-3}{36}\\ \frac {8}{36} & \frac {4}{36}\\ \end {pmatrix}\) =\(\begin {pmatrix} \frac {1}{12} & \frac {-1}{12}\\ \frac {2}{9} & \frac {1}{9}\\ \end {pmatrix}\)
We know that,
AX = B
X = A-1B
or, X =\(\begin {pmatrix} \frac {1}{12} & \frac {-1}{12}\\ \frac {2}{9} & \frac {1}{9}\\ \end {pmatrix}\)\(\begin {pmatrix} 11\\ 5\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {11}{12} & \frac {-5}{12}\\ \frac {22}{9} & \frac {5}{9}\\ \end {pmatrix}\)
or, X = \(\begin {pmatrix} \frac {11 - 5}{12}\\ \frac {22 + 5}{9}\\ \end {pmatrix}\)
or, X = \(\begin {pmatrix} \frac {6}{12}\\ \frac {27}{9}\\ \end {pmatrix}\)
∴X = \(\begin {pmatrix} \frac {1}{2}\\ 3\\ \end {pmatrix}\)
i.e.\(\begin {pmatrix} \frac 1x\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} \frac {1}{2}\\ 3\\ \end {pmatrix}\)
Taking the corresponding part of the equal matrix:
\(\frac 1x\) = \(\frac 12\)
∴ x = 2
∴ y = 3
∴ x = 2 and y = 3 Ans
Solve by matrix method:
\(\frac 4x\) + 3y = 11
3xy - 8 - 5x = 0
Here,
\(\frac 4x\) + 3y = 11.................................(1)
3xy - 8 -5x = 0
or, \(\frac {3xy}{x}\) - \(\frac 8x\) - \(\frac {5x}{x}\) = 0
or, -\(\frac 8x\) + 3y = 5...................................(2)
Given equation in matrix form:
\(\begin {pmatrix} 4 & 3\\ -8 & 3\\ \end {pmatrix}\)\(\begin {pmatrix} \frac 1x\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} 11\\ 5\\ \end {pmatrix}\), AX = B
where, A =\(\begin {pmatrix} 4 & 3\\ -8 & 3\\ \end {pmatrix}\), X =\(\begin {pmatrix} \frac 1x\\ y\\ \end {pmatrix}\) and B =\(\begin {pmatrix} 11\\ 5\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 4 & 3\\ -8 & 3\\ \end {vmatrix}\) = 4 × 3 - (-8) × 3 = 12 + 24 = 36 ≠ 0
It has a unique solution.
A-1= \(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj.A = \(\frac {1}{36}\)\(\begin {pmatrix} 3 & -3\\ 8 & 4\\ \end {pmatrix}\) =\(\begin {pmatrix} \frac {3}{36} & \frac {-3}{36}\\ \frac {8}{36} & \frac {4}{36}\\ \end {pmatrix}\) =\(\begin {pmatrix} \frac {1}{12} & \frac {-1}{12}\\ \frac {2}{9} & \frac {1}{9}\\ \end {pmatrix}\)
We know that,
AX = B
X = A-1B
or, X =\(\begin {pmatrix} \frac {1}{12} & \frac {-1}{12}\\ \frac {2}{9} & \frac {1}{9}\\ \end {pmatrix}\)\(\begin {pmatrix} 11\\ 5\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {11}{12} & \frac {-5}{12}\\ \frac {22}{9} & \frac {5}{9}\\ \end {pmatrix}\)
or, X = \(\begin {pmatrix} \frac {11 - 5}{12}\\ \frac {22 + 5}{9}\\ \end {pmatrix}\)
or, X = \(\begin {pmatrix} \frac {6}{12}\\ \frac {27}{9}\\ \end {pmatrix}\)
∴X = \(\begin {pmatrix} \frac {1}{2}\\ 3\\ \end {pmatrix}\)
i.e.\(\begin {pmatrix} \frac 1x\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} \frac {1}{2}\\ 3\\ \end {pmatrix}\)
Taking the corresponding part of the equal matrix:
\(\frac 1x\) = \(\frac 12\)
∴ x = 2
∴ y = 3
∴ x = 2 and y = 3 Ans
Solve by matrix method:
4x + \(\frac y5\) = 7
3x + \(\frac y4\) = 5
Here,
4x + \(\frac y5\) = 7
or, \(\frac {20x + y}{5}\) = 7
or, 20x + y = 35.........................................(1)
3x + \(\frac y4\) = 5
or, \(\frac {12x + y}{4}\) = 5
or, 12x + y = 20..........................................(2)
Given equation in matrix form:
\(\begin {pmatrix} 20 & 1\\ 12 & 1\\ \end {pmatrix}\)\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} 35\\ 20\\ \end {pmatrix}\), AX =B
where, A =\(\begin {pmatrix} 20 & 1\\ 12 & 1\\ \end {pmatrix}\), X =\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) and B =\(\begin {pmatrix} 35\\ 20\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 20 & 1\\ 12 & 1\\ \end {vmatrix}\) = 20 × 1 - 12 × 1 = 20 - 12 = 8 ≠ 0
It has a unique solution.
A-1= \(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj.A = \(\frac {1}{8}\)\(\begin {pmatrix} 1 & -1\\ -12 & 20\\ \end {pmatrix}\) =\(\begin {pmatrix} \frac {1}{8} & \frac {-1}{8}\\ \frac {-12}{8} & \frac {20}{8}\\ \end {pmatrix}\) = \(\begin {pmatrix} \frac {1}{8} & \frac {-1}{8}\\ \frac {-3}{2} & \frac {5}{2}\\ \end {pmatrix}\)
We know that,
AX = B
or, X = A-1B
or, X =\(\begin {pmatrix} \frac {1}{8} & \frac {-1}{8}\\ \frac {-3}{2} & \frac {5}{2}\\ \end {pmatrix}\)\(\begin {pmatrix} 35\\ 20\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {35}{8} - \frac {20}{8}\\ \frac {-105}{2} + \frac {100}{2}\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {35 - 20}{8}\\ \frac {-105 + 100}{2}\\ \end {pmatrix}\)
∴ X =\(\begin {pmatrix} \frac {15}{8}\\ \frac {-5}{2}\\ \end {pmatrix}\)
∴ x = \(\frac {15}{2}\) and y = \(\frac {-5}{2}\) Ans
Solve by the matrix method:
\(\frac 3x\) + \(\frac 4y\) = 2
\(\frac 9x\) - \(\frac 2y\) = \(\frac 52\)
Here,
\(\frac 3x\) + \(\frac 4y\) = 2..........................................(1)
\(\frac 9x\) - \(\frac 2y\) = \(\frac 52\)....................(2)
Given equation in the form of matrix:
\(\begin {pmatrix} 3 & 4\\ 9 & -2\\ \end {pmatrix}\)\(\begin {pmatrix} \frac 1x\\ \frac1y\\ \end {pmatrix}\) =\(\begin {pmatrix} 2\\ \frac52\\ \end {pmatrix}\), AX = B
where, A =\(\begin {pmatrix} 3 & 4\\ 9 & -2\\ \end {pmatrix}\), X =\(\begin {pmatrix} \frac 1x\\ \frac1y\\ \end {pmatrix}\) and B =\(\begin {pmatrix} 2\\ \frac52\\ \end {pmatrix}\)
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 3 & 4\\ 9 & -2\\ \end {vmatrix}\) = 3 × (-2) - 9 × 4 = -6 - 36 = -42 ≠ 0
It has a unique solution.
A-1= \(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj.A = \(\frac {1}{-42}\)\(\begin {pmatrix} -2 & -4\\ -9 & 3\\ \end {pmatrix}\) = \(\begin {pmatrix} \frac {-2}{-42} & \frac {-4}{-42}\\ \frac {-9}{-42} & \frac {3}{-42}\\ \end {pmatrix}\) =\(\begin {pmatrix} \frac {1}{21} & \frac {2}{21}\\ \frac {9}{42} & \frac {3}{42}\\ \end {pmatrix}\)
We know that:
AX = B
or, X = A-1B
or, X =\(\begin {pmatrix} \frac {1}{21} & \frac {2}{21}\\ \frac {9}{42} & \frac {3}{42}\\ \end {pmatrix}\)\(\begin {pmatrix} 2\\ \frac52\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {2}{21} + \frac {5}{21}\\ \frac {9}{21} - \frac {15}{84}\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {2 + 5}{21}\\ \frac {36 - 15}{84}\\ \end {pmatrix}\)
or, X =\(\begin {pmatrix} \frac {7}{21}\\ \frac {21}{84}\\ \end {pmatrix}\)
∴ X =\(\begin {pmatrix} \frac {1}{3}\\ \frac {1}{4}\\ \end {pmatrix}\)
i.e. \(\begin {pmatrix} \frac 1x\\ \frac1y\\ \end {pmatrix}\) = \(\begin {pmatrix} \frac {1}{3}\\ \frac {1}{4}\\ \end {pmatrix}\)
Taking the corresponding parts of the equal matrix:
\(\frac 1x\) = \(\frac 13\)
∴ x = 3
\(\frac 1y\) = \(\frac 14\)
∴ y = 4
∴ x = 3 and y = 4 Ans
Solve by matrix method:
In a company the men get Rs. 25 a day and the women get Rs. 20 a day. 50 people are employed and the total wages are Rs. 1150 a day. Find the member of men and women employed in the company?
Suppose the number of men = x
Suppose the number of women = y
Now,
1 men get Rs. 25 a day.
x men get Rs. 25x a day.
1 women get Rs. 20 a day.
y women get Rs. 20y a day.
According to the question,
x + y = 50.....................................(1)
25x + 20y = 1150...................(2)
Equation (1) and (2) in matrix form:
\(\begin {bmatrix} 1 & 1\\ 25 & 20\\ \end {bmatrix}\)\(\begin {bmatrix} x\\ y\\ \end {bmatrix}\) =\(\begin {bmatrix} 50\\ 1150\\ \end {bmatrix}\)
where:
A =\(\begin {bmatrix} 1 & 1\\ 25 & 20\\ \end {bmatrix}\) is a coefficient matrix.
X =\(\begin {bmatrix} x\\ y\\ \end {bmatrix}\) is a variable matrix.
B =\(\begin {bmatrix} 50\\ 1150\\ \end {bmatrix}\) is a constant matrix.
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 1 & 1\\ 25 & 20\\ \end {vmatrix}\) = 1 × 20 - 25 × 1 = 20 - 25 = -5 ≠ 0
It has a unique solution.
A-1= \(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj.A = \(\frac {1}{-5}\)\(\begin {pmatrix} 20 & -1\\ -25 & 1\\ \end {pmatrix}\) =\(\begin {pmatrix} \frac {20}{-5} & \frac {-1}{-5}\\ \frac {-25}{-5} & \frac {1}{-5}\\ \end {pmatrix}\) = \(\begin {pmatrix} -4 & \frac {1}{5}\\ 5& \frac {1}{-5}\\ \end {pmatrix}\)
We know that,
AX = B
or, X = A-1B
or, X =\(\begin {pmatrix} -4 & \frac {1}{5}\\ 5& \frac {1}{-5}\\ \end {pmatrix}\)\(\begin {bmatrix} 50\\ 1150\\ \end {bmatrix}\)
or, X =\(\begin {bmatrix} -200 + 230\\ 250 - 230\\ \end {bmatrix}\)
∴ X =\(\begin {bmatrix} 30\\ 20\\ \end {bmatrix}\)
∴ x = 30 and y = 20 Ans
Solve by matrix method;
1 kg potatoes and 1 kg onions together cost Rs. 50.5 kg potatoes and 3 kg onions together cost Rs. 190. Find the cost of per kg potatoes and onions?
Suppose the cost of potatoes for 1 kg be Rs. x and the cost of onions for 1 kg be Rs. y.
Here,
The cost of 1 kg potatoes be Rs. x.
The cost of 5 kg potatoes be Rs. 5x.
The cost of 1 kg onions be Rs. y.
The cost of 1 kg onions be Rs. 3y.
According to question:
x + y = 50...................................(1)
5x + 3y = 190..........................(2)
Equation (1) and (2) can be written in matrix form.
\(\begin {bmatrix} 1 & 1\\ 5 & 3\\ \end {bmatrix}\)\(\begin {bmatrix} x\\ y\\ \end {bmatrix}\) =\(\begin {bmatrix} 50\\ 190\\ \end {bmatrix}\)
where:
A =\(\begin {bmatrix} 1 & 1\\ 5 & 3\\ \end {bmatrix}\) is a coefficient matrix.
X =\(\begin {bmatrix} x\\ y\\ \end {bmatrix}\) is a variable matrix.
B =\(\begin {bmatrix} 50\\ 190\\ \end {bmatrix}\) is a constant matrix.
\(\begin {vmatrix} A\\ \end {vmatrix}\) =\(\begin {vmatrix} 1 & 1\\ 5 & 3\\ \end {vmatrix}\) = 1 × 3 - 5 × 1 = 3 - 5 = -2 ≠ 0
So, the system has unique solution.
A-1= \(\frac {1}{\begin {vmatrix} A\\ \end {vmatrix}}\) Adj.A = \(\frac {1}{-2}\)\(\begin {pmatrix} 3 & -1\\ -5 & 1\\ \end {pmatrix}\) = \(\begin {pmatrix} \frac {-3}{2} & \frac {1}{2}\\ \frac {5}{2} & \frac {-1}{2}\\ \end {pmatrix}\)
We know that,
AX = B
X = A-1B
or, X =\(\begin {pmatrix} \frac {-3}{2} & \frac {1}{2}\\ \frac {5}{2} & \frac {-1}{2}\\ \end {pmatrix}\)\(\begin {bmatrix} 50\\ 190\\ \end {bmatrix}\)
or, X =\(\begin {bmatrix} -75 + 95\\ 125 - 95\\ \end {bmatrix}\)
∴ X =\(\begin {bmatrix} 20\\ 30\\ \end {bmatrix}\)
∴x = 20 and y = 30
Hence, the cost per kg potatoes is Rs. 20 and the cost per kg onions is Rs. 30. Ans
Quiz
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