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Note on Matrix

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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.

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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 =\(\begin{bmatrix}b_{11}&b_{12}\\b_{21}&b_{22}\\ \end{bmatrix}\)

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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

example for Singular and non-singular matrices
example for 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.

 

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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Very Short Questions

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 =\(\begin{pmatrix} a & d \\ b & e \\ c & f \\ \end{pmatrix}\)

\(\begin {pmatrix} 4 & 1 \\ 7 & -3\\ \end {pmatrix}\) \(\begin {pmatrix} 2 & -1 \\ 1 & 3\\ \end {pmatrix}\)= \(\begin {pmatrix} x & -1 \\ 11 & y\\ \end {pmatrix}\)

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

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

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

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

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 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.

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

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

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

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 = \(\begin {bmatrix} 7 & 0 \\ 2 & 5 \\ \end {bmatrix}\) ..................(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 =\(\begin {bmatrix} 2 & 0 \\ 1 & 1 \\ \end {bmatrix}\)

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

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 = \(\begin {pmatrix} 1 & x \\ 0 & 1 \\ \end {pmatrix}\) , 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

\(\begin {pmatrix} 2x + 3 \\ y - 2 \\ 3z - y \\ \end {pmatrix}\) = \(\begin {pmatrix} 5 & -8 \\ 12 & -40 \\ -3 & 4 \\ \end {pmatrix}\)\(\begin {pmatrix} 7 \\ 2 \\ \end {pmatrix}\)

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

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

\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) = \(\begin {pmatrix} 3 & 5\\ 4 & 6\\ \end {pmatrix}\)\(\begin {pmatrix} 1\\ 2\\ \end {pmatrix}\)

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

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

\(\begin {pmatrix} 1 & 1\\ 3 & y\\ \end {pmatrix}\) \(\begin {pmatrix} x\\ 1\\ \end {pmatrix}\) =\(\begin {pmatrix} 4\\ 1\\ \end {pmatrix}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

0%
  •  If A = egin{pmatrix} 3 & 3\ 5 & 0\ end{pmatrix} and B = egin{pmatrix} -3 & 4\ 2 & -5\ end{pmatrix} then find the value of 2A - 3B.

    egin{pmatrix} 3 & 3\ 5 & 0\ end{pmatrix}


    egin{pmatrix} 4 & -6\ 15 & 0\ end{pmatrix}


    egin{pmatrix} 15 & -6\ 4 & 15\ end{pmatrix}


    egin{pmatrix} 0 & 5\ 5 & 0\ end{pmatrix}


  • If X = egin{pmatrix} 3 & -2 & 7 \ 5 & 6 & 4\ end{pmatrix} and Y = egin{pmatrix} 1 & -3 & 0 \ 4 & 2 & 8 \ end{pmatrix} , find the  matrix (7X - 4Y) .

    egin{pmatrix} 17 & -2 & 49 \ 19 & 34 & -4 \ end{pmatrix}


    egin{pmatrix} -2 & -2 & 17 \ -4 & 19 & 34 \ end{pmatrix}


    egin{pmatrix} 71 & -4 & 48 \ 19 & 34 & -4 \ end{pmatrix}


    egin{pmatrix} 81 & 1 & 54 \ 20 & 36 & -4 \ end{pmatrix}


  • A = egin{pmatrix} 19 & 5 \  14 & 30 \ end{pmatrix}  and A - 3B = egin{pmatrix}  1 & -1 \ -7 & 3 \ end{pmatrix} , What will be the matrix B. ?

     

    egin{pmatrix} 6 & 2 \ 7 & 9 \ end{pmatrix}


    egin{pmatrix} 6 & 7 \ 2 & 9 \ end{pmatrix}


    egin{pmatrix} 9 & 2 \ 7 & 6 \ end{pmatrix}


    egin{pmatrix} 1 & 6 \ 2 & 9 \ end{pmatrix}


  • Which matrix pre-multiplies to the matrix egin{pmatrix}  1 & 1 \ 3 & 4 \ end{pmatrix} to get a matrix  egin{pmatrix} 4 & 5 \ 6 & 2 \ end{pmatrix}  ?

     

    egin{pmatrix} 3 \ 12 end{pmatrix}


    egin{pmatrix} 2 \ 3 end{pmatrix}


    egin{pmatrix} 9 \ 2 end{pmatrix}


    egin{pmatrix} 9 \ 9 end{pmatrix}


  • A = egin{pmatrix} 19 & 5 \ 14 & 30 \ end{pmatrix} and A-3B =  egin{pmatrix} 1 & -1 \ -7  &  3 \ end{pmatrix}. Find the matrix B>

    egin{pmatrix} 9 & 2 \ 7 & 6 \ end{pmatrix}


    egin{pmatrix} 2 & 6 \ 9 & 7 \ end{pmatrix}


    egin{pmatrix} 6 & 2 \ 7 & 9 \ end{pmatrix}


    egin{pmatrix} 6 & 7 \ 2 & 9 \ end{pmatrix}


  • Which matrix multiplies to the matrix egin{pmatrix} 1 & 0 \ 1 & 3 \ end{pmatrix} to get matrix egin{pmatrix} 9 \ 15 \ end{pmatrix}. ? Find it .

    egin{pmatrix} 2 \  9 \ end{pmatrix}


    egin{pmatrix} 0 \ 1 \ end{pmatrix}


    egin{pmatrix} 9 \ 1 \ end{pmatrix}


    egin{pmatrix} 9 \ 2 \ end{pmatrix}


  • Find the values of x and y or a and  b. 
    egin{pmatrix} 4 & 1 \ 7 & -3 \ end{pmatrix}  egin{pmatrix} 2 & -1 \ 1 & 9 \ end{pmatrix}  =  egin{pmatrix} x & -1 \ 11 & y \ end{pmatrix}

    egin{pmatrix} 7 & -41 \ 11 & 9 \ end{pmatrix}


    egin{pmatrix} 2 & -1 \ 1 & 9 \ end{pmatrix}


    egin{pmatrix} 2 & -14 \ 1 & -11 \ end{pmatrix}


    egin{pmatrix} 7 & 9 \ -11 & -14 \ end{pmatrix}


  • If A = egin{pmatrix} 2 & 4 \ -1 & -1 \ end{pmatrix} , B = egin{pmatrix} -1 \  4 end{pmatrix} and C = egin{pmatrix} -3 \ -1 \ end{pmatrix} .  What will be A + 2B - 3C .

    egin{pmatrix} 10 \ 13  \ end{pmatrix}


    egin{pmatrix}  -1 \ 9 \ end{pmatrix}


    egin{pmatrix} 2  \ 1  \ end{pmatrix}


    egin{pmatrix} -1 \ 3 \ end{pmatrix}


  • egin{pmatrix} x & 4 \ 1 & 2 \ end{pmatrix}  egin{pmatrix} 2  \ -1  \ end{pmatrix} = egin{pmatrix} 4 \ 0 \ end{pmatrix} . What will be the value of X . 

    -1


    1


    2


    4


  • egin{pmatrix} 1 & 3 \ 2 & 4 \ end{pmatrix} egin{pmatrix} 1 \ 2 \ end{pmatrix} = egin{pmatrix} x \ y \ end{pmatrix} . Find X and Y . 

    1 , 2


    2 , 3


    1 , 3


    7 , 10


  • If A = egin{pmatrix} 2 & -3 \ -4 & 5 \ end{pmatrix} , B = egin{pmatrix} -3 & 2 \ 0 & 1 \ end{pmatrix} and C = egin{pmatrix} 1 & 0 \ 3 & 4 \ end{pmatrix} . Find the n (AB) (^t) + (BC) (^t) . 

    - (frac{2}{3})  , -1


    - (frac{2}{1})  , -3


    (frac{-2}{-3})  , 1


     (frac{3}{2})  , 1


  • If A = egin{pmatrix} 2 & 4 \ 3  &  2 \ end{pmatrix} and B = egin{pmatrix} 1 & -5 \ 6  &  2 \ end{pmatrix} , what will be the  determination of A - 3B . 

    189


    egin{pmatrix} 2 & 2 \ 2 & 3 \ end{pmatrix}


    289


    A - 3B


  •  If the determination of matrix A =  egin{pmatrix} 3p -2 \ 4  &  3 \ end{pmatrix} is 70 , what will be the value of p?

    6.89


    6.23


    9.89


    6.88


  • egin{pmatrix} 2m & 7 \ 5  &  9 \ end{pmatrix} and egin{pmatrix} 9 & n \ -5  &  4 \ end{pmatrix} .   What will be the value of m and n.

    -2 , 7


    2 , -7


    7 , 2


     2 , 7


  • A = egin{pmatrix} 1 & -2 \ 0  &  3 \ end{pmatrix} and B = egin{pmatrix} 3 & 1 \ 5  &  3 \ end{pmatrix} ; |A + B + 1| . What will be the value of determination .

    40


    04


    egin{pmatrix} 2 & -1 \ 3  &  0 \ end{pmatrix}


    41


  • You scored /15


    Take test again

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Jason

What is identity matrix? With example


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aisha

i was able to remove my confusions thanks to kullabs


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