Operations on Matrices

Subject: Optional Mathematics

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Overview

If A and B are two matrices of the same order, then A and B are said to be Conformable or Compatible for addition. If A and B are two matrices of the same order, then they are said to be conformable for subtraction. If A is any matrix and K is any constant or a scalar, then the matrix obtained by multiplying each element of A by K is denoted by KA and it is called the scalar multiple of A by K.

Operations on Matrices

Addition of Matrices

Addition of Matrices
Addition of Matrices
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If A and B are two matrices of the same order, then A and B are said to be Conformable or Compatible for addition. The sum of A and B is denoted by A+B and it is obtained by adding corresponding elements of A & B.

The matrix A+B will be of the same order as each of the matrices A and B is

If A= \(\begin{bmatrix} a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\end{bmatrix}\) and B= \(\begin{bmatrix} b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\end{bmatrix}\)

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then A+B =\(\begin{bmatrix} a_{11}+b_{11}&a_{12}+b_{12}&a_{13}+b_{13}\\a_{21}+_b{21}&a_{22}+b_{22}&a_{23}+b_{23}\end{bmatrix}\)

Foe example, if A= \(\begin{bmatrix} 1&2&3\\4&-3&7\end{bmatrix}\) and B= \(\begin{bmatrix} -3&4&7\\8&9&-3\end{bmatrix}\)

Then,

A+B =\(\begin{bmatrix} 1&2&3\\4&-3&7\end{bmatrix}\) + \(\begin{bmatrix} -3&4&7\\8&9&-3\end{bmatrix}\)

= \(\begin{bmatrix} 1-3&2+4&3+7\\4+8&-3+9&7-3\end{bmatrix}\)

= \(\begin{bmatrix} -2&6&10\\12&6&4\end{bmatrix}\)


Subtraction of Matrices

Subtraction of Matrices
Subtraction of Matrices Source:calculator.mathcaptain.com

If A and B are two matrices of the same order, then they are said to be conformable for subtraction. The difference of the matrix B from A is denoted by A-B and it is obtained by subtracting the elements of B from the corresponding elements of A.

The order of the matrix A-B is same as the order of A or B.

If A = \(\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22} &a_{23}\end{bmatrix}\) and B = \(\begin{bmatrix}b_{11}&b_{12}&b_{13}
\\b_{21}&b_{22}&b_{23}\end{bmatrix}\)

then A-B = \(\begin{bmatrix} a_{11}-b_{11}&a_{12}-b_{12}&a_{13}-b_{13}\\a_{21}-b_{21}&a_{22}-b_{22}&a_{23}-b_{23}\end{bmatrix}\)

Foe example, if A =\(\begin{bmatrix} 1&2&3\\4&-3&7\end{bmatrix}\) and B= \(\begin{bmatrix} -3&4&7\\8&9&-3\end{bmatrix}\)

Then

A-B=\(\begin{bmatrix} 1&2&3\\4&-3&7\end{bmatrix}\) and B= \(\begin{bmatrix} -3&4&7\\8&9&-3\end{bmatrix}\)

=\(\begin{bmatrix} 1+3&2-1&3-7\\4-8&-3-9&7+3\end{bmatrix}\)

=\(\begin{bmatrix} 4&-2&-4\\-4&-12&10\end{bmatrix}\)

B-A =\(\begin{bmatrix} -3&4&7\\8&9&-3\end{bmatrix}\) -\(\begin{bmatrix} 1&2&3\\4&-3&7\end{bmatrix}\)

=\(\begin{bmatrix} -3-1&4-2&7-3\\8-4&9+3&-3-7\end{bmatrix}\)

=\(\begin{bmatrix} -4&2&4\\4&-12&-10\end{bmatrix}\)

 

Multiplication of a matrix by a scalar (real number)

Multiplication of Matrices
Multiplication of Matrices Source:thejuniverse.org

If A is any matrix and K is any constant or a scalar, then the matrix obtained by multiplying each element of A by K is denoted by KA and it is called the scalar multiple of A by K. 

Algebraic properties of Matrix Addition

Addition of matrices satisfy the following properties:

Closure property

If A and B are two matrices of the same order, then their A+B is also a matrix of the same order as that of A or B.

If A=\(\begin{bmatrix} 2&4\\3&5\end{bmatrix}\) and B=\(\begin{bmatrix} 6&5\\2&1\end{bmatrix}\)

Now,

A+B = \(\begin{bmatrix} 2&4\\3&5\end{bmatrix}\) + \(\begin{bmatrix} 6&5\\2&1\end{bmatrix}\)

= \(\begin{bmatrix} 8&9\\5&6\end{bmatrix}\) which is again is again a2x2 matrix. Hence the closer property is satisfied.

Commutative property

If A and B are two matrices of the same order, then A+B=B+A.

Let, A = \(\begin{bmatrix} 2&4\\6&8\end{bmatrix}\) + \(\begin{bmatrix} 7&-2\\8&4\end{bmatrix}\)

Then A+B =\(\begin{bmatrix} 2&4\\6&8\end{bmatrix}\) +\(\begin{bmatrix} 7&-2\\8&4\end{bmatrix}\)

=\(\begin{bmatrix} 9&2\\14&12\end{bmatrix}\)

B+A =\(\begin{bmatrix} 7&-2\\8&4\end{bmatrix}\) +\(\begin{bmatrix} 2&4\\6&8\end{bmatrix}\)

=\(\begin{bmatrix}9&2\\14&12\end{bmatrix}\)

∴ A+B=B+A

Hence the commutative property is satisfied.

Associative property

If A, B and C three matrices of the same order, then (A+B) + C = A +(B+C)

Existence of identity element

If A is any matrix, then three exists a null matrix 0 of the same order such that A+0 = 0+A=A.

Existence of additive inverse

If A is a matrix of any order, then there exists another matrix -A of same order such that A+(-A) =(-A) +A = 0, then additive identity.

If A and B are the matrices of the same order and K is a scalar, then

K(A+B) = KA+KB.

If A is a matrix and C,K are any two scalars, then (C+K) A=CA+KA

If C,K are any two scalars and A is a matrix, then C(KA) = (CK)A.

Solving Matrix Equations

To solve the matrix equation A + X =B where A and B are two given matrices of the same order and X is unknown, we proceed in a manner similar to numbers.

Here, A + X = B

Adding the matrix (-A)to both sides, we get

(-A)+ A+ X = (-A) + B

or, (-A+A) +X = B-A

or, 0 + X = B-A

or, X = B-A, which is the required solution.

Transpose of matrix

Transpose of Matrix
Transpose of Matrix
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Let A be a matrix. Then the new matrix obtained by interchanging the corresponding rows and column of A is called the transpose of A. It is denoted by A' or At.

If A = \(\begin{bmatrix} a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\end{bmatrix}\)

Then At= \(\begin{bmatrix} a_{11}&a_{12}\\a_{13}&a_{21}\\a_{22}&a_{23}\end{bmatrix}\)

Here, the order of A' is 2 x 3 and that of At is 3 x 2. Hence, if the order of matrix A is mxn, then the order of At will be nxm. If A' is a square, matrix of order n, then At is also a square matrix of order n. If A' is a row matrix, then At is a column matrix.

Properties of transpose

  1. The transpose of the transpose of a matrix is itself. .i.e. (a') = A.
  2. The transpose of the sum of two matrices is the sum of their transposes. i.e (A+)' = A' +B'
  3. If A is any matrix and K is any number, then (KA)' = KA'.
Things to remember
  • The sum of A and B is denoted by A+B and it is obtained by adding corresponding elements of A &B.
  • The difference of the matrix B from A is denoted by A-B and it is obtained by subtracting the elements of B from the corresponding elements of A.
  • It includes every relationship which established among the people.
  • There can be more than one community in a society. Community smaller than society.
  • It is a network of social relationships which cannot see or touched.
  • common interests and common objectives are not necessary for society.
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Questions and Answers

Soln:

(a) here given,\(\begin{bmatrix} x+y \\ x-y \\ \end{bmatrix}\)=\(\begin{bmatrix} 4 \\ 2 \\ \end{bmatrix}\)so,

x+y=4.........(i)

and x-y=2 ........(ii)

Adding two equation, we get,

2x=6 ∴ x=\(\frac{6}{2}\)=3

Putting x=3 in equation (i), we get, x

3+y=4 or, y=4-3 or, y=1

∴x=3 and y=1 Ans.

here given,

\(\begin{bmatrix} x & x+y \\ 2 & 3\\ \end{bmatrix}\)=\(\begin{bmatrix} 4 & 5 \\ 5 & 3 \\ \end{bmatrix}\)

x=4 .........(i)

and x+y=5 ..........(ii)

putting x=4 in equation (ii) we get, 4+y=5,or,y=5-4=1

∴ x=4,y=1. Ans.

Soln:

Here given,

A =\(\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{bmatrix}\) and \(\begin {bmatrix} -1 & 0 & 3 \\ 1 & 0 & 5 \\ \end{bmatrix}\)

We have, commutative law of addition A+B=B+A

Now, A+B=\(\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{bmatrix}\)+\(\begin {bmatrix} -1 & 0 & 3 \\ 1 & 0 & 5 \\ \end{bmatrix}\)

=\(\begin{bmatrix} 1+ (-2) & 2+0 & 3+3 \\ 4+1 & 5+0 & 6+5 \\ \end{bmatrix}\)=\(\begin{bmatrix} 0 & 2 & 6 \\ 5 & 5 & 11 \\ \end{bmatrix}\)....(i)

Again, B+A=\(\begin {bmatrix} -1 & 0 & 3 \\ 1 & 0 & 5 \\ \end{bmatrix}\)+\(\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{bmatrix}\)=\(\begin{bmatrix} -1 + 1 & 0+2 & 3+3 \\ 1+4 & 0+5 & 5+6 \\ \end{bmatrix}\)

=\(\begin{bmatrix} 0 & 2 & 6 \\ 5 & 5 & 11 \\ \end{bmatrix}\)........(ii)

So, from (i) and (ii) A+B=B+A proved

Soln:

here,

\(\begin{bmatrix} x \\ -3 \\ \end{bmatrix}\)+\(\begin{bmatrix} 5 \\ y \\ \end{bmatrix}\)=\(\begin{bmatrix} 3 \\ 8 \\ \end{bmatrix}\)

or, \(\begin{bmatrix} x+5 \\ -3+y \\ \end{bmatrix}\)=\(\begin{bmatrix} 3 \\ 8 \\ \end{bmatrix}\)

or, x+5=3 and -3+y=8

or, x=3-5 and y=8+3

∴ x=-2 and y=11 Ans.

soln:

here given,

\(\begin{bmatrix} 4 & 4 \\ m & 0 \\ \end{bmatrix}\)= \(\begin{bmatrix} n & p \\ 3 & 4 \\ \end{bmatrix}\)+\(\begin{bmatrix} 2 & 3 \\ 3 & r \\ \end{bmatrix}\)

or,\(\begin{bmatrix} 4 & 4 \\ m & 0 \\ \end{bmatrix}\)=\(\begin{bmatrix} n+2 & p+4 \\ 3+3 & 4+r \\ \end{bmatrix}\)

or, \(\begin{bmatrix} 4 & 4 \\ m & 0 \\ \end{bmatrix}\)=\(\begin{bmatrix} n+2 & p+4 \\ 6 & 4+r \\ \end{bmatrix}\)

here, n+2=4..........(i) p+4=4..........(ii) m=9........(iii)

4+r=0..........(iv)

∴n=4-2=2, p=4-4=0, m=6 and r=-4

∴n=2, p=0, m=6 and r=-4 Ans.

soln:

here given, \(\begin{bmatrix} 3x-2 & 5y+4 \\ 2 & 4+2z \\ \end{bmatrix}\)=\(\begin{bmatrix} x+2 & y-4 \\ 2 & z-2 \\ \end{bmatrix}\)

or, 3x-2=x+2, 5y+4=y-4 and 4+2z=z-2

or, 3x-x=2+2 , 5y-y=-4-4 and 2z-z=-2-4

or, 2x=4, 4y=-8 and z=-6

or,x=\(\frac{4}{2}\), y=\(\frac{-8}{4}\) and z=-6

∴x=2, y=-2 and z=-6 Ans.

soln:

here given,

A=\(\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix}\)

(a) here, A'=\(\begin{bmatrix} a & d & g \\ b & e & h \\ c & f & i \\ \end{bmatrix}\)=\(\begin{bmatrix} a & d & g \\ b & e & h \\ c & f & i \\ \end{bmatrix}\) .Ans.

and (A')'=\(\begin{bmatrix} a & d & g \\ b & e & h \\ c & f & i \\ \end{bmatrix}\)= A=\(\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix}\)Ans.

(b) A and (A')' are equal.

i.e. (A')'=A

(c) here, B=\(\begin{bmatrix} 2 & 3 & -4 \\ -5 & 0 & 6 \\ \end{bmatrix}\)

∴B'=\(\begin{bmatrix} 2 & 3 & -4 \\ -5 & 0 & 6 \\ \end{bmatrix}\)'=\(\begin{bmatrix} 2 & -5 \\ 3 & 0 \\ -4 & 6 \\ \end{bmatrix}\).Ans.

and (B')'=\(\begin{bmatrix} 2 & -5 \\ 3 & 0 \\ -4 & 6 \\ \end{bmatrix}\)'=\(\begin{bmatrix} 2 & 3 & -4 \\ -5 & 0 & 6 \\ \end{bmatrix}\).Ans.

∴ (B')'=B

here, the relation of B and (B')' is same as A and (A')' in (b)

soln:

here given,

A=\(\begin{bmatrix} 2 & 3 \\ -4 & -5 \\ \end{bmatrix}\) and B=\(\begin{bmatrix} 0 & -3 \\ 2 & 5 \\ \end{bmatrix}\)

Here, A'=\(\begin{bmatrix} 2 & 3 \\ -4 & -5 \\ \end{bmatrix}\)=\(\begin{bmatrix} 2 & -4 \\ 3 & -5 \\ \end{bmatrix}\)

and, B'=\(\begin{bmatrix} 0 & -3 \\ 2 & 5 \\ \end{bmatrix}\)=\(\begin{bmatrix} 0 & 2 \\ -3 & 5 \\ \end{bmatrix}\)

(a) Now, A'+B'=\(\begin{bmatrix} 2 & -4 \\ 3 & -5 \\ \end{bmatrix}\)+\(\begin{bmatrix} 0 & 2 \\ -3 & 5 \\ \end{bmatrix}\)=\(\begin{bmatrix} 2+0-4+2 \\ 3-3-5+5 \\ \end{bmatrix}\)=\(\begin{bmatrix} 2-2 \\ 0 & 0 \\ \end{bmatrix}\)Ans.

(b) B'+A'=\(\begin{bmatrix} 0 & 2 \\ -3 & 5 \\ \end{bmatrix}\)+\(\begin{bmatrix} 2 & -4 \\ 3 & -5 \\ \end{bmatrix}\)=\(\begin{bmatrix} 0+2 & 2-4 \\ -3+3 & 5-5 \\ \end{bmatrix}\)=\(\begin{bmatrix} 2 & -2 \\ 0 & 0 \\ \end{bmatrix}\).Ans.

(c) From (a) and (b) , we get,

A'+B'=B'+A'

Here,

Given,

P=[1 0 2]

∴Transpose of P (P')=[1 0 2]'=\(\begin{bmatrix} 1 \\ 0 \\ 2 \\ \end{bmatrix}\)Ans.

 

Here,

Given,

 

Q=\(\begin{bmatrix} 3 & 4 & 0 \\ 5 & 6 & -2 \\ \end{bmatrix}\)

∴Transpose of Q (Q')=Q=\(\begin{bmatrix} 3 & 4 & 0 \\ 5 & 6 & -2 \\ \end{bmatrix}\)=\(\begin{bmatrix} 3 & 5 \\ 4 & 6 \\ 0 & -2 \\ \end{bmatrix}\)Ans.

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