Question

$\begin{bmatrix}{ 2 } & { 3 } \\ { 5 } & { 4 } \end{bmatrix} \begin{bmatrix}{ 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{bmatrix}$​

Answer 1

$\large\underline{\sf{Solution-}}$

Given matrix is

$\rm \: \begin{bmatrix}{ 2 } & { 3 } \\ { 5 } & { 4 } \end{bmatrix} \begin{bmatrix}{ 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{bmatrix}$

Let assume that

$\rm \: A = \begin{bmatrix}{ 2 } & { 3 } \\ { 5 } & { 4 } \end{bmatrix}$

and

$\rm \: B = \begin{bmatrix}{ 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{bmatrix}$

It means, Order of matrix A is 2 × 2 and order of matrix B is 2 × 3.

Now, we have to find AB.

We know, matrix multiplication of two matrices are defined if number of columns of pre multiplier is equal to number of rows of post multiplier.

Now, number of columns of matrix A is 2 and number of rows of matrix B is 2.

So, it implies, AB exist.

So,

$\rm \: AB$

$\rm \: = \: \begin{bmatrix}{ 2 } & { 3 } \\ { 5 } & { 4 } \end{bmatrix} \begin{bmatrix}{ 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{bmatrix}$

$\rm \: =  \: \begin{bmatrix}{2(3) + 3( - 1) } & { 2(0) + 3(1) } & { 2(3) + 3(5) } \\ {5(2) + 4( - 1) } & { 5(0) + 4(1) } & {5(3) + 4( 5) } \end{bmatrix}$

$\rm \: =  \: \begin{bmatrix}{ 4 - 3 } & { 0 + 3} & {6 + 15} \\ {10 -4} & {0 + 4 } & { 15 + 20} \end{bmatrix}$

$\rm \: =  \: \begin{bmatrix}{ 1 } & {3} & {21} \\ {6} & {4 } & {35} \end{bmatrix}$

Hence,

$\rm\implies \boxed{\sf{\rm \: \begin{bmatrix}{ 2 } & { 3 } \\ { 5 } & { 4 } \end{bmatrix} \begin{bmatrix}{ 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{bmatrix}=  \: \begin{bmatrix}{ 1 } & {3} & {21} \\ {6} & {4 } & {35} \end{bmatrix} \: }}$

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Additional Information :-

1. Matrix multiplication may or may not be Commutative.

2. Matrix multiplication is Associative, i.e. (AB)C = A(BC)

3. Matrix multiplication is Distributive, i.e A(B + C) = AB + AC