Question

What is the solution of
the given matrix
equation?
[3 1
5 2]
X =
[-1 2
3 1]​

Answer 1

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

Given that,

$\rm :\longmapsto\:\: \begin{bmatrix} 3 & 1\\ 5 & 2\end{bmatrix}X = \: \begin{bmatrix} - 1 & 2\\ 3 & 1\end{bmatrix} - - (1)$

Let assume that,

$\rm :\longmapsto\:A = \begin{bmatrix} 3 & 1\\ 5 & 2\end{bmatrix}$

and

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

So, equation (1) can be rewritten as

$\rm :\longmapsto\:AX = B$

$\red{ \bf \: Pre - multiply \: by \: {A}^{ - 1} \: on \: both \: sides}$

$\rm :\longmapsto\: {A}^{ - 1} AX = {A}^{ - 1} B$

$\rm :\longmapsto\: X = {A}^{ - 1} B - - (2)$

Now,

As

$\rm :\longmapsto\:A = \begin{bmatrix} 3 & 1\\ 5 & 2\end{bmatrix}$

$\rm :\longmapsto\: |A| = 6 - 5 = 1$

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

We know,

$\rm :\longmapsto\: {A}^{ - 1} = \dfrac{1}{ |A| } adjA$

Hence,

$\rm :\longmapsto\: {A}^{ - 1} = \dfrac{1}{1} \begin{bmatrix} 2 & - 1\\ - 5 & 3\end{bmatrix}$

$\bf\implies \: {A}^{ - 1} = \begin{bmatrix} 2 & - 1\\ - 5 & 3\end{bmatrix} - - - (3)$

Now substituting all the values in equation (2), we get

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

$\rm :\longmapsto\:X = \begin{bmatrix} - 2 - 3 & 4- 1\\ 5 + 9 & - 10 + 3\end{bmatrix}\:$

$\bf :\longmapsto\:X = \begin{bmatrix} -5 & 3\\ 14 & - 7\end{bmatrix}\:$

Additional Information :-

$\boxed{ \sf \: {AA}^{ - 1} = {A}^{ - 1}A =I}$

$\boxed{ \sf \: A \: (adjA) = (adjA) \: A = |A|I}$

$\boxed{ \sf \: |adjA| = { |A| }^{n - 1}}$

$\boxed{ \sf \: |AB| = |A| |B| }$

Answer 2

Step-by-step explanation:

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

Given that,

$\rm :\longmapsto\:\: \begin{bmatrix} 3 & 1\\ 5 & 2\end{bmatrix}X = \: \begin{bmatrix} - 1 & 2\\ 3 & 1\end{bmatrix} - - (1)$

Let assume that,

$\rm :\longmapsto\:A = \begin{bmatrix} 3 & 1\\ 5 & 2\end{bmatrix}$

and

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

So, equation (1) can be rewritten as

$\rm :\longmapsto\:AX = B$

$\red{ \bf \: Pre - multiply \: by \: {A}^{ - 1} \: on \: both \: sides}$

$\rm :\longmapsto\: {A}^{ - 1} AX = {A}^{ - 1} B$

$\rm :\longmapsto\: X = {A}^{ - 1} B - - (2)$

Now,

As

$\rm :\longmapsto\:A = \begin{bmatrix} 3 & 1\\ 5 & 2\end{bmatrix}$

$\rm :\longmapsto\: |A| = 6 - 5 = 1$

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

We know,

$\rm :\longmapsto\: {A}^{ - 1} = \dfrac{1}{ |A| } adjA$

Hence,

$\rm :\longmapsto\: {A}^{ - 1} = \dfrac{1}{1} \begin{bmatrix} 2 & - 1\\ - 5 & 3\end{bmatrix}$

$\bf\implies \: {A}^{ - 1} = \begin{bmatrix} 2 & - 1\\ - 5 & 3\end{bmatrix} - - - (3)$

Now substituting all the values in equation (2), we get

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

$\rm :\longmapsto\:X = \begin{bmatrix} - 2 - 3 & 4- 1\\ 5 + 9 & - 10 + 3\end{bmatrix}\:$

$\bf :\longmapsto\:X = \begin{bmatrix} -5 & 3\\ 14 & - 7\end{bmatrix}\:$

Additional Information :-

$\boxed{ \sf \: {AA}^{ - 1} = {A}^{ - 1}A =I}$

$\boxed{ \sf \: A \: (adjA) = (adjA) \: A = |A|I}$

$\boxed{ \sf \: |adjA| = { |A| }^{n - 1}}$

$\boxed{ \sf \: |AB| = |A| |B| }$