Question

Find the matrix A satisfying the matrix equation $\left[ \begin{matrix} 2 & 1 \\\ 3 & 2 \\\ \end{matrix} \right]A\left[ \begin{matrix} -3 & 2 \\\ 5 & -3 \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\\ 0 & 1 \\\ \end{matrix} \right]$.

Answer 1

We first need to solve the matrix equation by multiplying the inverse of the matrices X and Y which gives only matrix A on one side. Now we need to find the inverse of the matrix by finding the adjoint of the matrices. We use the inverse matrices and the matrix $\left[ \begin{matrix} 1 & 0 \\\ 0 & 1 \\\ \end{matrix} \right]$ to form a multiplication to find the matrix A.

We have been given a matrix equation $\left[ \begin{matrix} 2 & 1 \\\ 3 & 2 \\\ \end{matrix} \right]A\left[ \begin{matrix} -3 & 2 \\\ 5 & -3 \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\\ 0 & 1 \\\ \end{matrix} \right]$.
We assume $X=\left[ \begin{matrix} 2 & 1 \\\ 3 & 2 \\\ \end{matrix} \right]$, $Y=\left[ \begin{matrix} -3 & 2 \\\ 5 & -3 \\\ \end{matrix} \right]$ and $Z=\left[ \begin{matrix} 1 & 0 \\\ 0 & 1 \\\ \end{matrix} \right]$.
So, the equation becomes $XAY=Z$. We need to find the matrix A.

Complete step by step answer:
We first multiply with ${{X}^{-1}}$ to get
$\begin{aligned} & {{X}^{-1}}XAY={{X}^{-1}}Z \\\ & \Rightarrow IAY={{X}^{-1}}Z \\\ & \Rightarrow AY={{X}^{-1}}Z \\\ \end{aligned}$
We then multiply with ${{Y}^{-1}}$ to get
$\begin{aligned} & AY={{X}^{-1}}Z \\\ & \Rightarrow AY{{Y}^{-1}}={{X}^{-1}}Z{{Y}^{-1}} \\\ & \Rightarrow AI={{X}^{-1}}Z{{Y}^{-1}} \\\ & \Rightarrow A={{X}^{-1}}Z{{Y}^{-1}} \\\ \end{aligned}$
We found the matrix A. We now need to find the inverse matrix of X and Y. For inverse we need to find the adjoint of that matrix divided by its determinant value. So, for matrix M the inverse matrix will be ${{M}^{-1}}=\dfrac{adj\left( M \right)}{\left| M \right|}$.
In case of $X=\left[ \begin{matrix} 2 & 1 \\\ 3 & 2 \\\ \end{matrix} \right]$, $adj\left( X \right)=\left[ \begin{matrix} 2 & -1 \\\ -3 & 2 \\\ \end{matrix} \right]$ and $\left| X \right|=4-3=1$.
So, ${{X}^{-1}}=\dfrac{adj\left( X \right)}{\left| X \right|}=\left[ \begin{matrix} 2 & -1 \\\ -3 & 2 \\\ \end{matrix} \right]$.
In case of $Y=\left[ \begin{matrix} -3 & 2 \\\ 5 & -3 \\\ \end{matrix} \right]$, $adj\left( Y \right)=\left[ \begin{matrix} -3 & -2 \\\ -5 & -3 \\\ \end{matrix} \right]$ and $\left| Y \right|=9-10=-1$.
So, ${{Y}^{-1}}=\dfrac{adj\left( Y \right)}{\left| Y \right|}=\dfrac{1}{-1}\left[ \begin{matrix} -3 & -2 \\\ -5 & -3 \\\ \end{matrix} \right]=\left[ \begin{matrix} 3 & 2 \\\ 5 & 3 \\\ \end{matrix} \right]$.
Now we need to find multiplication value of $A={{X}^{-1}}Z{{Y}^{-1}}$.
So, $A=\left[ \begin{matrix} 2 & -1 \\\ -3 & 2 \\\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 \\\ 0 & 1 \\\ \end{matrix} \right]\left[ \begin{matrix} 3 & 2 \\\ 5 & 3 \\\ \end{matrix} \right]$. We do the multiplications one by one from left.
$A=\left[ \begin{matrix} 2 & -1 \\\ -3 & 2 \\\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 \\\ 0 & 1 \\\ \end{matrix} \right]\left[ \begin{matrix} 3 & 2 \\\ 5 & 3 \\\ \end{matrix} \right]=\left[ \begin{matrix} 2+0 & 0-1 \\\ -3+0 & 0+2 \\\ \end{matrix} \right]\left[ \begin{matrix} 3 & 2 \\\ 5 & 3 \\\ \end{matrix} \right]=\left[ \begin{matrix} 2 & -1 \\\ -3 & 2 \\\ \end{matrix} \right]\left[ \begin{matrix} 3 & 2 \\\ 5 & 3 \\\ \end{matrix} \right]$.
Now we do the second multiplication and get
$A=\left[ \begin{matrix} 2 & -1 \\\ -3 & 2 \\\ \end{matrix} \right]\left[ \begin{matrix} 3 & 2 \\\ 5 & 3 \\\ \end{matrix} \right]=\left[ \begin{matrix} 6-5 & 4-3 \\\ -9+10 & -6+6 \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 1 \\\ 1 & 0 \\\ \end{matrix} \right]$.
Therefore, the matrix A is $A=\left[ \begin{matrix} 1 & 1 \\\ 1 & 0 \\\ \end{matrix} \right]$.

Note: The easiest way to remember the adjoint of a $\left( 2\times 2 \right)$ matrix is to interchange the position of its diagonal elements and to change the sign of the other two elements. So, if a $\left( 2\times 2 \right)$ matrix is A where $A=\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]$, then $adj\left( A \right)=\left[ \begin{matrix} d & -b \\\ -c & a \\\ \end{matrix} \right]$.