Question

If $A=\left[ \begin{matrix} 0 & 1 & 3 \\\ 1 & 2 & x \\\ 2 & 3 & 1 \\\ \end{matrix} \right]$ and ${{A}^{-1}}=\left[ \begin{matrix} \dfrac{1}{2} & -4 & \dfrac{5}{2} \\\ \dfrac{-1}{2} & 3 & \dfrac{-3}{2} \\\ \dfrac{1}{2} & y & \dfrac{1}{2} \\\ \end{matrix} \right]$, find $x$ and $y$.

Answer 1

To solve this question we need to have the knowledge of matrix and its properties. We will use the formula ${{\left[ A \right]}^{-1}}=\dfrac{1}{\left| A \right|}\text{Adj}\text{.}\left[ A \right]$ , where $\text{Adj}\text{.}$ is the Adjoint Matrix. The first step will be to find the minor matrix of the given matrix $A$. The second step will be to find a cofactor factors matrix of $A$, then we are supposed to find the Adjoint matrix and lastly find the determinant. Then we will equate the R.H.S of the formula with the given matrix ${{\left[ A \right]}^{-1}}$ finding $x$ and $y$.

Complete step by step answer:
The question ask us to find the value of $x$ and $y$when$A=\left[ \begin{matrix} 0 & 1 & 3 \\\ 1 & 2 & x \\\ 2 & 3 & 1 \\\ \end{matrix} \right]$ and ${{A}^{-1}}=\left[ \begin{matrix} \dfrac{1}{2} & -4 & \dfrac{5}{2} \\\ \dfrac{-1}{2} & 3 & \dfrac{-3}{2} \\\ \dfrac{1}{2} & y & \dfrac{1}{2} \\\ \end{matrix} \right]$.
The question will be solved using the formula ${{\left[ A \right]}^{-1}}=\dfrac{1}{\left| A \right|}\text{Adj}\text{.}\left[ A \right]$. We will find the R.H.S and then equate it to the given value of ${{\left[ A \right]}^{-1}}$. The first step in this process will be to find the Minor of matrix of $A$ , which is:
${{M}_{A}}=\left[ \begin{matrix} 2-3x & 1-2x & 3-4 \\\ 1-9 & 0-6 & 0-2 \\\ x-6 & 0-3 & 0-1 \\\ \end{matrix} \right]$
On solving the above matrix we get:
${{M}_{A}}=\left[ \begin{matrix} 2-3x & 1-2x & -1 \\\ -8 & -6 & -2 \\\ x-6 & -3 & -1 \\\ \end{matrix} \right]$
The second step is to find the cofactors matrix of $A$ which can be found using the below formula.
$C_{ij}=(-1)^{i+j}M_{ij}$
${{C}_{A}}=\left[ \begin{matrix} 2-3x & -\left( 1-2x \right) & -1 \\\ -\left( -8 \right) & -6 & -\left( -2 \right) \\\ x-6 & -\left( -3 \right) & -1 \\\ \end{matrix} \right]$
On calculating the above expression we get:
${{C}_{A}}=\left[ \begin{matrix} 2-3x & 2x-1 & -1 \\\ 8 & -6 & 2 \\\ x-6 & 3 & -1 \\\ \end{matrix} \right]$
The third step is to find the Adjoint of matrix $A$, which is transpose of cofactor matrix:
$\text{Adj }\left[ A \right]=\left[ \begin{matrix} 2-3x & 8 & x-6 \\\ 2x-1 & -6 & 3 \\\ -1 & 2 & -1 \\\ \end{matrix} \right]$
The fourth step is to find the determinant of the matrix.
$\left| A \right|=\left| \begin{matrix} 0 & 1 & 3 \\\ 1 & 2 & x \\\ 2 & 3 & 1 \\\ \end{matrix} \right|$
On calculation we get:
$\Rightarrow \left| A \right|=0\left( 2-3x \right)-1\left( 1-2x \right)+3\left( 3-4 \right)$
$\Rightarrow \left| A \right|=-1\left( 1-2x \right)+3\left( -1 \right)$
$\Rightarrow \left| A \right|=2x-1-3$
$\Rightarrow \left| A \right|=2x-4$
$\Rightarrow \left| A \right|=2\left( x-2 \right)$
In the above four steps we found the value for the R.H.S of the formula of the ${{\left[ A \right]}^{-1}}$, which says ${{\left[ A \right]}^{-1}}=\dfrac{1}{\left| A \right|}\text{Adj}\text{.}\left[ A \right]$. On substituting the value in the formula we get:
$\left[ \begin{matrix} \dfrac{1}{2} & -4 & \dfrac{5}{2} \\\ \dfrac{-1}{2} & 3 & \dfrac{-3}{2} \\\ \dfrac{1}{2} & y & \dfrac{1}{2} \\\ \end{matrix} \right]=\dfrac{1}{2\left( x-2 \right)}\left[ \begin{matrix} 2-3x & 8 & x-6 \\\ 2x-1 & -6 & 3 \\\ -1 & 2 & -1 \\\ \end{matrix} \right]$
Now we will be comparing the terms in L.H.S to the terms in R.H.S. We will find the value of $x$. For that we will equate the first term of both the matrices. On doing this we get:
$\Rightarrow \dfrac{1}{2}=\dfrac{2-3x}{2\left( x-2 \right)}$
$\Rightarrow \dfrac{1}{1}=\dfrac{2-3x}{\left( x-2 \right)}$
$\Rightarrow x-2=2-3x$
On calculating further we get:
$\Rightarrow x+3x=2+2$
$\Rightarrow 4x=4$
On dividing both side by $4$ , we get:
$\Rightarrow x=1$
Now we will solve for the value of $y$, the equation for solving this is:
$\Rightarrow y=\dfrac{2}{2\left( x-2 \right)}$
Since we know the value of $x$ , so after substitution the value of $y$ will be:
$\Rightarrow y=\dfrac{1}{\left( 1-2 \right)}$
$\Rightarrow y=-1$
$\therefore$ The value of $x$ and $y$ are $1$ and $-1$ respectively.

Note: Each element in a square matrix has its own minor. The minor is the value of the determinant of the matrix that results from crossing out the row and column of the element under consideration. To find the Adjoint of a matrix we have to find the minor of the matrix and then the cofactor of the matrix.