Question

Find the value of $\alpha $ if \[A=\left[ \begin{matrix}
   \begin{matrix}
   2 \\
   1 \\
\end{matrix} & \begin{matrix}
   3 \\
   -2 \\
\end{matrix} \\
\end{matrix} \right]\] and ${{A}^{-1}}=\alpha A$ .
A. 7
B. -7
C. $\dfrac{1}{7}$
D. $-\dfrac{1}{7}$

Answer 1

Hint: To find the value of $\alpha $ in ${{A}^{-1}}=\alpha A$ , we will first use the formula ${{A}^{-1}}=\dfrac{adj(A)}{\left| A \right|}$ . After finding $adj(A)$ and $\left| A \right|$ , we will substitute these in the previous formula. After few simplifications and rearrangement, this equation will be of the form ${{A}^{-1}}=\alpha A$ . Hence by comparing, we get the value of $\alpha $ .

Complete step by step answer:
It is given that \[A=\left[ \begin{matrix}
   \begin{matrix}
   2 \\
   1 \\
\end{matrix} & \begin{matrix}
   3 \\
   -2 \\
\end{matrix} \\
\end{matrix} \right]\] and ${{A}^{-1}}=\alpha A$ . We need to find the value of $\alpha $ .
Let us consider ${{A}^{-1}}=\alpha A$ .
We know that ${{A}^{-1}}=\dfrac{adj(A)}{\left| A \right|}...(i)$ .
Let us first find $adj(A)$ . Consider a $2\times 2$ matrix as shown below:
\[A=\left[ \begin{matrix}
   \begin{matrix}
   {{a}_{11}} \\
   {{a}_{21}} \\
\end{matrix} & \begin{matrix}
   {{a}_{12}} \\
   {{a}_{22}} \\
\end{matrix} \\
\end{matrix} \right]\]
To find the adjoint of a $2\times 2$ matrix, we should interchange ${{a}_{11}}\text{ and }{{a}_{21}}$ while changing the signs of ${{a}_{12}}\text{ and }{{a}_{21}}$ . Changing signs means + becomes – and – becomes +.
Hence, \[adj(A)=\left[ \begin{matrix}
   \begin{matrix}
   {{a}_{22}} \\
   -{{a}_{21}} \\
\end{matrix} & \begin{matrix}
   -{{a}_{12}} \\
   {{a}_{12}} \\
\end{matrix} \\
\end{matrix} \right]\] .
Hence, we can find the adjoint of \[A=\left[ \begin{matrix}
   \begin{matrix}
   2 \\
   1 \\
\end{matrix} & \begin{matrix}
   3 \\
   -2 \\
\end{matrix} \\
\end{matrix} \right]\] and is shown below.
$adj(A)=\left[ \begin{matrix}
   \begin{matrix}
   -2 \\
   -1 \\
\end{matrix} & \begin{matrix}
   -3 \\
   2 \\
\end{matrix} \\
\end{matrix} \right]...(ii)$
Now, let us find the determinant of A.
Let us consider the determinant of matrix \[A=\left[ \begin{matrix}
   \begin{matrix}
   {{a}_{11}} \\
   {{a}_{21}} \\
\end{matrix} & \begin{matrix}
   {{a}_{12}} \\
   {{a}_{22}} \\
\end{matrix} \\
\end{matrix} \right]\]
\[\left| A \right|=\left| \begin{matrix}
   \begin{matrix}
   {{a}_{11}} \\
   {{a}_{21}} \\
\end{matrix} & \begin{matrix}
   {{a}_{12}} \\
   {{a}_{22}} \\
\end{matrix} \\
\end{matrix} \right|={{a}_{11}}{{a}_{22}}-{{a}_{12}}{{a}_{21}}\]
Now let us find the determinant of the matrix \[A=\left[ \begin{matrix}
   \begin{matrix}
   2 \\
   1 \\
\end{matrix} & \begin{matrix}
   3 \\
   -2 \\
\end{matrix} \\
\end{matrix} \right]\] .
\[\left| A \right|=\left| \begin{matrix}
   \begin{matrix}
   2 \\
   1 \\
\end{matrix} & \begin{matrix}
   3 \\
   -2 \\
\end{matrix} \\
\end{matrix} \right|=\left( 2\times -2 \right)-\left( 3\times 1 \right)\]
This can be solved to give
\[\left| A \right|=-4-3=-7...(iii)\]
Now let us substitute (ii) and (iii) in (i). We will get
${{A}^{-1}}=-\dfrac{1}{7}\left[ \begin{matrix}
   \begin{matrix}
   -2 \\
   -1 \\
\end{matrix} & \begin{matrix}
   -3 \\
   2 \\
\end{matrix} \\
\end{matrix} \right]$
Let us take ‘-‘ common from the matrix to outside. Then we get
\[{{A}^{-1}}=\dfrac{1}{7}\left[ \begin{matrix}
   \begin{matrix}
   2 \\
   1 \\
\end{matrix} & \begin{matrix}
   3 \\
   -2 \\
\end{matrix} \\
\end{matrix} \right]\]
Now, this is of the form
\[{{A}^{-1}}=\dfrac{1}{7}A\]
We can now compare this with ${{A}^{-1}}=\alpha A$ .
Hence, we get $\alpha =\dfrac{1}{7}$

So, the correct answer is “Option C”.

Note: Be careful with the formulas used. The adjoint calculated as explained before , is for a $2\times 2$ matrix. The entire procedure will be different for higher order matrices. When finding the adjoint of $2\times 2$ matrix, \[A=\left[ \begin{matrix}
   \begin{matrix}
   {{a}_{11}} \\
   {{a}_{21}} \\
\end{matrix} & \begin{matrix}
   {{a}_{12}} \\
   {{a}_{22}} \\
\end{matrix} \\
\end{matrix} \right]\] , do not change the signs of ${{a}_{11}}\text{ and }{{a}_{21}}$ and interchange ${{a}_{12}}\text{ and }{{a}_{21}}$ . Do the opposite of this.