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

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:

\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.