Question

The inverse of the matrix $\left[ \begin{matrix}
   3 & {-2} \\
  1 & 4 \\
\end{matrix} \right]$ is
A. $\left[ \begin{matrix}
   {\dfrac {4}{14}} & {\dfrac {2}{14}} \\
   {\dfrac {-1}{14}} & {\dfrac {3}{14}} \\
\end{matrix} \right]$

B. $\left[ \begin{matrix}
   {\dfrac {3}{14}} & {\dfrac {-2}{14}} \\
   {\dfrac {1}{14}} & {\dfrac {4}{14}} \\
\end{matrix} \right]$

C. $\left[ \begin{matrix}
   {\dfrac {4}{14}} & {\dfrac {-2}{14}} \\
   {\dfrac {1}{14}} & {\dfrac {3}{14}} \\
\end{matrix} \right]$

D. $\left[ \begin{matrix}
   {\dfrac {3}{14}} & {\dfrac {2}{14}} \\
   {\dfrac {1}{14}} & {\dfrac {4}{14}} \\
\end{matrix} \right]$

Answer 1

Hint:
We have to find ${{A}^{-1}}$ of this 2×2 matrix, so for that we will solve for $adjA$ and $|A|$. And then dividing both of them will get our required solution. As for finding ${{A}^{-1}}$ the formula suggests ${{A}^{-1}}=\dfrac{adjA}{|A|}$.



Formula Used:
${{A}^{-1}}=\dfrac{adjA}{|A|}$
|A|=${{a}_{11}}{{a}_{22}}-{{a}_{12}}{{a}_{21}}$



Complete step-by-step answer:
We are given matrix $A$= $\left[ \begin{matrix}
   3 & {-2} \\
  1 & 4 \\
\end{matrix} \right]$ and we have to find ${{A}^ {-1}} $
So first of all as the formula suggests we have to find $adjA$,
$adjA$ of matrix $\left[ \begin{matrix}
   3 & {-2} \\
  1 & 4 \\
\end{matrix} \right]$ is $adjA=\left[ \begin{matrix}
   4 & 2 \\
   -1 & 3 \\
\end{matrix} \right]$.
As we have to interchange element ${{a}_{11}}$to element ${{a}_{22}}$
And change signs of element ${{a}_{12}}$and ${{a}_{21}}$
Now find the value of determinant of A,
$|A|= 4(3) -(-2) (1) $
$|A| =12+2$
$\Rightarrow |A|=14$
Now as ${{A}^ {-1}} =\dfrac{adjA}{|A|} $
$\Rightarrow {{A}^{-1}}=\dfrac{\left[ \begin{matrix}
   \begin{matrix}
   4 \\
   -1 \\
\end{matrix} & \begin{matrix}
   2 \\
   3 \\
\end{matrix} \\
\end{matrix} \right]}{14}$
$\Rightarrow {{A}^{-1}}=\dfrac{1}{14}\left[ \begin{matrix}
   \begin{matrix}
   4 \\
   -1 \\
\end{matrix} & \begin{matrix}
   2 \\
   3 \\
\end{matrix} \\
\end{matrix} \right]$
$\Rightarrow {{A}^{-1}}=\left[ \begin{matrix}
   \begin{matrix}
   \dfrac{4}{14} \\
   \dfrac{-1}{14} \\
\end{matrix} & \begin{matrix}
   \dfrac{2}{14} \\
   \dfrac{3}{14} \\
\end{matrix} \\
\end{matrix} \right]$
Hence option A is correct.

Note:
${{A}^{-1}}$ exists only when $|A|\ne 0$. We have to remember the formula for ${{A}^{-1}}$ and for that we have to solve $adjA$ and $|A|$ for a matrix. Sometimes students make mistakes while solving $adjA$ and $|A|$ as they don’t interchange elements in the formula of $adjA$ and $|A|$ also we have to cross multiply very carefully and take care of signs as well.