If A= [ rr 2 & -1 1 & 3 ], then A ^ -1 = ? - Vidyadip

MCQ

If $A=\left[\begin{array}{rr}2 & -1 \\ 1 & 3\end{array}\right]$, then $A ^{-1}= ?$

✓
$\left[\begin{array}{cc}\frac{3}{7} & \frac{1}{7} \\ \frac{-1}{7} & \frac{2}{7}\end{array}\right]$
B
$\left[\begin{array}{cc}\frac{3}{7} & \frac{-1}{7} \\ \frac{1}{7} & \frac{2}{7}\end{array}\right]$
C
$\left[\begin{array}{ll}\frac{1}{5} & \frac{2}{7} \\ \frac{1}{7} & \frac{3}{7}\end{array}\right]$
D
$\left[\begin{array}{ll}\frac{1}{3} & \frac{1}{7} \\ \frac{1}{7} & \frac{2}{7}\end{array}\right]$

✓

Answer

Correct option: A.

$\left[\begin{array}{cc}\frac{3}{7} & \frac{1}{7} \\ \frac{-1}{7} & \frac{2}{7}\end{array}\right]$

$A^{-1}=\frac{1}{|A|} \text{adj } A \ldots (i)$
$ |A|=3 \times 2-(1) \times(-1)$
$=7$
$C_{11}=3, C_{12}=-1$
$C_{21}=1, C_{22}=2$
Co$-$factor matrix $A=\left(\begin{array}{ll}2 & 1 \\ 4 & 3\end{array}\right)$
$\text{Adj } A=\left(\begin{array}{ll}3 & -1 \\ 1 & 2\end{array}\right)^{\prime}$
$=\left(\begin{array}{cc}3 & 1 \\-1 & 2\end{array}\right)$
Putting in $1$
$\begin{array}{l} A^{-1}=\frac{1}{|7|}\left(\begin{array}{cc}3 & 1 \\-1 & 2\end{array}\right) \\ =\left(\begin{array}{cc} \frac{3}{7} & \frac{1}{7} \\ \frac{-1}{7} & \frac{2}{7} \end{array}\right) \end{array}$

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