If A is solution of x^2-3 x-7=0 and A= [ cc 5 & 3 -1 & -2 ], then… - Vidyadip

MCQ

If A is solution of $x^2-3 x-7=0$ and $A=\left[\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right]$, then $A^{-1}$ equals

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

✓

Answer

Correct option: A.

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

(A) $A^2-3 A-7 I=0$
$\Rightarrow A -3 I -7 A^{-1}=0 \Rightarrow A^{-1}=\frac{1}{7}(A-3 I )$
$\therefore \quad A ^{-1}=\frac{1}{7}\left\{\left[\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right]-\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]\right\}$
$=\left[\begin{array}{cc}\frac{2}{7} & \frac{3}{7} \\ -\frac{1}{7} & -\frac{5}{7}\end{array}\right]$

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