Question

If $\alpha $ and $\beta $ are the roots of the equation $\left[1 5\right]\begin{bmatrix} 1 & 3 \\ -4 & 7 \end{bmatrix}^{2}\begin{bmatrix} \frac{7}{19} & -\frac{3}{19} \\ \frac{4}{19} & \frac{1}{19} \end{bmatrix}^{4}$ $\begin{bmatrix} 1 & 3 \\ -4 & 7 \end{bmatrix}^{2}\begin{bmatrix} x^{2}-5x+5 \\ -3 \end{bmatrix}=\left[- 4\right],$ then the value of $\left(2 - \alpha \right)\left(2 - \beta \right)$ is

• A. $51$
• B. $-12$
• C. $12$
• D. $-7$

Answer 1

Answer: (B)

Let, $A=\begin{bmatrix} 1 \\ -4 3 \\ 7 \end{bmatrix}$ then $A^{- 1}=\begin{bmatrix} \frac{7}{19} & -\frac{3}{19} \\ \frac{4}{19} & \frac{1}{19} \end{bmatrix}$
Since, $AA^{- 1}=I=A^{- 1}A$
Hence, $\begin{bmatrix} 1 & 3 \\ -4 & 7 \end{bmatrix}\begin{bmatrix} \frac{7}{19} & -\frac{3}{19} \\ \frac{4}{19} & \frac{1}{19} \end{bmatrix}=I=\begin{bmatrix} \frac{7}{19} & -\frac{3}{19} \\ \frac{4}{19} & \frac{1}{19} \end{bmatrix}\begin{bmatrix} 1 & 3 \\ -4 & 7 \end{bmatrix}$
$\Rightarrow \begin{bmatrix} 1 & 5 \end{bmatrix}\begin{bmatrix} x^{2}-5x+5 \\ -3 \end{bmatrix}=\left[- 4\right]$
$\Rightarrow x^{2}+5x+5-15=-4$
$\Rightarrow x^{2}-5x-6=0\Rightarrow x=6,-1$
Hence, $\left(2 - \alpha \right)\left(2 - \beta \right)=-4\times 3=-12$