Question

If $B =\begin{bmatrix}5&2\alpha&1\\ 0&2&1\\ \alpha&3&-1\end{bmatrix}$is the inverse of a $3 \times3$ matrix $A$ then the sum of all values of $\alpha$ for which det $\left(A\right) +1=0$ is

Answer 1

$\because B =A^{-1} =\Rightarrow \left| B\right| = \frac{1}{\left| A\right|}$
Now, $\left| B\right| =\begin{vmatrix}5&2\alpha&1\\ 0&2&1\\ \alpha&3&-1\end{vmatrix} = 2\alpha^{2} -2\alpha -25$
Given, det $\left(A\right) + 1 =0$
$\Rightarrow \frac{1}{2\alpha^{2} -2\alpha -25} +1 =0$
$\Rightarrow \frac{2\alpha^{2} -2\alpha -24}{2\alpha^{2} -2\alpha -25} =0$
$\Rightarrow \alpha =4 -3$
$\Rightarrow $ sum of values of $\alpha =1$