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

Option 1)
$0$

Option 2)
$1$

Option 3)
$2$

Option 4)
$-1$

Answers (2)

$B=\begin{bmatrix} 5 &2\alpha &1 \\ 0 &2 &1 \\ \alpha & 3 &-1 \end{bmatrix}$

$\left | B \right |=5\left ( -5 \right )-2\alpha \left ( -\alpha \right )-2\alpha$

$=2\alpha ^{2}-2\alpha -25$

$1+\left | A \right |=2\alpha ^{2}-2\alpha -25+1=0$

$=2\alpha ^{2}-2\alpha -24=0$

$=\alpha ^{2}-\alpha -12=0$

$=\alpha ^{2}-4\alpha +3\alpha -12=0$

$=\left ( \alpha -4 \right )\left ( \alpha +3 \right )=0$

$root=$ $4,-3$

$sum\: \: of\: \: root=4-3$

$=1$

Option 1)

$0$

Option 2)

$1$

Option 3)

$2$

Option 4)

$-1$

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