Question

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

• A. $2$
• B. $1$
• C. $0$
• D. $-1$

Answer 1

The correct option is B $1$

As $B$ is the inverse of matrix $A$
Therefore $det\left(A\right)=\frac{1}{det\left(B\right)}det\left(B\right)=5(-2-3)+\alpha (2\alpha -2)=2{\alpha }^2-2\alpha -25$
Also $det\left(A\right)+1=0$
$\Rightarrow {\alpha }^2-\alpha -12=0$
$\therefore \alpha =4,-3$
Sum$=4-3=1$