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Prove the following

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 $(\mathrm{A})+1=0$, is :

  1. 0

  2. 2

  3. 1

  4. -1

Correct Option: , 3

Solution:

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

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

$1+|\mathrm{A}|=0$

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

Sum $=1$

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