Question

If |5 2α 1|
B = |0 2 1|
|α 3 -1|
is the inverse of a 3 × 3 matrix A, then the sum of all values of α for which det (A) + 1 = 0, is :
(A) 0 (B) –1
(C) 1 (D) 2

Answer 1

$\textbf{Given:}$

$B=\left(\begin{array}{ccc}5&2\alpha&1\\0&2&1\\\alpha&3&-1\end{array}\right)$

$det(B)=\left|\begin{array}{ccc}5&2\alpha&1\\0&2&1\\\alpha&3&-1\end{array}\right|$

$det(B)=5(-2-3)-2\alpha(0-\alpha)+1(0-2\alpha)$

$det(B)=-25+2{\alpha}^2-2\alpha$

$\text{Since B is the inverse of A, we have }AB=I$

$\implies\,det(AB)=det(I)$

$\implies\,det(A)det(B)=1$

$\implies\,det(A)=\frac{1}{det(B)}$

$\implies\,det(A)=\frac{1}{2{\alpha}^2-2\alpha-25}$

$\text{Also, }det(A)+1=0$

$\frac{1}{2{\alpha}^2-2\alpha-25}+1=0$

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

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

$\implies(\alpha-4)(\alpha+3)=0$

$\implies\alpha=4,-3$

$\therefore\textbf{Sum of the values of \alpha =4+(-3)=1}$

$\implies\textbf{Option (C) is correct}$

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