Matrices and Determinants 3 Question 1
1. 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 $\operatorname{det}(A)+1=0$, is
(a) 0
(b) -1
(c) 1
(d) 2
(2019 Main, 12 April I)
Show Answer
Answer:
Correct Answer: 1. (c)
Solution:
- Given matrix $B$ is the inverse matrix of $3 \times 3$ matrix $A$,
where $B=\left[\begin{array}{ccc}5 & 2 \alpha & 1 \\ 0 & 2 & 1 \\ \alpha & 3 & -1\end{array}\right]$
We know that, $ \operatorname{det}(A)=\frac{1}{\operatorname{det}(B)} \quad\left[\because \operatorname{det}\left(A^{-1}\right)=\frac{1}{\operatorname{det}(A)}\right] $
Since, $\quad \operatorname{det}(A)+1=0$ (given)
$ \begin{aligned} & \frac{1}{\operatorname{det}(B)}+1=0 \\ & \Rightarrow \quad \operatorname{det}(B)=-1 \\ & \Rightarrow \quad 5(-2-3)-2 \alpha(0-\alpha)+1(0-2 \alpha)=-1 \\ & \Rightarrow \quad-25+2 \alpha^2-2 \alpha=-1 \\ & \Rightarrow \quad 2 \alpha^2-2 \alpha-24=0 \\ & \Rightarrow \quad \alpha^2-\alpha-12=0 \\ & \Rightarrow \quad(\alpha-4)(\alpha+3)=0 \\ & \Rightarrow \quad \alpha=-3,4 \\ \end{aligned} $
So, required sum of all values of $\alpha$ is $4-3=1$
