If $A=\left[\begin{array}{ccc}2 & 1 & 3 \\ -1 & 2 & 0 \\ 4 & 1 & 3\end{array}\right],…
If $A=\left[\begin{array}{ccc}2 & 1 & 3 \\ -1 & 2 & 0 \\ 4 & 1 & 3\end{array}\right], B=\left[\begin{array}{lll}3 & 2 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 0\end{array}\right]$, then $\operatorname{det}\left(2 B^{-1} A^{-1}\right)$ is equal to
- $\frac{1}{6}$
- $\frac{-1}{24}$
- $\frac{1}{3}$
- $\frac{-1}{6}$
Solution
$
\text { (d) } \begin{aligned}
A & =\left[\begin{array}{ccc}
2 & 1 & 3 \\
-1 & 2 & 0 \\
4 & 1 & 3
\end{array}\right], B=\left[\begin{array}{lll}
3 & 2 & 1 \\
1 & 2 & 3 \\
3 & 1 & 0
\end{array}\right] \\
A^{-1} & =\frac{1}{|A|} \operatorname{adj} A, B^{-1}=\frac{1}{|B|} \operatorname{adj} B \\
|A| & =2(6)-1(-3)+3(-1-8),|B|=4 \\
& =12+3-27=-12
\end{aligned}
$
$
\begin{aligned}
\operatorname{adj} A & =\left[\begin{array}{ccc}
6 & 0 & -6 \\
3 & -6 & -3 \\
-9 & 2 & 5
\end{array}\right], \operatorname{adj} B=\left[\begin{array}{ccc}
-3 & 1 & 4 \\
9 & -3 & -8 \\
-5 & 3 & 4
\end{array}\right] \\
\therefore A^{-1} & =\left[\begin{array}{ccc}
-1 / 2 & 0 & 1 / 2 \\
-1 / 4 & 1 / 2 & 1 / 4 \\
3 / 4 & -1 / 6 & -5 / 12
\end{array}\right], \\
B^{-1} & =\left[\begin{array}{ccc}
-3 / 4 & 1 / 4 & 1 \\
9 / 4 & -3 / 4 & -2 \\
-5 / 4 & 3 / 4 & 1
\end{array}\right]
\end{aligned}
$
Now, $2 B^{-1} A^{-1}=2\left[\begin{array}{ccc}-3 / 4 & 1 / 4 & 1 \\ 9 / 4 & -3 / 4 & -2 \\ -5 / 4 & 3 / 4 & 1\end{array}\right]$
$\left[\begin{array}{ccc}-1 / 2 & 0 & 1 / 2 \\ -1 / 4 & 1 / 2 & 1 / 4 \\ 3 / 4 & -1 / 6 & -5 / 12\end{array}\right]$
$=\left[\begin{array}{ccc}17 / 8 & -1 / 12 & -35 / 24 \\ -39 / 8 & -1 / 12 & 85 / 24 \\ 19 / 8 & 5 / 12 & -41 / 24\end{array}\right]$
$\therefore \operatorname{det}\left(2 B^{-1} A^{-1}\right)=\left|\begin{array}{ccc}17 / 8 & -1 / 12 & -35 / 24 \\ -39 / 8 & -1 / 12 & 85 / 24 \\ 19 / 8 & 5 / 12 & -41 / 24\end{array}\right|$
$=-\frac{1}{6}$
Asked in: AP EAMCET 2021 (23 Aug Shift 1)
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