Let $A$ and $B$ be two square matrices of order 3 such that $|A|=3$ and $|B|=2$. Then…

Let $A$ and $B$ be two square matrices of order 3 such that $|A|=3$ and $|B|=2$. Then $\left|\mathrm{A}^{\mathrm{T}} \mathrm{A}(\operatorname{adj}(2 \mathrm{~A}))^{-1}(\operatorname{adj}(4 \mathrm{~B}))(\operatorname{adj}(\mathrm{AB}))^{-1} \mathrm{AA}^{\mathrm{T}}\right|$ is equal to :
  1. 108
  2. 32
  3. 81
  4. 64

Solution

$\begin{aligned} & |\mathrm{A}|=3,|\mathrm{~B}|=2 \\ & \left|\mathrm{~A}^{\mathrm{T}} \mathrm{A}(\operatorname{adj}(2 \mathrm{~A}))^{-1}(\operatorname{adj}(4 \mathrm{~B}))(\operatorname{adj}(\mathrm{AB}))^{-1} \mathrm{AA}^{\mathrm{T}}\right|\end{aligned}$ \(\begin{array}{ccc} =3 \times 3 \times \mid \left(\operatorname{adj}(2 \mathrm{~A})^{-1}|\times| \operatorname{adj}(4 \mathrm{~B})| \times|(\operatorname{adj}(\mathrm{AB}))^{-1}\mid \times 3 \times 3\right. \\ \downarrow \qquad\qquad \downarrow \qquad\qquad \downarrow \\ \frac{1}{|\operatorname{adj}(2 \mathrm{~A})|} \qquad 2^{12} \times 2^2 \qquad \frac{1}{|\operatorname{adj}(\mathrm{AB})|} \end{array}\)
$\begin{aligned} & =\frac{1}{2^6|\operatorname{adj} \mathrm{A}|} \quad=\frac{1}{|\operatorname{adj} \mathrm{B} \cdot \operatorname{adj} \mathrm{A}|} \\ & =\frac{1}{2^6 \cdot 3^2} \quad=\frac{1}{2^2 \cdot 3^2} \\ & =3^4 \cdot \frac{1}{2^6 \cdot 3^2} \cdot 2^{12} \cdot 2^2 \cdot \frac{1}{2^2 \cdot 3^2}=64 \\ & \end{aligned}$

Asked in: JEE Main 2024 (05 Apr Shift 1)

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