Q. Given that matrix$A = \begin{bmatrix} x & 3 & 2 \\ 1 & y & 4 \\ 2 & 2 & \mathcal{z} \end{bmatrix}$. If $xy\mathcal{z} = 60$ and $8x + 4y + 3\mathcal{z} = 20$, then $A(adjA)$ is equal to
A) $\begin{bmatrix}
64 & 0 & 0 \\
0 & 64 & 0 \\
0 & 0 & 64
\end{bmatrix}$
B) $\begin{bmatrix}
88 & 0 & 0 \\
0 & 88 & 0 \\
0 & 0 & 88
\end{bmatrix}$
C) $\begin{bmatrix}
68 & 0 & 0 \\
0 & 68 & 0 \\
0 & 0 & 68
\end{bmatrix}$
D) $\begin{bmatrix}
34 & 0 & 0 \\
0 & 34 & 0 \\
0 & 0 & 34
\end{bmatrix}$
Explanation
(c)
$A$ adj $A|A|I$
$|A| = xy\mathcal{z -}8x - 3\left( \mathcal{z -}8 \right) + 2(2 - 2y)$
$|A| = xy\mathcal{z -}\left( 8x + 3\mathcal{z} + 4y \right) + 28$
$= 60 - 20 + 28$
$= 68$
$\Rightarrow A(adjA) = 68\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix} = \begin{bmatrix}
68 & 0 & 0 \\
0 & 68 & 0 \\
0 & 0 & 68
\end{bmatrix}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.