MHT CET · Maths · Matrices
Given \(\mathrm{A}=\left[\begin{array}{lll}x & 3 & 2 \\ 1 & y & 4 \\ 2 & 2 & z\end{array}\right], x y z=60\) and \(8 x+4 y+3 z=20\), then \(\mathrm{A} \cdot(\operatorname{adjA})\) is equal to
- A \(\left[\begin{array}{ccc}60 & 0 & 0 \\ 0 & 60 & 0 \\ 0 & 0 & 60\end{array}\right]\)
- B \(\left[\begin{array}{ccc}20 & 0 & 0 \\ 0 & 20 & 0 \\ 0 & 0 & 20\end{array}\right]\)
- C \(\left[\begin{array}{ccc}68 & 0 & 0 \\ 0 & 68 & 0 \\ 0 & 0 & 68\end{array}\right]\)
- D \(\left[\begin{array}{ccc}108 & 0 & 0 \\ 0 & 108 & 0 \\ 0 & 0 & 108\end{array}\right]\)
Answer & Solution
Correct Answer
(C) \(\left[\begin{array}{ccc}68 & 0 & 0 \\ 0 & 68 & 0 \\ 0 & 0 & 68\end{array}\right]\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& |A|=x(y z-8)+3(8-z)+2(2-2 y) \\
& =x y z-8 x+24-3 z+4-4 y \\
& =x y z-(8 x+4 y+3 z)+28 \\
& =60-20+28 \\
& =68
\end{aligned}
\)
we have, \(A \cdot \operatorname{adj}(A)=|A| I=\left[\begin{array}{ccc}68 & 0 & 0 \\ 0 & 68 & 0 \\ 0 & 0 & 68\end{array}\right]\)
\begin{aligned}
& |A|=x(y z-8)+3(8-z)+2(2-2 y) \\
& =x y z-8 x+24-3 z+4-4 y \\
& =x y z-(8 x+4 y+3 z)+28 \\
& =60-20+28 \\
& =68
\end{aligned}
\)
we have, \(A \cdot \operatorname{adj}(A)=|A| I=\left[\begin{array}{ccc}68 & 0 & 0 \\ 0 & 68 & 0 \\ 0 & 0 & 68\end{array}\right]\)
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