JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
Let \(A\) be a matrix such that \(A.\,\left[ {\begin{array}{*{20}{c}}
1&2\\
0&3
\end{array}} \right]\) is a scalar matrix and \(\left| {3A} \right| = 108\) . Then \(A^2\) equals
- A \(\left[ {\begin{array}{*{20}{c}}
4&{ - 32}\\
0&{36}
\end{array}} \right]\) - B \(\left[ {\begin{array}{*{20}{c}}
4&0\\
{ - 32}&{36}
\end{array}} \right]\) - C \(\left[ {\begin{array}{*{20}{c}}
{36}&0\\
{ - 32}&4
\end{array}} \right]\) - D \(\left[ {\begin{array}{*{20}{c}}
{36}&{ - 32}\\
0&4
\end{array}} \right]\)
Answer & Solution
Correct Answer
(D) \(\left[ {\begin{array}{*{20}{c}}
{36}&{ - 32}\\
0&4
\end{array}} \right]\)
Step-by-step Solution
Detailed explanation
\((d)\) Since \(A.\left[ {\begin{array}{*{20}{c}} 1&2\\ 0&3 \end{array}} \right]\) is a scalar matrix and \(\left| {3A} \right| = 108\) Suppose the scalar matrix is \(\left[ {\begin{array}{*{20}{c}} k&0\\ 0&k \end{array}} \right]\)…
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