Question:
If the matrices \(A = \begin{bmatrix}1&1&2\\1&3&4\\1&-1&3\end{bmatrix}\), \(B = \mathrm{adj}A\) and \(C = 3A\), then \(\frac{|\mathrm{adj}B|}{|C|}\) is equal to :
If the matrices \(A = \begin{bmatrix}1&1&2\\1&3&4\\1&-1&3\end{bmatrix}\), \(B = \mathrm{adj}A\) and \(C = 3A\), then \(\frac{|\mathrm{adj}B|}{|C|}\) is equal to :
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Use \(|\mathrm{adj}\,M| = |M|^{n-1}\) and \(|kM| = k^n|M|\) for a \(3\times3\) matrix.
Updated On: Nov 5, 2026
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The Correct Option is A
Solution and Explanation
\(|A| = 1(9+4) - 1(3-4) + 2(-1-3) = 13 + 1 - 8 = 6\).
\(|B| = |\mathrm{adj}A| = |A|^2 = 36\).
\(|\mathrm{adj}B| = |B|^2 = 36^2 = 1296\).
\(|C| = |3A| = 3^3|A| = 27\times 6 = 162\).
\(\frac{|\mathrm{adj}B|}{|C|} = \frac{1296}{162} = 8\).
Answer: (1) 8
\(|B| = |\mathrm{adj}A| = |A|^2 = 36\).
\(|\mathrm{adj}B| = |B|^2 = 36^2 = 1296\).
\(|C| = |3A| = 3^3|A| = 27\times 6 = 162\).
\(\frac{|\mathrm{adj}B|}{|C|} = \frac{1296}{162} = 8\).
Answer: (1) 8
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