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If \( A=\left[\begin{array}{lll}1 & 1 & 2 \\ 1 & 3 & 4 \\ 1 & -1 & 3\end{array}\right] \), and \( B=\operatorname{adj} A \quad \& \quad C=3 A \), then \( \mid \frac{|\operatorname{adj} B|}{|C|}= \) (a) 72 (b) 2 (c) 8 (d) 16 |
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Answer» The correct option is (c) 8. |A| = \(\begin{vmatrix}1&1&2\\1&3&4\\1&-1&3\end{vmatrix}\) = \(=1\begin{vmatrix}3&4\\-1&3\end{vmatrix}-1\begin{vmatrix}1&4\\1&3\end{vmatrix}+2\begin{vmatrix}1&3\\1&-1\end{vmatrix}\) = 1(9 + 4) - 1(3 - 4) + 2(-1 - 3) = 13 + 1 - 8 = 6 |B| = |adj A| = |A|3-1 = |A|2 = 62 = 36 |adj B| = |B|3-1 = |B|2 = 362 = 62 x 62 |C| = |3A| = 33|A| = 33 x 6 \(\therefore\) \(\frac{|adj B|}{|C|} = \frac{6^2\times6^2}{3^3\times6}\) = \(\frac{6\times2^2}3\) = 23 = 8 |
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