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For A = \(\begin{bmatrix} 3& 1 \\[0.3em] -1 & 2 \\[0.3em] \end{bmatrix}\) then 14A-1 is given by :A = [(3,1)(-1,2)] (a) \(14\begin{bmatrix} 2& -1 \\[0.3em] 1 & 3 \\[0.3em] \end{bmatrix}\) (b) \(\begin{bmatrix} 4& -2 \\[0.3em] 2 & 6 \\[0.3em] \end{bmatrix}\) (c) \(2\begin{bmatrix} 2& -1 \\[0.3em] 1 & -3 \\[0.3em] \end{bmatrix}\) (d) \(2\begin{bmatrix} -3& -1 \\[0.3em] 1 & -2 \\[0.3em] \end{bmatrix}\) |
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Answer» Option : (b) |A| = 7, adj A \(=\begin{bmatrix}2&-1\\1&3\end{bmatrix}\) \(\therefore14A^{-1}=14\frac{adj A}{|A|}=2\,adj A=\begin{bmatrix}4&-2\\2&6\end{bmatrix}\) |
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