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Given that A = \(\begin{bmatrix} \alpha & \beta \\[0.3em] \gamma & -\alpha\\[0.3em] \end{bmatrix}\) and A2 = 3I, then :A = [(α,β)(γ,-α)](a) 1 + α2 + βγ = 0(b) 1 - α2 - βγ = 0(c) 3 - α2 - βγ = 0(d) 3 + α2 + βγ = 0 |
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Answer» Option : (c) \(A^2=3I,\) \(\Rightarrow\begin{bmatrix}\alpha^2+\beta \gamma&0\\0&\beta \gamma+\alpha^2\end{bmatrix}=\begin{bmatrix}3&0\\0&3\end{bmatrix}\) \(\Rightarrow3-\alpha^2-\beta \gamma=0\) |
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