Let $A = \begin{pmatrix} 4 & -2 \ \alpha & \beta \end{pmatrix}$. If $A^2 + \gamma A + 18I = O$, then $\det(A)$ is equal to _______.

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Aicharya · Accepted Answer

$A=\begin{bmatrix} 4 & -2 \ \alpha & \beta \end{bmatrix} \Rightarrow 	ext{det}A=4\beta+2\alpha$ Characteristic equation of the matrix is $\begin{bmatrix} 4-\lambda & -2 \ \alpha & \beta-\lambda \end{bmatrix}=0$ $\Rightarrow 4\beta+\lambda^2-(\beta+4)\lambda+2\alpha=0$ $\Rightarrow \lambda^2-(\beta+4)\lambda+2\alpha+4\beta=0$ Comparing with given equation $A^2+\gamma A+18I=0$, we get $\Rightarrow \gamma=-(\beta+4) 	ext{ and } 2\alpha+4\beta=18$ Hence $	ext{det}A=4\beta+2\alpha=18$

Let $A = \begin{pmatrix} 4 & -2 \\ \alpha & \beta \end{pmatrix}$. If $A^2 + \gamma A + 18I = O$, then…

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