Let P = begin{bmatrix} 3 & -1 & -2 2 & 0 & a | vedclass

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(A) Given $PQ = kI_3$. Taking the determinant on both sides,$|P||Q| = |kI_3| = k^3|I_3| = k^3$. Given $|Q| = \frac{k^2}{2}$,so $|P| \cdot \frac{k^2}{2

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$17$

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(A) Given $PQ = kI_3$. Taking the determinant on both sides,$|P||Q| = |kI_3| = k^3|I_3| = k^3$. Given $|Q| = \frac{k^2}{2}$,so $|P| \cdot \frac{k^2}{2} = k^3$. Since $k 
eq 0$,$|P| = 2k$. Calculating $|P|$: $|P| = 3(0 - (-5\alpha)) - (-1)(0 - 3\alpha) + (-2)(-10 - 0) = 3(5\alpha) + (-3\alpha) + 20 = 15\alpha - 3\alpha + 20 = 12\alpha + 20$. Thus,$12\alpha + 20 = 2k$,or $k = 6\alpha + 10$. Since $PQ = kI_3$,$Q = kP^{-1} = k \cdot \frac{	ext{adj}(P)}{|P|} = k \cdot \frac{	ext{adj}(P)}{2k} = \frac{1}{2} 	ext{adj}(P)$. The element $q_{23}$ is the $(2,3)$ entry of $\frac{1}{2} 	ext{adj}(P)$. The $(2,3)$ entry of $	ext{adj}(P)$ is the cofactor $C_{32} = -\begin{vmatrix} 3 & -2 \ 2 & \alpha \end{vmatrix} = -(3\alpha - (-4)) = -(3\alpha + 4)$. So,$q_{23} = \frac{-(3\alpha + 4)}{2} = -\frac{k}{8}$. This implies $4(3\alpha + 4) = k$. Substituting $k = 6\alpha + 10$: $12\alpha + 16 = 6\alpha + 10 \Rightarrow 6\alpha = -6 \Rightarrow \alpha = -1$. Then $k = 6(-1) + 10 = 4$. Finally,$\alpha^2 + k^2 = (-1)^2 + 4^2 = 1 + 16 = 17$.

Let $P = \begin{bmatrix} 3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0 \end{bmatrix}$ where $\alpha \in R$. Suppose $Q = [q_{ij}]$ is a matrix satisfying $PQ = kI_3$ for some non-zero $k \in R$. If $q_{23} = -\frac{k}{8}$ and $|Q| = \frac{k^2}{2}$,then $\alpha^2 + k^2$ is equal to?

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