Let x in R and let P = begin{bmatrix} 1 & 1 & | vedclass

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(A) Given $R = PQP^{-1}$.$\det(R) = \det(PQP^{-1}) = \det(P) \det(Q) \det(P^{-1}) = \det(Q)$.$\det(Q) = 2(24 - 0) - x(0 - 0) + x(0 - 4x) = 48 - 4x^2$.

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$3, 4$

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(A) Given $R = PQP^{-1}$.$\det(R) = \det(PQP^{-1}) = \det(P) \det(Q) \det(P^{-1}) = \det(Q)$.$\det(Q) = 2(24 - 0) - x(0 - 0) + x(0 - 4x) = 48 - 4x^2$.Option $(1)$: For $x = 1$,$\det(R) = 48 - 4(1)^2 = 44 
eq 0$. Since $\det(R) 
eq 0$,the system $R \begin{bmatrix} \alpha \ \beta \ \gamma \end{bmatrix} = \mathbf{0}$ has only the trivial solution $\alpha = \beta = \gamma = 0$. $A$ unit vector requires $\alpha^2 + \beta^2 + \gamma^2 = 1$,which is impossible. Thus,$(1)$ is incorrect.Option $(2)$: $PQ = QP \iff PQP^{-1} = Q \iff R = Q$. This implies $PQP^{-1} = Q$. For this specific $P$,$R$ is not equal to $Q$ for any $x$. Thus,$(2)$ is incorrect.Option $(3)$: $\det \begin{bmatrix} 2 & x & x \ 0 & 4 & 0 \ x & x & 5 \end{bmatrix} + 8 = [2(20 - 0) - x(0 - 0) + x(0 - 4x)] + 8 = (40 - 4x^2) + 8 = 48 - 4x^2 = \det(R)$. Thus,$(3)$ is correct.Option $(4)$: For $x = 0$,$Q = \begin{bmatrix} 2 & 0 & 0 \ 0 & 4 & 0 \ 0 & 0 & 6 \end{bmatrix}$ and $P = \begin{bmatrix} 1 & 1 & 1 \ 0 & 2 & 2 \ 0 & 0 & 3 \end{bmatrix}$. $P^{-1} = \begin{bmatrix} 1 & -1/2 & 0 \ 0 & 1/2 & -1/3 \ 0 & 0 & 1/3 \end{bmatrix}$.$R = PQP^{-1} = \begin{bmatrix} 2 & 1 & 2/3 \ 0 & 4 & 4/3 \ 0 & 0 & 6 \end{bmatrix}$.Solving $(R - 6I) \begin{bmatrix} 1 \ a \ b \end{bmatrix} = 0$ gives $\begin{bmatrix} -4 & 1 & 2/3 \ 0 & -2 & 4/3 \ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 \ a \ b \end{bmatrix} = 0$.$-2a + 4b/3 = 0 \implies a = 2b/3$. $-4 + a + 2b/3 = 0 \implies -4 + 2b/3 + 2b/3 = 0 \implies 4b/3 = 4 \implies b = 3, a = 2$.$a + b = 2 + 3 = 5$. Thus,$(4)$ is correct.

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