Question 5

Let $R = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ where $a, b, c, d$ are non-zero real numbers. Given $QR = RP$: $\begin{pmatrix} x & y \\ z & 4 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$ $\begin{pmatrix} ax + yc & bx + yd \\ az + 4c & bz + 4d \end{pmatrix} = \begin{pmatrix} 2a & 3b \\ 2c & 3d \end{pmatrix}$

Comparing entries:

1. $ax + yc = 2a \implies yc = a(2 - x)$
2. $bx + yd = 3b \implies yd = b(3 - x)$
3. $az + 4c = 2c \implies 2c = -az \implies c = -\frac{az}{2}$
4. $bz + 4d = 3d \implies d = -bz$

Substitute $c$ and $d$ in (1) and (2): From (1): $y(-\frac{az}{2}) = a(2 - x) \implies -\frac{yz}{2} = 2 - x \implies yz = 2x - 4$ From (2): $y(-bz) = b(3 - x) \implies -yz = 3 - x \implies yz = x - 3$

Equating the expressions for $yz$: $2x - 4 = x - 3 \implies x = 1$ Substituting $x = 1$ back: $yz = 1 - 3 = -2$

Thus, $Q = \begin{pmatrix} 1 & y \\ z & 4 \end{pmatrix}$ with $yz = -2$.

Evaluating options: (A) $\det(Q - 2I) = \begin{vmatrix} 1-2 & y \\ z & 4-2 \end{vmatrix} = \begin{vmatrix} -1 & y \\ z & 2 \end{vmatrix} = -2 - yz = -2 - (-2) = 0$. (True) (B) $\det(Q - 6I) = \begin{vmatrix} 1-6 & y \\ z & 4-6 \end{vmatrix} = \begin{vmatrix} -5 & y \\ z & -2 \end{vmatrix} = 10 - yz = 10 - (-2) = 12$. (True) (C) $\det(Q - 3I) = \begin{vmatrix} 1-3 & y \\ z & 4-3 \end{vmatrix} = \begin{vmatrix} -2 & y \\ z & 1 \end{vmatrix} = -2 - yz = -2 - (-2) = 0$. (False) (D) $yz = -2$. (False)