Question:

Let $R = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ where $a, b, c, d \neq 0$. Equate $QR$ and $RP$: $$QR = \begin{pmatrix} x & y \\ z & 4 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} ax+yc & bx+yd \\ az+4c & bz+4d \end{pmatrix}$$ $$RP = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 2a & 3b \\ 2c & 3d \end{pmatrix}$$ Comparing components gives: 1) $ax + yc = 2a \implies yc = a(2-x)$ 2) $bx + yd = 3b \implies yd = b(3-x)$ 3) $az + 4c = 2c \implies az = -2c$ 4) $bz + 4d = 3d \implies bz = -d$ From 3) and 4), $z = \dfrac{-2c}{a} = \dfrac{-d}{b} \implies ad = 2bc$. Now divide 1) by 2): $$\dfrac{c}{d} = \dfrac{a(2-x)}{b(3-x)} \implies \dfrac{ad}{bc} = \dfrac{3-x}{2-x}$$ Substitute $ad/bc = 2$: $$2 = \dfrac{3-x}{2-x} \implies 4 - 2x = 3 - x \implies x = 1$$ Substitute $x=1$ back into 1) and 3): $$y = a/c, \quad z = -2c/a \implies yz = \left(\dfrac{a}{c}\right)\left(\dfrac{-2c}{a}\right) = -2$$ Thus, $yz = -2$. With $Q = \begin{pmatrix} 1 & y \\ z & 4 \end{pmatrix}$, we have: $$\det(Q - kI) = (1-k)(4-k) - yz = k^2 - 5k + 4 - (-2) = k^2 - 5k + 6$$ - For $k = 2$, $\det(Q - 2I) = 4 - 10 + 6 = 0$ (Option A is True). - For $k = 6$, $\det(Q - 6I) = 36 - 30 + 6 = 12$ (Option B is True). - For $k = 3$, $\det(Q - 3I) = 9 - 15 + 6 = 0 \neq 15$ (Option C is False). - $yz = -2 \neq 2$ (Option D is False).