Let I = begin{bmatrix} 1 & 0 0 & 1 e | vedclass

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(A) Given $QR = RP$ where $R = \begin{bmatrix} r_1 & r_2 \ r_3 & r_4 \end{bmatrix}$ with $r_i 
eq 0$.
Multiplying $Q$ and $R$ gives $\begin

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$A, B$

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(A) Given $QR = RP$ where $R = \begin{bmatrix} r_1 & r_2 \ r_3 & r_4 \end{bmatrix}$ with $r_i 
eq 0$.
Multiplying $Q$ and $R$ gives $\begin{bmatrix} xr_1 + yr_3 & xr_2 + yr_4 \ zr_1 + 4r_3 & zr_2 + 4r_4 \end{bmatrix} = \begin{bmatrix} 2r_1 & 3r_2 \ 2r_3 & 3r_4 \end{bmatrix}$.
Comparing elements:
$1$) $xr_1 + yr_3 = 2r_1 \Rightarrow (x-2)r_1 = -yr_3 \Rightarrow \frac{r_3}{r_1} = \frac{2-x}{y}$.
$2$) $zr_1 + 4r_3 = 2r_3 \Rightarrow zr_1 = -2r_3 \Rightarrow \frac{r_3}{r_1} = -\frac{z}{2}$.
Equating these,$\frac{2-x}{y} = -\frac{z}{2} \Rightarrow 4-2x = -yz \Rightarrow yz = 2x-4$.
$3$) $xr_2 + yr_4 = 3r_2 \Rightarrow (x-3)r_2 = -yr_4 \Rightarrow \frac{r_4}{r_2} = \frac{3-x}{y}$.
$4$) $zr_2 + 4r_4 = 3r_4 \Rightarrow zr_2 = -r_4 \Rightarrow \frac{r_4}{r_2} = -z$.
Equating these,$\frac{3-x}{y} = -z \Rightarrow 3-x = -yz \Rightarrow yz = x-3$.
Equating $yz$: $2x-4 = x-3 \Rightarrow x = 1$. Then $yz = 1-3 = -2$.
The characteristic polynomial of $Q$ is $|Q - \lambda I| = \begin{vmatrix} 1-\lambda & y \ z & 4-\lambda \end{vmatrix} = (1-\lambda)(4-\lambda) - yz = \lambda^2 - 5\lambda + 4 - (-2) = \lambda^2 - 5\lambda + 6 = (\lambda-2)(\lambda-3)$.
$(A)$ $|Q-2I| = (2-2)(2-3) = 0$. True.
$(B)$ $|Q-6I| = (6-2)(6-3) = 4 	imes 3 = 12$. True.
$(C)$ $|Q-3I| = (3-2)(3-3) = 0 
eq 15$. False.
$(D)$ $yz = -2 
eq 2$. False.
Thus,$(A)$ and $(B)$ are true.

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