Let $$I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ and $$P = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$$. Let $$Q = \begin{pmatrix} x & y \\ z & 4 \end{pmatrix}$$ for some non-zero real numbers $$x$$, $$y$$, and $$z$$, for which there is a $$2 \times 2$$ matrix $$R$$ with all entries being non-zero real numbers, such that $$QR = RP$$.

Then which of the following statements is (are) TRUE?

The condition $$QR = RP$$ will be used to connect the unknown entries of $$Q$$ with the (non-zero) entries of $$R = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$.

Compute both products:

$$QR = \begin{pmatrix} x & y \\ z & 4 \end{pmatrix} \!\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} xa + yc & xb + yd \\ za + 4c & zb + 4d \end{pmatrix}$$

$$RP = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \!\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 2a & 3b \\ 2c & 3d \end{pmatrix}$$

Equating corresponding entries of $$QR$$ and $$RP$$ yields four linear equations:

$$\begin{aligned} xa + yc &= 2a \quad &-(1)\\ xb + yd &= 3b \quad &-(2)\\ za + 4c &= 2c \quad &-(3)\\ zb + 4d &= 3d \quad &-(4) \end{aligned}$$

From $$(3)$$ and $$(4)$$ obtain $$z$$ in two ways:

$$z = \frac{-2c}{a} \quad -(5), \qquad z = \frac{-d}{b} \quad -(6)$$

Equating $$(5)$$ and $$(6)$$ gives the relation $$2bc = ad \quad -(7)$$.

Now express $$x$$ from $$(1)$$ and $$(2)$$:

$$x = 2 - \frac{c}{a}\,y \quad -(8), \qquad x = 3 - \frac{d}{b}\,y \quad -(9)$$

Equating $$(8)$$ and $$(9)$$ gives $$y\Bigl(\frac{d}{b} - \frac{c}{a}\Bigr) = 1 \quad -(10)$$

Use $$(7)$$ to rewrite $$\frac{d}{b} = \frac{2c}{a}$$, so the bracket in $$(10)$$ becomes $$\frac{2c}{a} - \frac{c}{a} = \frac{c}{a}.$$ Hence $$y \cdot \frac{c}{a} = 1 \;\;\Longrightarrow\;\; y = \frac{a}{c} \quad -(11)$$

Substitute $$(11)$$ into $$(8)$$: $$x = 2 - \frac{c}{a}\cdot\frac{a}{c} = 2 - 1 = 1 \quad -(12)$$

Finally, from $$(5)$$ and $$(11)$$ obtain $$z = \frac{-2c}{a}, \qquad yz = \frac{a}{c}\cdot\frac{-2c}{a} = -2 \quad -(13)$$

Thus the matrix $$Q$$ is $$Q = \begin{pmatrix} 1 & y \\ z & 4 \end{pmatrix}, \quad y\neq 0,\; z\neq 0,\; yz=-2.$$ (The specific non-zero values of $$a,b,c,d$$ simply scale $$y$$ and $$z$$ while keeping $$yz=-2$$.)

Determinant of $$Q - 2I$$
$$Q-2I = \begin{pmatrix} 1-2 & y \\ z & 4-2 \end{pmatrix} = \begin{pmatrix} -1 & y \\ z & 2 \end{pmatrix}$$ $$\det(Q-2I) = (-1)(2) - yz = -2 - (-2) = 0.$$ Statement A is true.

Determinant of $$Q - 6I$$
$$Q-6I = \begin{pmatrix} 1-6 & y \\ z & 4-6 \end{pmatrix} = \begin{pmatrix} -5 & y \\ z & -2 \end{pmatrix}$$ $$\det(Q-6I) = (-5)(-2) - yz = 10 - (-2) = 12.$$ Statement B is true.

Determinant of $$Q - 3I$$
$$Q-3I = \begin{pmatrix} 1-3 & y \\ z & 4-3 \end{pmatrix} = \begin{pmatrix} -2 & y \\ z & 1 \end{pmatrix}$$ $$\det(Q-3I) = (-2)(1) - yz = -2 - (-2) = 0\neq 15.$$ Statement C is false.

Product $$yz$$
From $$(13)$$, $$yz=-2\neq 2,$$ so Statement D is false.

Hence the correct statements are:
Option A  and  Option B.