$\text{Let}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\text{and}\mathrm{ P}=\left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right].\text{Let}\mathrm{ Q}=\left[\begin{array}{cc}x& y\\ z& 4\end{array}\right]\text{for some non-zero real numbers}x,y\text{and}\mathrm{ z}\text{ for which there}$

$\text{is a}\mathrm{ 2}\times 2\text{ matrix}\mathrm{ R}\text{ with all entries being non-zero real numbers, such that }QR=RP.\text{Then which of the}$

$\text{following statements is (are) TRUE?}$

The correct statements are:
1. The determinant of Q - 2I is zero
2. The determinant of Q - 6I is 12

Step-by-step Explanation:

Let the given matrices be:
$I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ ,
$P=\left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right]$ ,
$Q=\left[\begin{array}{cc}x& y\\ z& 4\end{array}\right]$

where $x,y,z$ are non-zero real numbers. We are given that there exists a $2\times 2$ matrix $R$ with all non-zero real entries such that:

$QR=RP$

Let $R=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ where $a,b,c,d\ne 0$.

Substituting $Q$, $R$, and $P$ into the equation $QR=RP$ yields:
$\left[\begin{array}{cc}x& y\\ z& 4\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right]$

By matrix multiplication, we obtain:
$\left[\begin{array}{cc}xa+yc& xb+yd\\ za+4c& zb+4d\end{array}\right]=\left[\begin{array}{cc}2a& 3b\\ 2c& 3d\end{array}\right]$

Comparing corresponding elements, we get the following system of equations:
(1) $xa+yc=2a$
(2) $xb+yd=3b$
(3) $za+4c=2c\Rightarrow za=-2c$
(4) $zb+4d=3d\Rightarrow zb=-d$

From equations (3) and (4), since $a,b\ne 0$:
$z=\frac{-2c}{a}=\frac{-d}{b}$
This gives:
$\frac{c}{a}=\frac{d}{2b}$

Let $\frac{c}{a}=k$. Then $c=ka$ and $d=2kb$.
Consequently, we can write $z$ as:
$z=-2k$

Now, let's substitute these expressions into equations (1) and (2):
From (1):
$(x-2)a+y\left(ka\right)=0\Rightarrow x-2+yk=0\Rightarrow x-2=-yk$
From (2):
$(x-3)b+y\left(2kb\right)=0\Rightarrow x-3+2yk=0\Rightarrow x-3=-2yk$

Let $yk=m$. The system becomes:
$x-2=-m$
$x-3=-2m$

Solving for $m$ by subtracting the second equation from the first:
$1=m$
Thus, $yk=1$.
Substituting $m=1$ back, we find:
$x=2-1=1$

Now, let's determine the value of $yz$:
$yz=y(-2k)=-2\left(yk\right)=-2\left(1\right)=-2$

Hence, the matrix $Q$ is:
$Q=\left[\begin{array}{cc}1& y\\ z& 4\end{array}\right]$ with $yz=-2$.

Checking Statement 1:
We evaluate $Q-2I$:
$Q-2I=\left[\begin{array}{cc}1-2& y\\ z& 4-2\end{array}\right]=\left[\begin{array}{cc}-1& y\\ z& 2\end{array}\right]$
The determinant is:
$det(Q-2I)=(-1)\left(2\right)-yz=-2-(-2)=0$
Thus, Statement 1 is TRUE.

Checking Statement 2:
We evaluate $Q-6I$:
$Q-6I=\left[\begin{array}{cc}1-6& y\\ z& 4-6\end{array}\right]=\left[\begin{array}{cc}-5& y\\ z& -2\end{array}\right]$
The determinant is:
$det(Q-6I)=(-5)(-2)-yz=10-(-2)=12$
Thus, Statement 2 is TRUE.