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?

Given: I = 1 0 0 1 , P = 2 0 0 3 Q = x y z 4 , x, y, z ≠ 0

$\exists\, R \ (\text{all entries non-zero}) \text{ with } QR = RP$

$\text{To find: which statements (1)--(4) are TRUE}$

R = a b c d , a, b, c, d ≠ 0

$QR = RP$

$x a + y c = 2a$

$x b + y d = 3b$

$z a + 4c = 2c \Rightarrow z a = -2c$

$z b + 4d = 3d \Rightarrow z b = -d$

$c = -\dfrac{za}{2}, \quad d = -zb$

$x - \dfrac{yz}{2} = 2, \qquad x - yz = 3$

$2 + \dfrac{yz}{2} = 3 + yz \Rightarrow yz = -2$

$x = 3 + yz = 1$

Q = 1 y z 4 , yz = -2

$\det(Q - \lambda I) = (1-\lambda)(4-\lambda) - yz$

$\det(Q - 2I) = (-1)(2) - yz = -2 + 2 = 0 \Rightarrow (1)\ \text{TRUE}$

$\det(Q - 6I) = (-5)(-2) - yz = 10 + 2 = 12 \Rightarrow (2)\ \text{TRUE}$

$\det(Q - 3I) = (-2)(1) - yz = -2 + 2 = 0 \ne 15 \Rightarrow (3)\ \text{FALSE}$

$yz = -2 \ne 2 \Rightarrow (4)\ \text{FALSE}$

$\text{TRUE: } (1)\ \text{and}\ (2)$

$\boxed{(1)\ \text{and}\ (2)}$