Matrix Equation QR Equals RP with Nonzero Entries and Determinants

To determine which of the statements are true, we use the given matrix equation: $QR = RP$

Let the matrix $R$ be represented as: $R = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ where $a, b, c, d$ are non-zero real numbers.

Given matrices are: $P = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}, \quad Q = \begin{pmatrix} x & y \\ z & 4 \end{pmatrix}$

Compute $QR$: $QR = \begin{pmatrix} x & y \\ z & 4 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} ax + cy & bx + dy \\ az + 4c & bz + 4d \end{pmatrix}$

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

Equating corresponding entries of $QR$ and $RP$:

1. First column equations: $ax + cy = 2a \implies (x - 2)a + cy = 0$ $az + 4c = 2c \implies az + 2c = 0 \implies \frac{c}{a} = -\frac{z}{2}$

Substituting $\frac{c}{a} = -\frac{z}{2}$ into the first equation: $(x - 2) + \left(-\frac{z}{2}\right)y = 0 \implies x - 2 = \frac{yz}{2} \implies yz = 2x - 4 \quad \text{--- (1)}$

1. Second column equations: $bx + dy = 3b \implies (x - 3)b + dy = 0$ $bz + 4d = 3d \implies bz + d = 0 \implies \frac{d}{b} = -z$

Substituting $\frac{d}{b} = -z$ into the first equation: $(x - 3) + (-z)y = 0 \implies x - 3 = yz \quad \text{--- (2)}$

Equating the two expressions for $yz$ from equations (1) and (2): $2x - 4 = x - 3 \implies x = 1$

Substituting $x = 1$ back to find $yz$: $yz = 1 - 3 = -2$

Thus, the matrix $Q$ is: $Q = \begin{pmatrix} 1 & y \\ z & 4 \end{pmatrix} \quad \text{with } yz = -2$

Now, we evaluate each of the given options:

• Option (A): $Q - 2I = \begin{pmatrix} 1 - 2 & y \\ z & 4 - 2 \end{pmatrix} = \begin{pmatrix} -1 & y \\ z & 2 \end{pmatrix}$ $\det(Q - 2I) = (-1)(2) - yz = -2 - (-2) = 0$ Hence, statement (A) is TRUE.

• Option (B): $Q - 6I = \begin{pmatrix} 1 - 6 & y \\ z & 4 - 6 \end{pmatrix} = \begin{pmatrix} -5 & y \\ z & -2 \end{pmatrix}$ $\det(Q - 6I) = (-5)(-2) - yz = 10 - (-2) = 12$ Hence, statement (B) is TRUE.

• Option (C): $Q - 3I = \begin{pmatrix} 1 - 3 & y \\ z & 4 - 3 \end{pmatrix} = \begin{pmatrix} -2 & y \\ z & 1 \end{pmatrix}$ $\det(Q - 3I) = (-2)(1) - yz = -2 - (-2) = 0 \neq 15$ Hence, statement (C) is FALSE.

• Option (D): $yz = -2 \neq 2$ Hence, statement (D) is FALSE.

Therefore, the correct statements are (A) and (B).