We need to solve the system of simultaneous equations using Cramer's rule: $p + 2q - r = 9$ $2p - q + 3r = -2$ $3p + 2q + 3r = 9$

1. Problem Description

We need to solve the system of simultaneous equations using Cramer's rule:

$p + 2q - r = 9$

$2p - q + 3r = -2$

$3p + 2q + 3r = 9$

2. Solution Steps

First, we calculate the determinant of the coefficient matrix:

$D = \begin{vmatrix} 1 & 2 & -1 \\ 2 & -1 & 3 \\ 3 & 2 & 3 \end{vmatrix} = 1 \cdot \begin{vmatrix} -1 & 3 \\ 2 & 3 \end{vmatrix} - 2 \cdot \begin{vmatrix} 2 & 3 \\ 3 & 3 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 2 & -1 \\ 3 & 2 \end{vmatrix} = 1(-3-6) - 2(6-9) - 1(4-(-3)) = -9 - 2(-3) - 7 = -9 + 6 - 7 = -10$

Next, we calculate $D_p$ by replacing the first column of $D$ with the constants on the right-hand side of the equations:

$D_p = \begin{vmatrix} 9 & 2 & -1 \\ -2 & -1 & 3 \\ 9 & 2 & 3 \end{vmatrix} = 9 \cdot \begin{vmatrix} -1 & 3 \\ 2 & 3 \end{vmatrix} - 2 \cdot \begin{vmatrix} -2 & 3 \\ 9 & 3 \end{vmatrix} + (-1) \cdot \begin{vmatrix} -2 & -1 \\ 9 & 2 \end{vmatrix} = 9(-3-6) - 2(-6-27) - 1(-4-(-9)) = 9(-9) - 2(-33) - 1(5) = -81 + 66 - 5 = -20$

Then, we calculate $D_q$ by replacing the second column of $D$ with the constants on the right-hand side of the equations:

$D_q = \begin{vmatrix} 1 & 9 & -1 \\ 2 & -2 & 3 \\ 3 & 9 & 3 \end{vmatrix} = 1 \cdot \begin{vmatrix} -2 & 3 \\ 9 & 3 \end{vmatrix} - 9 \cdot \begin{vmatrix} 2 & 3 \\ 3 & 3 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 2 & -2 \\ 3 & 9 \end{vmatrix} = 1(-6-27) - 9(6-9) - 1(18-(-6)) = -33 - 9(-3) - 24 = -33 + 27 - 24 = -30$

Then, we calculate $D_r$ by replacing the third column of $D$ with the constants on the right-hand side of the equations:

$D_r = \begin{vmatrix} 1 & 2 & 9 \\ 2 & -1 & -2 \\ 3 & 2 & 9 \end{vmatrix} = 1 \cdot \begin{vmatrix} -1 & -2 \\ 2 & 9 \end{vmatrix} - 2 \cdot \begin{vmatrix} 2 & -2 \\ 3 & 9 \end{vmatrix} + 9 \cdot \begin{vmatrix} 2 & -1 \\ 3 & 2 \end{vmatrix} = 1(-9-(-4)) - 2(18-(-6)) + 9(4-(-3)) = 1(-5) - 2(24) + 9(7) = -5 - 48 + 63 = 10$

Using Cramer's rule, we have:

$p = \frac{D_p}{D} = \frac{-20}{-10} = 2$

$q = \frac{D_q}{D} = \frac{-30}{-10} = 3$

$r = \frac{D_r}{D} = \frac{10}{-10} = -1$

3. Final Answer

$p = 2$, $q = 3$, $r = -1$