Question

Solve the system of equations by matrix method: \[ 3x - 2y + 3z = 8, 2x + y - z = 1, 4x - 3y + 2z = 4 \]

Hint:

For $3\times 3$ systems, Cramer's Rule is the fastest way if $\det(A)\neq 0$.

Answer 1

Step 1: Write system in matrix form. \[ AX = B \] where \[ A = \begin{bmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, B = \begin{bmatrix} 8 \\ 1 \\ 4 \end{bmatrix} \] Step 2: Solve using inverse method. \[ X = A^{-1}B \] Step 3: Find $\det(A)$. \[ \det(A) = 3\begin{vmatrix} 1 & -1 \\ -3 & 2 \end{vmatrix} - (-2)\begin{vmatrix} 2 & -1 \\ 4 & 2 \end{vmatrix} + 3\begin{vmatrix} 2 & 1 \\ 4 & -3 \end{vmatrix} \] \[ = 3(1\cdot 2 - (-1)(-3)) + 2(2\cdot 2 - (-1)(4)) + 3(2(-3)-1(4)) \] \[ = 3(2-3) + 2(4+4) + 3(-6-4) = 3(-1) + 16 + (-30) = -17 \] So, $\det(A) \neq 0$, hence solution exists. Step 4: Use Cramer's Rule. \[ x = \frac{\det(A_{1})}{\det(A)}, y = \frac{\det(A_{2})}{\det(A)}, z = \frac{\det(A_{3})}{\det(A)} \] - Replace first column of $A$ with $B$ for $A_{1}$: \[ A_{1} = \begin{bmatrix} 8 & -2 & 3 \\ 1 & 1 & -1 \\ 4 & -3 & 2 \end{bmatrix}, \det(A_{1}) = 17 \] - Replace second column of $A$ with $B$ for $A_{2}$: \[ A_{2} = \begin{bmatrix} 3 & 8 & 3 \\ 2 & 1 & -1 \\ 4 & 4 & 2 \end{bmatrix}, \det(A_{2}) = -34 \] - Replace third column of $A$ with $B$ for $A_{3}$: \[ A_{3} = \begin{bmatrix} 3 & -2 & 8 \\ 2 & 1 & 1 \\ 4 & -3 & 4 \end{bmatrix}, \det(A_{3}) = -51 \] Step 5: Compute values. \[ x = \frac{17}{-17} = -1, y = \frac{-34}{-17} = 2, z = \frac{-51}{-17} = 3 \]

Final Answer: \[ \boxed{x = -1,\; y = 2,\; z = 3} \]