Question

Solve by matrix method the system of equations: \[\begin{aligned} 2x - 3y + 5z &= 11 \\ 3x + 2y - 4z &= -5 \\ x + y - 2z &= -3 \end{aligned}\]

Hint:

Cramer's rule is the quickest way to solve 3-variable equations if determinant is non-zero.

Answer 1

Step 1: Write in matrix form. \\ \[ AX = B \] where \[ A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, B = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}. \] Step 2: Find $\det(A)$. \\ \[ \det(A) = \begin{vmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{vmatrix}. \] Expanding along first row: \[ = 2 \begin{vmatrix} 2 & -4 \\ 1 & -2 \end{vmatrix} - (-3) \begin{vmatrix} 3 & -4 \\ 1 & -2 \end{vmatrix} + 5 \begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix}. \] \[ = 2((-4) - (-4)) + 3((-6+4)) + 5((3-2)). \] \[ = 2(0) + 3(-2) + 5(1) = -6 + 5 = -1. \] So, $\det(A) = -1 \neq 0$, hence system has unique solution. Step 3: Use $X = A^{-1B$.} \\ Using Cramer's rule: \[ x = \frac{\det(A_x)}{\det(A)}, y = \frac{\det(A_y)}{\det(A)}, z = \frac{\det(A_z)}{\det(A)}. \] Step 4: Compute $\det(A_x)$. \\ Replace first column by $B$: \[ A_x = \begin{bmatrix} 11 & -3 & 5 \\ -5 & 2 & -4 \\ -3 & 1 & -2 \end{bmatrix}. \] \[ \det(A_x) = 11\begin{vmatrix} 2 & -4 \\ 1 & -2 \end{vmatrix} - (-3)\begin{vmatrix} -5 & -4 \\ -3 & -2 \end{vmatrix} + 5\begin{vmatrix} -5 & 2 \\ -3 & 1 \end{vmatrix}. \] \[ = 11((-4) - (-4)) + 3((-10 + 12)) + 5((-5+6)). \] \[ = 11(0) + 3(2) + 5(1) = 6 + 5 = 11. \] So, \[ x = \frac{11}{-1} = -11. \] Step 5: Compute $\det(A_y)$. \\ Replace second column by $B$: \[ A_y = \begin{bmatrix} 2 & 11 & 5 \\ 3 & -5 & -4 \\ 1 & -3 & -2 \end{bmatrix}. \] \[ \det(A_y) = 2\begin{vmatrix} -5 & -4 \\ -3 & -2 \end{vmatrix} - 11\begin{vmatrix} 3 & -4 \\ 1 & -2 \end{vmatrix} + 5\begin{vmatrix} 3 & -5 \\ 1 & -3 \end{vmatrix}. \] \[ = 2((-10+12)) - 11((-6+4)) + 5((-9+5)). \] \[ = 2(2) - 11(-2) + 5(-4). \] \[ = 4 + 22 - 20 = 6. \] So, \[ y = \frac{6}{-1} = -6. \] Step 6: Compute $\det(A_z)$. \\ Replace third column by $B$: \[ A_z = \begin{bmatrix} 2 & -3 & 11 \\ 3 & 2 & -5 \\ 1 & 1 & -3 \end{bmatrix}. \] \[ \det(A_z) = 2\begin{vmatrix} 2 & -5 \\ 1 & -3 \end{vmatrix} - (-3)\begin{vmatrix} 3 & -5 \\ 1 & -3 \end{vmatrix} + 11\begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix}. \] \[ = 2((-6+5)) + 3((-9+5)) + 11((3-2)). \] \[ = 2(-1) + 3(-4) + 11(1). \] \[ = -2 -12 + 11 = -3. \] So, \[ z = \frac{-3}{-1} = 3. \]

Final Answer: \[ \boxed{x=-11,\; y=-6,\; z=3} \]