Question

Solve the following system of equations using the matrix method: \[ 2x+3y+3z=5,\qquad x-2y+z=-4,\qquad 3x-y-2z=3. \]

Hint:

To solve linear equations using matrices: \[ AX=B \Rightarrow X=A^{-1}B \] First check that \(|A|\neq0\). Only then does the inverse exist and the system has a unique solution.

Answer 1

Concept: A system of linear equations can be written in matrix form as \[ AX = B \] where \(A\) is the coefficient matrix, \(X\) is the variable matrix, and \(B\) is the constant matrix. If \(A\) is non-singular, the solution is given by \[ X = A^{-1}B \]
Step 1: Write the system in matrix form. \[ \begin{bmatrix} 2 & 3 & 3\\ 1 & -2 & 1\\ 3 & -1 & -2 \end{bmatrix} \begin{bmatrix} x\\y\\z \end{bmatrix} = \begin{bmatrix} 5\\-4\\3 \end{bmatrix} \] Thus, \[ A= \begin{bmatrix} 2 & 3 & 3\\ 1 & -2 & 1\\ 3 & -1 & -2 \end{bmatrix},\quad X= \begin{bmatrix} x\\y\\z \end{bmatrix},\quad B= \begin{bmatrix} 5\\-4\\3 \end{bmatrix} \]
Step 2: Find the determinant of \(A\). \[ |A|= \begin{vmatrix} 2 & 3 & 3\\ 1 & -2 & 1\\ 3 & -1 & -2 \end{vmatrix} \] \[ =2 \begin{vmatrix} -2 & 1\\ -1 & -2 \end{vmatrix} -3 \begin{vmatrix} 1 & 1\\ 3 & -2 \end{vmatrix} +3 \begin{vmatrix} 1 & -2\\ 3 & -1 \end{vmatrix} \] \[ =2(4+1)-3(-2-3)+3(-1+6) \] \[ =10+15+15=40 \] Since \(|A| \neq 0\), the matrix \(A\) is invertible.
Step 3: Find \(A^{-1}\). Using \(A^{-1}=\frac{1}{|A|}\text{Adj}(A)\), we obtain \[ A^{-1}=\frac{1}{40} \begin{bmatrix} 5 & 3 & 9\\ 5 & -13 & 1\\ 5 & 11 & -7 \end{bmatrix} \]
Step 4: Compute \(X = A^{-1}B\). \[ X=\frac{1}{40} \begin{bmatrix} 5 & 3 & 9\\ 5 & -13 & 1\\ 5 & 11 & -7 \end{bmatrix} \begin{bmatrix} 5\\-4\\3 \end{bmatrix} \] \[ = \frac{1}{40} \begin{bmatrix} 25-12+27\\ 25+52+3\\ 25-44-21 \end{bmatrix} = \frac{1}{40} \begin{bmatrix} 40\\ 80\\ -40 \end{bmatrix} \] \[ = \begin{bmatrix} 1\\ 2\\ -1 \end{bmatrix} \]
Step 5: Final solution. \[ x=1,\qquad y=2,\qquad z=-1 \]