Question

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

Hint:

To solve a system of linear equations using the matrix method, write the system as a matrix equation \( A \cdot X = B \), find \( A^{-1} \), and solve for \( X \) using \( X = A^{-1} \cdot B \).

Answer 1

We can solve this system of linear equations using the matrix method.
Step 1: Write the system as a matrix equation We write the system of equations as: \[ \begin{bmatrix} 2 & 1 & -1 \\ 3 & -2 & 3 \\ 4 & -3 & 2 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 8 \\ 4 \end{bmatrix}. \] Let \( A = \begin{bmatrix} 2 & 1 & -1 \\ 3 & -2 & 3 \\ 4 & -3 & 2 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), and \( B = \begin{bmatrix} 1 \\ 8 \\ 4 \end{bmatrix} \). The matrix equation is \( A \cdot X = B \).
Step 2: Find \( A^{-1} \) We already computed \( A^{-1} \) in part (a), and it is: \[ A^{-1} = \begin{bmatrix} 7 & -1 & 1 \\ -1 & 1 & -1 \\ 1 & -1 & 7 \end{bmatrix}. \]
Step 3: Solve for \( X \) Now, solve for \( X \) by multiplying both sides of the equation \( A \cdot X = B \) by \( A^{-1} \): \[ X = A^{-1} \cdot B. \] Substitute the values: \[ X = \begin{bmatrix} 7 & -1 & 1 \\ -1 & 1 & -1 \\ 1 & -1 & 7 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 8 \\ 4 \end{bmatrix}. \] Performing the matrix multiplication: \[ X = \begin{bmatrix} 7(1) + (-1)(8) + (1)(4) \\ (-1)(1) + (1)(8) + (-1)(4) \\ (1)(1) + (-1)(8) + 7(4) \end{bmatrix} = \begin{bmatrix} 7 - 8 + 4 \\ -1 + 8 - 4 \\ 1 - 8 + 28 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 21 \end{bmatrix}. \] Thus, the solution to the system of equations is: \[ x = 3, y = 3, z = 21. \]
Conclusion: The solution to the system of equations is: \[ \boxed{x = 3, \, y = 3, \, z = 21}. \]