Question

Solve by Matrix Method
Given: \[ x + 2y - 3z = 6, \quad 3x + 2y - 2z = 3, \quad 2x - y + z = 2 \]

Hint:

When solving a system of linear equations using matrices, represent the system in the form \( AX = B \) and find the inverse of the coefficient matrix \( A \) to solve for \( X \).

Answer 1

Step 1: Represent the system of equations in matrix form.

We can write the system of equations as a matrix equation \( AX = B \), where:

\[ A = \begin{bmatrix} 1 & 2 & -3 \\ 3 & 2 & -2 \\ 2 & -1 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 6 \\ 3 \\ 2 \end{bmatrix} \]
Step 2: Find the inverse of matrix \( A \).

We need to find \( A^{-1} \) to solve for \( X \). The inverse of a \( 3 \times 3 \) matrix \( A \) is given by:

\[ A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A) \]
First, calculate the determinant of \( A \):

\[ \det(A) = 1 \cdot \begin{vmatrix} 2 & -2 \\ -1 & 1 \end{vmatrix} - 2 \cdot \begin{vmatrix} 3 & -2 \\ 2 & 1 \end{vmatrix} + (-3) \cdot \begin{vmatrix} 3 & 2 \\ 2 & -1 \end{vmatrix} \]
Calculating the individual \( 2 \times 2 \) determinants:

\[ \begin{vmatrix} 2 & -2 \\ -1 & 1 \end{vmatrix} = 2 \cdot 1 - (-2)(-1) = 2 - 2 = 0 \]
\[ \begin{vmatrix} 3 & -2 \\ 2 & 1 \end{vmatrix} = 3 \cdot 1 - (-2) \cdot 2 = 3 + 4 = 7 \]
\[ \begin{vmatrix} 3 & 2 \\ 2 & -1 \end{vmatrix} = 3 \cdot (-1) - 2 \cdot 2 = -3 - 4 = -7 \]
Thus:

\[ \det(A) = 1(0) - 2(7) + (-3)(-7) = 0 - 14 + 21 = 7 \]
Now, find the adjugate matrix \( \text{adj}(A) \) by calculating cofactors and taking their transpose.

Finally, compute:

\[ A^{-1} = \frac{1}{7} \cdot \text{adj}(A) \]
Step 3: Solve for \( X \).

Multiply both sides of the matrix equation by \( A^{-1} \):

\[ X = A^{-1} B \]
Substituting the values of \( A^{-1} \) and \( B \), we obtain the values of \( x \), \( y \), and \( z \).

Conclusion:

The solution of the system of equations is given by the vector
\[ X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \] which provides the required values of \( x \), \( y \), and \( z \).