Question

Solve the following system of linear equations by matrix method:
\( x + y + z = 6 \)
\( y + 3z = 11 \)
\( x + z = 2y \)

Hint:

Before starting the matrix method, always rearrange the equations so that all variable terms are on the left and constants are on the right, with variables in the same order (e.g., x, then y, then z). This prevents errors when forming the coefficient matrix A.

Answer 1

Step 1: Understanding the Concept:
We can solve a system of linear equations by representing it in the matrix form \( AX = B \), where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix. The solution is then found using the formula \( X = A^{-1}B \).
Step 2: Key Formula or Approach:
1. Rewrite the equations in standard form and express the system as \( AX = B \).
2. Find the inverse of the coefficient matrix A.
3. Multiply \( A^{-1} \) by B to find the solution matrix X.
Step 3: Detailed Explanation or Calculation:
1. Formulate the matrix equation:
First, rewrite the third equation in standard form: \( x + z = 2y → x - 2y + z = 0 \).
The system is:
\( x + y + z = 6 \)
\( 0x + y + 3z = 11 \)
\( x - 2y + z = 0 \)
In matrix form \( AX = B \):
\[ A = \begin{bmatrix} 1 & 1 & 1
0 & 1 & 3
1 & -2 & 1 \end{bmatrix}, X = \begin{bmatrix} x
y
z \end{bmatrix}, B = \begin{bmatrix} 6
11
0 \end{bmatrix} \] 2. Find the inverse of A:
First, find the determinant of A:
\[ |A| = 1(1 - (-6)) - 1(0 - 3) + 1(0 - 1) = 7 + 3 - 1 = 9 \] Next, find the adjugate of A. The cofactor matrix is:
\[ C = \begin{bmatrix} 7 & 3 & -1
-3 & 0 & 3
2 & -3 & 1 \end{bmatrix} \] \[ \text{adj}(A) = C^T = \begin{bmatrix} 7 & -3 & 2
3 & 0 & -3
-1 & 3 & 1 \end{bmatrix} \] The inverse is:
\[ A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{9} \begin{bmatrix} 7 & -3 & 2
3 & 0 & -3
-1 & 3 & 1 \end{bmatrix} \] 3. Solve for X:
\[ X = A^{-1}B = \frac{1}{9} \begin{bmatrix} 7 & -3 & 2
3 & 0 & -3
-1 & 3 & 1 \end{bmatrix} \begin{bmatrix} 6
11
0 \end{bmatrix} \] \[ X = \frac{1}{9} \begin{bmatrix} (7)(6) + (-3)(11) + (2)(0)
(3)(6) + (0)(11) + (-3)(0)
(-1)(6) + (3)(11) + (1)(0) \end{bmatrix} = \frac{1}{9} \begin{bmatrix} 42 - 33
18
-6 + 33 \end{bmatrix} = \frac{1}{9} \begin{bmatrix} 9
18
27 \end{bmatrix} \] \[ X = \begin{bmatrix} 1
2
3 \end{bmatrix} \] Step 4: Final Answer:
The solution is \( x = 1, y = 2, z = 3 \).