Question

Solve the system of equations by matrix method:
\( 3x - 2y + 3z = 8 \)
\( 2x + y - z = 1 \)
\( 4x - 3y + 2z = 4 \)

Hint:

The matrix method for solving systems of equations is systematic but prone to arithmetic errors, especially when finding the adjugate matrix. Always be careful with the signs of the cofactors. After finding the solution, quickly substitute the values of x, y, and z back into one of the original equations to check if it holds true.

Answer 1

Step 1: Understanding the Concept:
A system of linear equations can be represented in matrix form as \( AX = B \), where A is the matrix of coefficients, X is the column matrix of variables, and B is the column matrix of constants. The solution can be found by pre-multiplying both sides by the inverse of A, which gives \( X = A^{-1}B \).
Step 2: Key Formula or Approach:
1. Express the system of equations in the matrix form \( AX = B \).
2. Find the inverse of the coefficient matrix A, using \( A^{-1} = \frac{1}{\det(A)}\text{adj}(A) \).
3. Solve for the variable matrix X by computing the product \( X = A^{-1}B \).
Step 3: Detailed Explanation or Calculation:
1. Formulate the matrix equation:
The given system is:
\( 3x - 2y + 3z = 8 \)
\( 2x + y - z = 1 \)
\( 4x - 3y + 2z = 4 \)
This can be written as \( AX=B \), where:
\[ A = \begin{bmatrix} 3 & -2 & 3
2 & 1 & -1
4 & -3 & 2 \end{bmatrix}, X = \begin{bmatrix} x
y
z \end{bmatrix}, B = \begin{bmatrix} 8
1
4 \end{bmatrix} \] 2. Find the inverse of A:
First, calculate the determinant of A:
\[ \det(A) = 3( (1)(2) - (-1)(-3) ) - (-2)( (2)(2) - (-1)(4) ) + 3( (2)(-3) - (1)(4) ) \] \[ \det(A) = 3(2 - 3) + 2(4 + 4) + 3(-6 - 4) \] \[ \det(A) = 3(-1) + 2(8) + 3(-10) = -3 + 16 - 30 = -17 \] Since \( \det(A) \neq 0 \), a unique solution exists.
Next, find the adjugate of A. The matrix of cofactors is:
\( C_{11} = -1 \), \( C_{12} = -8 \), \( C_{13} = -10 \)
\( C_{21} = -5 \), \( C_{22} = -6 \), \( C_{23} = 1 \)
\( C_{31} = -1 \), \( C_{32} = 9 \), \( C_{33} = 7 \)
\[ \text{Cofactor Matrix} = \begin{bmatrix} -1 & -8 & -10
-5 & -6 & 1
-1 & 9 & 7 \end{bmatrix} \] \[ \text{adj}(A) = C^T = \begin{bmatrix} -1 & -5 & -1
-8 & -6 & 9
-10 & 1 & 7 \end{bmatrix} \] \[ A^{-1} = \frac{1}{-17} \begin{bmatrix} -1 & -5 & -1
-8 & -6 & 9
-10 & 1 & 7 \end{bmatrix} \] 3. Solve for X:
\[ X = A^{-1}B = \frac{1}{-17} \begin{bmatrix} -1 & -5 & -1
-8 & -6 & 9
-10 & 1 & 7 \end{bmatrix} \begin{bmatrix} 8
1
4 \end{bmatrix} \] Perform the matrix multiplication:
\[ \begin{bmatrix} (-1)(8) + (-5)(1) + (-1)(4)
(-8)(8) + (-6)(1) + (9)(4)
(-10)(8) + (1)(1) + (7)(4) \end{bmatrix} = \begin{bmatrix} -8 - 5 - 4
-64 - 6 + 36
-80 + 1 + 28 \end{bmatrix} = \begin{bmatrix} -17
-34
-51 \end{bmatrix} \] Now, multiply by the scalar \( \frac{1}{-17} \):
\[ X = \frac{1}{-17} \begin{bmatrix} -17
-34
-51 \end{bmatrix} = \begin{bmatrix} \frac{-17}{-17}
\frac{-34}{-17}
\frac{-51}{-17} \end{bmatrix} = \begin{bmatrix} 1
2
3 \end{bmatrix} \] \[ \begin{bmatrix} x
y
z \end{bmatrix} = \begin{bmatrix} 1
2
3 \end{bmatrix} \] Step 4: Final Answer:
The solution to the system of equations is \( x = 1, y = 2, z = 3 \).