Question

Solve the following equations by the method of inversion: \( 2x-y+z=1 \), \( x+2y+3z=8 \), \( 3x+y-4z=1 \).

Hint:

The matrix inversion method is procedural. Be careful with arithmetic, especially with negative signs when calculating the determinant and cofactors. Always check your final answer by plugging it back into the original equations.

Answer 1

The system can be written in matrix form \( AX=B \), where: \[ A = \begin{bmatrix} 2 & -1 & 1
1 & 2 & 3
3 & 1 & -4 \end{bmatrix}, \quad X = \begin{bmatrix} x
y
z \end{bmatrix}, \quad B = \begin{bmatrix} 1
8
1 \end{bmatrix} \] The solution is \( X = A^{-1}B \).
Step 1: Find the determinant of A. \[ |A| = 2(-8-3) - (-1)(-4-9) + 1(1-6) = 2(-11) + 1(-13) - 5 = -22 - 13 - 5 = -40 \] Step 2: Find the adjugate of A.
The matrix of cofactors is: \[ C = \begin{bmatrix} -11 & 13 & -5
-3 & -11 & -5
-5 & -5 & 5 \end{bmatrix} \] The adjugate is the transpose of the cofactor matrix: \[ \text{adj}(A) = C^T = \begin{bmatrix} -11 & -3 & -5
13 & -11 & -5
-5 & -5 & 5 \end{bmatrix} \] Step 3: Find the inverse of A. \[ A^{-1} = \frac{1}{|A|} \text{adj}(A) = -\frac{1}{40} \begin{bmatrix} -11 & -3 & -5
13 & -11 & -5
-5 & -5 & 5 \end{bmatrix} \] Step 4: Calculate \( X = A^{-1}B \). \[ X = -\frac{1}{40} \begin{bmatrix} -11 & -3 & -5
13 & -11 & -5
-5 & -5 & 5 \end{bmatrix} \begin{bmatrix} 1
8
1 \end{bmatrix} = -\frac{1}{40} \begin{bmatrix} -11(1) - 3(8) - 5(1)
13(1) - 11(8) - 5(1)
-5(1) - 5(8) + 5(1) \end{bmatrix} \] \[ X = -\frac{1}{40} \begin{bmatrix} -11 - 24 - 5
13 - 88 - 5
-5 - 40 + 5 \end{bmatrix} = -\frac{1}{40} \begin{bmatrix} -40
-80
-40 \end{bmatrix} = \begin{bmatrix} 1
2
1 \end{bmatrix} \] Thus, \( x=1, y=2, z=1 \).