Question

Solve by matrix method the following system of equations:
\(2x + y + z = 1\)
\(x - 2y - z = \frac{3}{2}\)
\(3y - 5z = 9\)

Hint:

The matrix method is systematic but prone to calculation errors, especially with signs in cofactors. To avoid fractions inside the B matrix, you could have multiplied the second equation by 2 initially to get \(2x - 4y - 2z = 3\). The final answer would be the same, but the intermediate calculations might be simpler.

Answer 1

Step 1: Understanding the Concept:
The matrix method for solving a system of linear equations involves representing the system in the form \(AX = B\). The solution is then found by calculating \(X = A^{-1}B\), where \(A^{-1}\) is the inverse of the coefficient matrix \(A\). This method is applicable only if the determinant of \(A\) is non-zero.
Step 2: Key Formula or Approach:
1. Write the system of equations in the matrix form \(AX = B\).
2. Calculate the determinant of A, \(|A|\).
3. Find the adjugate of A, \(\text{adj}(A)\).
4. Calculate the inverse of A using the formula \(A^{-1} = \frac{1}{|A|}\text{adj}(A)\).
5. Find the solution by computing \(X = A^{-1}B\).
Step 3: Detailed Explanation:
The given system of equations can be written as: \[ 2x + 1y + 1z = 1 \] \[ 1x - 2y - 1z = \frac{3}{2} \] \[ 0x + 3y - 5z = 9 \] In matrix form \(AX = B\), we have: \[ A = \begin{bmatrix} 2 & 1 & 1
1 & -2 & -1
0 & 3 & -5 \end{bmatrix}, \quad X = \begin{bmatrix} x
y
z \end{bmatrix}, \quad B = \begin{bmatrix} 1
3/2
9 \end{bmatrix} \] First, calculate the determinant of A: \[ |A| = 2((-2)(-5) - (-1)(3)) - 1((1)(-5) - (-1)(0)) + 1((1)(3) - (-2)(0)) \] \[ |A| = 2(10 + 3) - 1(-5 - 0) + 1(3 - 0) \] \[ |A| = 2(13) - (-5) + 3 = 26 + 5 + 3 = 34 \] Since \(|A| = 34 \neq 0\), a unique solution exists.
Next, find the adjugate of A. The matrix of cofactors is: \[ C_{11} = (-1)^{1+1} \begin{vmatrix} -2 & -1
3 & -5 \end{vmatrix} = 10 - (-3) = 13 \] \[ C_{12} = (-1)^{1+2} \begin{vmatrix} 1 & -1
0 & -5 \end{vmatrix} = -(-5 - 0) = 5 \] \[ C_{13} = (-1)^{1+3} \begin{vmatrix} 1 & -2
0 & 3 \end{vmatrix} = 3 - 0 = 3 \] \[ C_{21} = (-1)^{2+1} \begin{vmatrix} 1 & 1
3 & -5 \end{vmatrix} = -(-5 - 3) = 8 \] \[ C_{22} = (-1)^{2+2} \begin{vmatrix} 2 & 1
0 & -5 \end{vmatrix} = -10 - 0 = -10 \] \[ C_{23} = (-1)^{2+3} \begin{vmatrix} 2 & 1
0 & 3 \end{vmatrix} = -(6 - 0) = -6 \] \[ C_{31} = (-1)^{3+1} \begin{vmatrix} 1 & 1
-2 & -1 \end{vmatrix} = -1 - (-2) = 1 \] \[ C_{32} = (-1)^{3+2} \begin{vmatrix} 2 & 1
1 & -1 \end{vmatrix} = -(-2 - 1) = 3 \] \[ C_{33} = (-1)^{3+3} \begin{vmatrix} 2 & 1
1 & -2 \end{vmatrix} = -4 - 1 = -5 \] The matrix of cofactors is \(\begin{bmatrix} 13 & 5 & 3
8 & -10 & -6
1 & 3 & -5 \end{bmatrix}\).
The adjugate of A is the transpose of the cofactor matrix: \[ \text{adj}(A) = \begin{bmatrix} 13 & 8 & 1
5 & -10 & 3
3 & -6 & -5 \end{bmatrix} \] Now, find the inverse of A: \[ A^{-1} = \frac{1}{|A|}\text{adj}(A) = \frac{1}{34} \begin{bmatrix} 13 & 8 & 1
5 & -10 & 3
3 & -6 & -5 \end{bmatrix} \] Finally, calculate the solution \(X = A^{-1}B\): \[ \begin{bmatrix} x
y
z \end{bmatrix} = \frac{1}{34} \begin{bmatrix} 13 & 8 & 1
5 & -10 & 3
3 & -6 & -5 \end{bmatrix} \begin{bmatrix} 1
3/2
9 \end{bmatrix} \] \[ \begin{bmatrix} x
y
z \end{bmatrix} = \frac{1}{34} \begin{bmatrix} 13(1) + 8(3/2) + 1(9)
5(1) - 10(3/2) + 3(9)
3(1) - 6(3/2) - 5(9) \end{bmatrix} \] \[ \begin{bmatrix} x
y
z \end{bmatrix} = \frac{1}{34} \begin{bmatrix} 13 + 12 + 9
5 - 15 + 27
3 - 9 - 45 \end{bmatrix} = \frac{1}{34} \begin{bmatrix} 34
17
-51 \end{bmatrix} \] \[ \begin{bmatrix} x
y
z \end{bmatrix} = \begin{bmatrix} 34/34
17/34
-51/34 \end{bmatrix} = \begin{bmatrix} 1
1/2
-3/2 \end{bmatrix} \] Step 4: Final Answer:
The solution to the system of equations is \(x = 1\), \(y = \frac{1}{2}\), and \(z = -\frac{3}{2}\).