Question

Solve the system of equations
x + y + z = 2,
2x + y - 3z = 0,
x - y + z - 4 = 0 by matrix method.

Hint:

After finding the solution using the matrix method, it's always a good practice to substitute the values of x, y, and z back into the original equations to verify that they are correct. This quick check can help you catch any calculation errors.

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 coefficient matrix, X is the column matrix of variables, and B is the column matrix of constants. The solution can be found using the matrix inverse method, where X = A\(^{-1}\)B, provided that A is non-singular (\(|\)A\(|\) \(\neq\) 0).
Step 2: Key Formula or Approach:
1. Write the system of equations in the matrix form AX = B.
2. Calculate the determinant of the coefficient matrix A, \(|\)A\(|\).
3. If \(|\)A\(|\) \(\neq\) 0, find the inverse of A, A\(^{-1}\), using A\(^{-1}\) = \(\frac{1}{|\text{A}|}\)adj(A).
4. Solve for the variables by calculating X = A\(^{-1}\)B.
Step 3: Detailed Explanation or Calculation:
1. Express in Matrix Form AX = B:
First, rewrite the third equation in standard form: \(x - y + z = 4\).
The system is: \[ x + y + z = 2 \] \[ 2x + y - 3z = 0 \] \[ x - y + z = 4 \] In matrix form: \[ A = \begin{bmatrix} 1 & 1 & 1
2 & 1 & -3
1 & -1 & 1 \end{bmatrix}, X = \begin{bmatrix} x
y
z \end{bmatrix}, B = \begin{bmatrix} 2
0
4 \end{bmatrix} \] 2. Calculate the Determinant \(|\)A\(|\): \[ |\text{A}| = 1(1 \cdot 1 - (-3) \cdot (-1)) - 1(2 \cdot 1 - (-3) \cdot 1) + 1(2 \cdot (-1) - 1 \cdot 1) \] \[ |\text{A}| = 1(1 - 3) - 1(2 + 3) + 1(-2 - 1) \] \[ |\text{A}| = 1(-2) - 1(5) + 1(-3) = -2 - 5 - 3 = -10 \] Since \(|\)A\(|\) = -10 \(\neq\) 0, a unique solution exists.
3. Find the Inverse A\(^{-1}\):
First, find the adjugate of A. The cofactor matrix C is: \[ C_{11} = +(1-3)=-2, C_{12} = -(2 - (-3))=-5, C_{13} = +(-2-1)=-3 \] \[ C_{21} = -(1 - (-1))=-2, C_{22} = +(1-1)=0, C_{23} = -(-1-1)=2 \] \[ C_{31} = +(-3-1)=-4, C_{32} = -(-3-2)=5, C_{33} = +(1-2)=-1 \] So, C = \(\begin{bmatrix} -2 & -5 & -3
-2 & 0 & 2
-4 & 5 & -1 \end{bmatrix}\).
The adjugate is adj(A) = C\(^T\): \[ \text{adj(A)} = \begin{bmatrix} -2 & -2 & -4
-5 & 0 & 5
-3 & 2 & -1 \end{bmatrix} \] The inverse is A\(^{-1}\) = \(\frac{1}{|\text{A}|}\)adj(A): \[ A^{-1} = -\frac{1}{10} \begin{bmatrix} -2 & -2 & -4
-5 & 0 & 5
-3 & 2 & -1 \end{bmatrix} \] 4. Solve for X:
\[ X = A^{-1}B = -\frac{1}{10} \begin{bmatrix} -2 & -2 & -4
-5 & 0 & 5
-3 & 2 & -1 \end{bmatrix} \begin{bmatrix} 2
0
4 \end{bmatrix} \] \[ X = -\frac{1}{10} \begin{bmatrix} (-2)(2) + (-2)(0) + (-4)(4)
(-5)(2) + (0)(0) + (5)(4)
(-3)(2) + (2)(0) + (-1)(4) \end{bmatrix} \] \[ X = -\frac{1}{10} \begin{bmatrix} -4 + 0 - 16
-10 + 0 + 20
-6 + 0 - 4 \end{bmatrix} = -\frac{1}{10} \begin{bmatrix} -20
10
-10 \end{bmatrix} \] \[ X = \begin{bmatrix} (-20)/(-10)
10/(-10)
(-10)/(-10) \end{bmatrix} = \begin{bmatrix} 2
-1
1 \end{bmatrix} \] Step 4: Final Answer:
The solution to the system of equations is x = 2, y = -1, and z = 1.