Question

Solve the following system of equations by matrix method:
\(x + y + 2z = 1\)
\(3x + 2y + z = 7\)
\(2x + y + 3z = 2\)

Hint:

After finding the solution (x, y, z), it is a very good practice to substitute these values back into one of the original equations to quickly verify your answer. For example, using the first equation: \(2 + 1 + 2(-1) = 3 - 2 = 1\). This confirms the solution is correct.

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 is found by pre-multiplying both sides by the inverse of A, which gives \(X = A^{-1}B\).
Step 2: Key Formula or Approach:
1. Write the system in the form \(AX = B\). 2. Calculate the inverse of the coefficient matrix, \(A^{-1}\). 3. Multiply \(A^{-1}\) by B to find the solution matrix X.
Step 3: Detailed Explanation:
1. Represent the system in matrix form: \[ A = \begin{bmatrix} 1 & 1 & 2
3 & 2 & 1
2 & 1 & 3 \end{bmatrix}, \quad X = \begin{bmatrix} x
y
z \end{bmatrix}, \quad B = \begin{bmatrix} 1
7
2 \end{bmatrix} \] The equation is \(AX = B\), and the solution is \(X = A^{-1}B\).
2. Find the inverse of A: First, calculate the determinant of A: \[ \det(A) = 1(2 \cdot 3 - 1 \cdot 1) - 1(3 \cdot 3 - 1 \cdot 2) + 2(3 \cdot 1 - 2 \cdot 2) \] \[ = 1(6 - 1) - 1(9 - 2) + 2(3 - 4) \] \[ = 1(5) - 1(7) + 2(-1) = 5 - 7 - 2 = -4 \] Since \(\det(A) \neq 0\), a unique solution exists.
Next, find the adjugate of A. The cofactors are:
\(C_{11} = +(6-1) = 5\)
\(C_{12} = -(9-2) = -7\)
\(C_{13} = +(3-4) = -1\)
\(C_{21} = -(1 \cdot 3 - 2 \cdot 1) = -(3-2) = -1\)
\(C_{22} = +(1 \cdot 3 - 2 \cdot 2) = +(3-4) = -1\)
\(C_{23} = -(1 \cdot 1 - 1 \cdot 2) = -(1-2) = 1\)
\(C_{31} = +(1 \cdot 1 - 2 \cdot 2) = +(1-4) = -3\)
\(C_{32} = -(1 \cdot 1 - 2 \cdot 3) = -(1-6) = 5\)
\(C_{33} = +(1 \cdot 2 - 1 \cdot 3) = +(2-3) = -1\)
The adjugate matrix is \(\text{adj}(A) = C^T = \begin{bmatrix} 5 & -1 & -3
-7 & -1 & 5
-1 & 1 & -1 \end{bmatrix}\).
The inverse is \(A^{-1} = \frac{1}{-4} \begin{bmatrix} 5 & -1 & -3
-7 & -1 & 5
-1 & 1 & -1 \end{bmatrix}\).
3. Solve for X: \[ X = A^{-1}B = -\frac{1}{4} \begin{bmatrix} 5 & -1 & -3
-7 & -1 & 5
-1 & 1 & -1 \end{bmatrix} \begin{bmatrix} 1
7
2 \end{bmatrix} \] \[ X = -\frac{1}{4} \begin{bmatrix} (5)(1) + (-1)(7) + (-3)(2)
(-7)(1) + (-1)(7) + (5)(2)
(-1)(1) + (1)(7) + (-1)(2) \end{bmatrix} \] \[ X = -\frac{1}{4} \begin{bmatrix} 5 - 7 - 6
-7 - 7 + 10
-1 + 7 - 2 \end{bmatrix} = -\frac{1}{4} \begin{bmatrix} -8
-4
4 \end{bmatrix} \] \[ X = \begin{bmatrix} (-8)/(-4)
(-4)/(-4)
(4)/(-4) \end{bmatrix} = \begin{bmatrix} 2
1
-1 \end{bmatrix} \] So, \(\begin{bmatrix} x
y
z \end{bmatrix} = \begin{bmatrix} 2
1
-1 \end{bmatrix}\).
Step 4: Final Answer:
The solution to the system of equations is x = 2, y = 1, and z = -1.