Question

Solve the following system of equations by matrix method :
2x + 3y + 3z = 5
x - 2y + z = -4
3x - y - 2z = 3

Hint:

After finding the solution (x, y, z), it is a very good practice to substitute these values back into one or all of the original equations to quickly verify your answer. For example, in the first equation: \( 2(1) + 3(2) + 3(-1) = 2 + 6 - 3 = 5 \). This confirms the result is correct.

Answer 1

Step 1: Understanding the Concept:
A system of linear equations can be represented in matrix form as \( AX = B \). The solution can be found using the formula \( X = A^{-1}B \), provided that the coefficient matrix A is invertible (i.e., its determinant is non-zero).
Step 2: Setting up the Matrix Equation:
The given system of equations can be written as \( AX = B \), where: \[ A = \begin{bmatrix} 2 & 3 & 3
1 & -2 & 1
3 & -1 & -2 \end{bmatrix}, \quad X = \begin{bmatrix} x
y
z \end{bmatrix}, \quad B = \begin{bmatrix} 5
-4
3 \end{bmatrix} \] Step 3: Finding the Inverse of the Coefficient Matrix A:
First, we calculate the determinant of A to check for invertibility. \[ \det(A) = |A| = 2((-2)(-2) - (1)(-1)) - 3((1)(-2) - (1)(3)) + 3((1)(-1) - (-2)(3)) \] \[ = 2(4+1) - 3(-2-3) + 3(-1+6) = 2(5) - 3(-5) + 3(5) = 10 + 15 + 15 = 40 \] Since \( |A| = 40 \neq 0 \), the inverse \( A^{-1} \) exists.
Next, we find the adjugate of A. The cofactor matrix C is: \[ C_{11} = 5, \quad C_{12} = -(-5) = 5, \quad C_{13} = 5 \] \[ C_{21} = -(-6+3) = 3, \quad C_{22} = -4-9 = -13, \quad C_{23} = -(-2-9) = 11 \] \[ C_{31} = 3-(-6) = 9, \quad C_{32} = -(2-3) = 1, \quad C_{33} = -4-3 = -7 \] So, the cofactor matrix is \( C = \begin{bmatrix} 5 & 5 & 5
3 & -13 & 11
9 & 1 & -7 \end{bmatrix} \).
The adjugate is the transpose of the cofactor matrix: \( \text{adj}(A) = C^T = \begin{bmatrix} 5 & 3 & 9
5 & -13 & 1
5 & 11 & -7 \end{bmatrix} \).
The inverse is \( A^{-1} = \frac{1}{|A|} \text{adj}(A) \): \[ A^{-1} = \frac{1}{40} \begin{bmatrix} 5 & 3 & 9
5 & -13 & 1
5 & 11 & -7 \end{bmatrix} \] Step 4: Solving for X:
Now we calculate \( X = A^{-1}B \): \[ \begin{bmatrix} x
y
z \end{bmatrix} = \frac{1}{40} \begin{bmatrix} 5 & 3 & 9
5 & -13 & 1
5 & 11 & -7 \end{bmatrix} \begin{bmatrix} 5
-4
3 \end{bmatrix} \] \[ = \frac{1}{40} \begin{bmatrix} (5)(5) + (3)(-4) + (9)(3)
(5)(5) + (-13)(-4) + (1)(3)
(5)(5) + (11)(-4) + (-7)(3) \end{bmatrix} \] \[ = \frac{1}{40} \begin{bmatrix} 25 - 12 + 27
25 + 52 + 3
25 - 44 - 21 \end{bmatrix} = \frac{1}{40} \begin{bmatrix} 40
80
-40 \end{bmatrix} = \begin{bmatrix} 1
2
-1 \end{bmatrix} \] Step 5: Final Answer:
By comparing the elements of the resulting matrix, we find the solution: \( x = 1, y = 2, \) and \( z = -1 \).