Solve the system of linear equations using the matrix method: x - y…

Question 47 of 58intermediate🔧 ApplyLong Answer4 marks

Solve the system of linear equations using the matrix method:

x - y + 2z = 7

3x + 4y - 5z = -5

2x - y + 3z = 12

Correct Answer

x = 2, y = 1, z = 3

Exercise: EXERCISE 4.5 | Q: 14 | (Page 22)

For More Understanding

Explanation

This question from Exercise 4.6 (Q.14) requires solving a system of three linear equations using the matrix method. The student must express the system in the form $AX = B$ , verify that $A$ is nonsingular by calculating $|A| \neq 0$ , find the inverse using $\text{adj}(A)/|A|$ , and finally compute $X = A^{-1}B$ to obtain the values of $x$ , $y$ , and $z$ .

Solution Steps

Step 1: Write the system in matrix form $AX = B$ , where
$A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix}$
Step 2: Calculate $|A|$ :
$|A| = 1(12 - 5) + 1(9 + 10) + 2(-3 - 8) = 7 + 19 - 22 = 4 \neq 0$
Since $|A| \neq 0$ , $A$ is nonsingular and $A^{-1}$ exists.
Step 3: Find cofactors:
$A_{11} = 7, \quad A_{12} = -19, \quad A_{13} = -11$ $A_{21} = 1, \quad A_{22} = -1, \quad A_{23} = -1$ $A_{31} = -3, \quad A_{32} = 11, \quad A_{33} = 7$
Step 4: Form $\text{adj}(A)$ :
$\text{adj}(A) = \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix}$
Then
$A^{-1} = \frac{1}{4} \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix}$
Step 5: Calculate $X = A^{-1}B$ :
$X = \frac{1}{4} \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix} \begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 8 \\ 4 \\ 12 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}$
Step 6: Hence, $x = 2$ , $y = 1$ , $z = 3$ .

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