Question 38 of 58intermediate🔧 ApplyLong Answer4 marks

Examine the consistency of the system of equations: 3x - y - 2z = 2 2y - z = -1 3x - 5y = 3

Correct Answer

The system of equations is inconsistent.

Exercise: EXERCISE 4.5 | Q: 5 | (Page 22)
For More Understanding

Explanation

To examine the consistency, we write the system in the form AX=BAX = B. We calculate the determinant of the coefficient matrix AA. If ∣A∣≠0|A| \neq 0, the system is consistent with a unique solution. If ∣A∣=0|A| = 0, we calculate (adjA)B(adj A)B. If (adjA)B≠0(adj A)B \neq 0, the system is inconsistent (solution does not exist). Here, ∣A∣=0|A| = 0 and (adjA)B≠0(adj A)B \neq 0, leading to the conclusion that the system is inconsistent.

Solution Steps

  1. Step 1: Write the system in matrix form AX=BAX = B. A=[3−1−202−13−50]A = \begin{bmatrix} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{bmatrix}, X=[xyz]X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, and B=[2−13]B = \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}.

  2. Step 2: Calculate the determinant of matrix AA (∣A∣|A|). Expanding along Row 1:

    ∣A∣=3(0−5)−(−1)(0−(−3))+(−2)(0−6)|A| = 3(0 - 5) - (-1)(0 - (-3)) + (-2)(0 - 6) ∣A∣=3(−5)+1(3)−2(−6)|A| = 3(-5) + 1(3) - 2(-6)

    ∣A∣=−15+3+12=0|A| = -15 + 3 + 12 = 0. Since ∣A∣=0|A| = 0, the matrix AA is singular, and the system does not have a unique solution.

  3. Step 3: Calculate the adjoint of AA (adjAadj A). First, find the cofactors of AA:

    C11=−5,C12=−3,C13=−6C_{11} = -5, C_{12} = -3, C_{13} = -6 C21=10,C22=6,C23=12C_{21} = 10, C_{22} = 6, C_{23} = 12 C31=5,C32=3,C33=6C_{31} = 5, C_{32} = 3, C_{33} = 6 adjA=[−5105−363−6126]adj A = \begin{bmatrix} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{bmatrix}
  4. Step 4: Calculate the product (adjA)B(adj A)B. (adjA)B=[−5105−363−6126][2−13]=[−10−10+15−6−6+9−12−12+18]=[−5−3−6]≠0(adj A)B = \begin{bmatrix} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{bmatrix} \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix} = \begin{bmatrix} -10 - 10 + 15 \\ -6 - 6 + 9 \\ -12 - 12 + 18 \end{bmatrix} = \begin{bmatrix} -5 \\ -3 \\ -6 \end{bmatrix} \neq 0.

  5. Step 5: Conclusion. Since ∣A∣=0|A| = 0 and (adjA)B≠0(adj A)B \neq 0, the solution does not exist. Therefore, the system is inconsistent.