Question 46 of 58intermediate🔧 ApplyLong Answer4 marks

Solve system of linear equations, using matrix method: 2x + 3y + 3z = 5 x - 2y + z = -4 3x - y - 2z = 3

Correct Answer

x = 1, y = 2, z = -1

Exercise: EXERCISE 4.5 | Q: 13 | (Page 22)
For More Understanding

Explanation

The solution follows the matrix method demonstrated in Example 17 of the provided context. The system of equations is expressed in the form AX = B. The determinant of matrix A is calculated to ensure it is non-singular. The cofactors and the adjoint of A are then computed to find the inverse of A. Finally, the solution is obtained by multiplying the inverse of A with matrix B.

Solution Steps

  1. The system of equations can be written in the form AX = B, where:

    A=[2331−213−1−2],X=[xyz] and B=[5−43]A = \begin{bmatrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \text{ and } B = \begin{bmatrix} 5 \\ -4 \\ 3 \end{bmatrix}

  2. We see that the determinant of A is:

    ∣A∣=2((−2)(−2)−(1)(−1))−3((1)(−2)−(1)(3))+3((1)(−1)−(−2)(3))|A| = 2((-2)(-2) - (1)(-1)) - 3((1)(-2) - (1)(3)) + 3((1)(-1) - (-2)(3)) ∣A∣=2(4+1)−3(−2−3)+3(−1+6)=2(5)−3(−5)+3(5)=10+15+15=40≠0|A| = 2(4 + 1) - 3(-2 - 3) + 3(-1 + 6) = 2(5) - 3(-5) + 3(5) = 10 + 15 + 15 = 40 \neq 0

    Hence, A is nonsingular and so its inverse exists.

  3. Now, we find the cofactors of matrix A:

    A11=5,A12=5,A13=5A_{11} = 5, A_{12} = 5, A_{13} = 5 A21=3,A22=−13,A23=11A_{21} = 3, A_{22} = -13, A_{23} = 11 A31=9,A32=1,A33=−7A_{31} = 9, A_{32} = 1, A_{33} = -7

  4. Therefore, the adjoint of A is:

    adj(A)=[5395−131511−7]adj(A) = \begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix}

    And the inverse is:

    A−1=1∣A∣adj(A)=140[5395−131511−7]A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{40} \begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix}

  5. So, X = A⁻¹B:

    X=140[5395−131511−7][5−43]=140[25−12+2725+52+325−44−21]=140[4080−40]=[12−1]X = \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} 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}

    Hence, x = 1, y = 2 and z = -1.

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