Question 45 of 58intermediate🔧 ApplyLong Answer4 marks

Solve system of linear equations, using matrix method: x - y + z = 4 2x + y - 3z = 0 x + y + z = 2

Correct Answer

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

A=[1−1121−3111],X=[xyz] and B=[402]A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \text{ and } B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}

We calculate the determinant of matrix A:

∣A∣=1(1+3)−(−1)(2+3)+1(2−1)=4+5+1=10|A| = 1(1 + 3) - (-1)(2 + 3) + 1(2 - 1) = 4 + 5 + 1 = 10

Since ∣A∣≠0|A| \neq 0, A is nonsingular and its inverse exists.

Now, we find the cofactors of the elements of A:

A11=4,A12=−5,A13=1A_{11} = 4, A_{12} = -5, A_{13} = 1 A21=2,A22=0,A23=−2A_{21} = 2, A_{22} = 0, A_{23} = -2 A31=2,A32=5,A33=3A_{31} = 2, A_{32} = 5, A_{33} = 3

Thus, the adjoint of matrix A is:

adj(A)=[422−5051−23]adj(A) = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}

Therefore, the inverse is:

A−1=1∣A∣adj(A)=110[422−5051−23]A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}

Now, calculating X = A−1A^{-1}B:

X=110[422−5051−23][402]=110[16+0+4−20+0+104+0+6]=110[20−1010]=[2−11]X = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 16 + 0 + 4 \\ -20 + 0 + 10 \\ 4 + 0 + 6 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}

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

Exercise: EXERCISE 4.5 | Q: 12 | (Page 22)
For More Understanding

Explanation

This solution follows the standard matrix method for solving a system of linear equations, as demonstrated in Example 17 of the provided context. The process involves expressing the system as AX=B, verifying the existence of the inverse by checking that the determinant |A| is non-zero, calculating the adjoint of A, and finally computing X = A−1A^{-1}B to find the values of the variables.

Solution Steps

  1. Step 1: Represent the system of equations in matrix form AX = B.

  2. Step 2: Calculate the determinant |A| to ensure the matrix is nonsingular.

  3. Step 3: Calculate the cofactors and the adjoint of matrix A.

  4. Step 4: Determine the inverse matrix A−1A^{-1} using the formula A−1A^{-1} = adj(A)/|A|.

  5. Step 5: Solve for X by multiplying A−1A^{-1} with B to get the final values for x, y, and z.

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