Question 41 of 58intermediate🔧 ApplyLong Answer4 marks

Solve the system of linear equations, using the matrix method:

2x−y=−22x - y = -2 3x+4y=33x + 4y = 3
Correct Answer

x = -5/11, y = 12/11

Exercise: EXERCISE 4.5 | Q: 8 | (Page 22)
For More Understanding

Explanation

This question from Exercise 4.5 requires solving a system of two linear equations using the matrix method. The solution follows the standard procedure: expressing the system as AX=BAX = B, finding the determinant to verify the inverse exists, calculating the adjoint, and then multiplying A−1A^{-1} with BB to get the solution matrix XX.

Solution Steps

  1. Step 1: Write the system in matrix form AX=BAX = B, where

    A=[2−134],X=[xy],B=[−23]A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \end{bmatrix}, \quad B = \begin{bmatrix} -2 \\ 3 \end{bmatrix}

  2. Step 2: Find ∣A∣|A|.

    ∣A∣=2(4)−(−1)(3)=8+3=11≠0|A| = 2(4) - (-1)(3) = 8 + 3 = 11 \neq 0

    Hence, AA is nonsingular and its inverse exists.

  3. Step 3: Find adj(A)\text{adj}(A).

    adj(A)=[41−32]\text{adj}(A) = \begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix}

  4. Step 4: Calculate A−1A^{-1}.

    A−1=1∣A∣×adj(A)=111×[41−32]A^{-1} = \frac{1}{|A|} \times \text{adj}(A) = \frac{1}{11} \times \begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix}

  5. Step 5: Find X=A−1BX = A^{-1}B.

    X=A−1B=111×[41−32]×[−23]=111×[−512]=[−5/1112/11]X = A^{-1}B = \frac{1}{11} \times \begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix} \times \begin{bmatrix} -2 \\ 3 \end{bmatrix} = \frac{1}{11} \times \begin{bmatrix} -5 \\ 12 \end{bmatrix} = \begin{bmatrix} -5/11 \\ 12/11 \end{bmatrix}

  6. Step 6: Hence, x=−5/11x = -5/11 and y=12/11y = 12/11.

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