Question 43 of 58intermediate🔧 ApplyLong Answer4 marks

Solve system of linear equations, using matrix method:

5x+2y=35x + 2y = 3 3x+2y=53x + 2y = 5
Correct Answer

x = -1, y = 4

Exercise: EXERCISE 4.5 | Q: 10 | (Page 22)
For More Understanding

Explanation

The system of linear equations is solved using the matrix method. First, express the equations in matrix form AX=BAX = B. Then find the determinant of AA to check if the inverse exists.

Since ∣A∣≠0|A| \neq 0, the matrix is nonsingular. Calculate the inverse using the adjoint matrix, then multiply A−1A^{-1} by BB to obtain the solution matrix XX.

Solution Steps

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

    A=[5232],X=[xy],B=[35]A = \begin{bmatrix} 5 & 2 \\ 3 & 2 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \end{bmatrix}, \quad B = \begin{bmatrix} 3 \\ 5 \end{bmatrix}

  2. Step 2: Find determinant

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

    Hence, AA is nonsingular and its inverse exists.

  3. Step 3: Find adjoint

    adj(A)=[2−2−35]\text{adj}(A) = \begin{bmatrix} 2 & -2 \\ -3 & 5 \end{bmatrix}

  4. Step 4: Calculate inverse

    A−1=1∣A∣×adj(A)=14×[2−2−35]A^{-1} = \frac{1}{|A|} \times \text{adj}(A) = \frac{1}{4} \times \begin{bmatrix} 2 & -2 \\ -3 & 5 \end{bmatrix}

  5. Step 5: Find X=A−1BX = A^{-1}B

    X=14×[2−2−35]×[35]=14×[6−10−9+25]=14×[−416]=[−14]X = \frac{1}{4} \times \begin{bmatrix} 2 & -2 \\ -3 & 5 \end{bmatrix} \times \begin{bmatrix} 3 \\ 5 \end{bmatrix} = \frac{1}{4} \times \begin{bmatrix} 6-10 \\ -9+25 \end{bmatrix} = \frac{1}{4} \times \begin{bmatrix} -4 \\ 16 \end{bmatrix} = \begin{bmatrix} -1 \\ 4 \end{bmatrix}

  6. Step 6: Hence, x=−1x = -1, y=4y = 4

    ∴ x=−1,  y=4\boxed{x = -1, \; y = 4}

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