Question 40 of 58intermediate🔧 ApplyLong Answer4 marks

Solve system of linear equations, using matrix method:

5x+2y=45x + 2y = 4 7x+3y=57x + 3y = 5
Correct Answer

x = 2, y = -3

Exercise: EXERCISE 4.5 | Q: 7 | (Page 22)
For More Understanding

Explanation

This question tests the matrix method for solving a system of linear equations. The student must express the system as AX=BAX = B, verify that AA is nonsingular by checking ∣A∣≠0|A| \neq 0, find the inverse A−1A^{-1}, and then compute X=A−1BX = A^{-1}B to obtain the solution.

Solution Steps

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

    A=[5273],X=[xy],B=[45].A = \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \end{bmatrix}, \quad B = \begin{bmatrix} 4 \\ 5 \end{bmatrix}.

  2. Step 2: Calculate the determinant:

    ∣A∣=5(3)−7(2)=15−14=1≠0|A| = 5(3) - 7(2) = 15 - 14 = 1 \neq 0

    Hence, AA is nonsingular and its inverse exists.

  3. Step 3: Find the inverse of AA. For a 2×22 \times 2 matrix,

    A−1=1∣A∣[3−2−75]=[3−2−75].A^{-1} = \frac{1}{|A|} \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}.

  4. Step 4: Calculate X=A−1BX = A^{-1}B:

    X=[3−2−75][45]=[3(4)+(−2)(5)(−7)(4)+5(5)]=[12−10−28+25]=[2−3].X = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix} \begin{bmatrix} 4 \\ 5 \end{bmatrix} = \begin{bmatrix} 3(4) + (-2)(5) \\ (-7)(4) + 5(5) \end{bmatrix} = \begin{bmatrix} 12 - 10 \\ -28 + 25 \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix}.

  5. Step 5: Therefore, ∴x=2andy=−3\therefore \quad x = 2 \quad \text{and} \quad y = -3.

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