Question

Let \( 2x + 5y - 1 = 0 \) and \( 3x + 2y - 7 = 0 \) represent the equations of two lines on which the ants are moving on the ground. Using matrix method, find a point common to the paths of the ants.

Hint:

Quick Tip: When solving a system of equations using the matrix method, always check if the determinant of the coefficient matrix is non-zero, ensuring that the system has a unique solution.

Answer 1

The given system of linear equations is: \[ 2x + 5y = 1 \quad \text{(1)} \] \[ 3x + 2y = 7 \quad \text{(2)} \] We can write this system in matrix form as: \[ \begin{pmatrix} 2 & 5 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \end{pmatrix} \] Let the matrix be \( A \), the column matrix of variables be \( \mathbf{x} \), and the constant matrix be \( \mathbf{b} \): \[ A = \begin{pmatrix} 2 & 5 \\ 3 & 2 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 1 \\ 7 \end{pmatrix} \] To solve for \( \mathbf{x} \), we find \( A^{-1} \). The inverse of matrix \( A \) is given by: \[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \] where \( \text{det}(A) = (2 \times 2) - (5 \times 3) = 4 - 15 = -11 \). So, \[ A^{-1} = \frac{1}{-11} \begin{pmatrix} 2 & -5 \\ -3 & 2 \end{pmatrix} = \begin{pmatrix} -\frac{2}{11} & \frac{5}{11} \\ \frac{3}{11} & -\frac{2}{11} \end{pmatrix} \] Now, multiply \( A^{-1} \) with \( \mathbf{b} \): \[ \mathbf{x} = A^{-1} \mathbf{b} = \begin{pmatrix} -\frac{2}{11} & \frac{5}{11} \\ \frac{3}{11} & -\frac{2}{11} \end{pmatrix} \begin{pmatrix} 1 \\ 7 \end{pmatrix} \] Performing the matrix multiplication: \[ \mathbf{x} = \begin{pmatrix} -\frac{2}{11} \times 1 + \frac{5}{11} \times 7 \\ \frac{3}{11} \times 1 + -\frac{2}{11} \times 7 \end{pmatrix} = \begin{pmatrix} \frac{33}{11} \\ -\frac{11}{11} \end{pmatrix} = \begin{pmatrix} 3 \\ -1 \end{pmatrix} \] Thus, the point common to the paths of the ants is \( (x, y) = (3, -1) \).