Question

Using matrices and determinants, find the value(s) of $k$ for which the pair of equations \[ 5x - ky = 2; \quad 7x - 5y = 3 \] has a unique solution.

Hint:

For a system of linear equations to have a unique solution, the determinant of the coefficient matrix must be non-zero. If the determinant is zero, the system either has no solution or infinitely many solutions.

Answer 1

The given pair of equations is: \[ 5x - ky = 2 \quad \text{(1)}, \] \[ 7x - 5y = 3 \quad \text{(2)}. \] We can represent this system of equations in matrix form as: \[ \begin{pmatrix} 5 & -k \\ 7 & -5 \end{pmatrix} \begin{pmatrix} x \ y \e\nd{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}. \] For the system to have a unique solution, the determinant of the coefficient matrix must be non-zero. The coefficient matrix is: \[ A = \begin{pmatrix} 5 & -k \\ 7 & -5 \end{pmatrix}. \] The determinant of $A$ is given by: \[ \text{det} = (5)(-5) - (7)(-k) = -25 + 7k. \] For a unique solution, we require: \[ \text{det} \neq 0, \] which gives: \[ -25 + 7k \neq 0, \] \[ 7k \neq 25, \] \[ k \neq \frac{25}{7}. \] Thus, the value of $k$ for which the system has a unique solution is any value of $k$ except $\frac{25}{7}$.