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 provided system of equations is:\[5x - ky = 2 \quad \text{(1)},\]\[7x - 5y = 3 \quad \text{(2)}.\]In matrix form, this system is:\[\begin{pmatrix} 5 & -k \\ 7 & -5 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}.\]A unique solution exists if the determinant of the coefficient matrix is non-zero. The coefficient matrix is:\[A = \begin{pmatrix} 5 & -k \\ 7 & -5 \end{pmatrix}.\]The determinant of $A$ is calculated as:\[\text{det}(A) = (5)(-5) - (7)(-k) = -25 + 7k.\]For a unique solution, the determinant must not be zero:\[\text{det}(A) eq 0,\]meaning:\[-25 + 7k eq 0,\]which simplifies to:\[7k eq 25,\]and thus:\[k eq \frac{25}{7}.\]Therefore, the system has a unique solution for all values of $k$ except $\frac{25}{7}$.