1. Tardigrade
  2. Question
  3. Mathematics
  4. The system of simultaneous equations kx + 2y - z = 1, (k - 1)y - 2z = 2 and (k + 2)z = 3 have a solution if k equals:

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The system of simultaneous equations kx + 2y - z = 1, (k - 1)y - 2z = 2 and (k + 2)z = 3 have a solution if k equals:

Determinants

Solution:

The matrix form of the given system of equation is:
$\begin{bmatrix}k&2&-1\\ 0&k-1&-2\\ 0&0&k+2\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}= \begin{bmatrix}1\\ 2\\ 3\end{bmatrix}$
Therefore, the coefficient matrix will have a solution iff it’s determinant $\neq$ 0
i.e. (k - 1) (k + 2) $\neq$ 0
i.e., k $\neq$ 0, 1, -2
Thus the given system of equation will have a solution only at k = - 1.