Question

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:

• A. 0
• B. -2
• C. 1
• D. -1

Answer 1

Answer: (D)

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