Question

The system of linear equations
$x+y+z=2$
$2x+y-z=3$
$3x+2y+kz=4$ has a unique solution if

• A. $k\ne 0$
• B. $-1<k<1$
• C. $-2<k<2$
• D. $k=0$

Answer 1

Answer: (A)

$k\ne 0$

Here, $D=\left|\begin{array}{ccc}1& 1& 1\\ 2& 1& -1\\ 3& 2& k\end{array}\right|$

On expanding the above determinant we get,
$D=1(k-2)-1(2k+3)+1(4-3)$

$D=-k$

For unique solution, $D\ne 0$ $\Rightarrow k\ne 0$