Question

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

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

Answer 1

Answer: (A)

$k\ne 0$

The given system of equations has a unique solution, if

$\left|\begin{array}{ccc}1& 1& 1\\ 2& 1& -1\\ 3& 2& k\end{array}\right|\ne 0$

$\Rightarrow 1(k+2)-1(2k+3)+1(4-3)\ne 0$

$\Rightarrow -k+0\ne 0$

$\Rightarrow k\ne 0$.