Question

For what values of k, the system of linear equations

$x+y+z=2$

$2x+y-z=3$

$3x+2y+kz=4$ has a unique solution

Answer 1

Given system of linear equations is

$x+y+z=2$

$2x+y-z=3$

$3x+2y+kz=4$

It can be written in matrix form as

$AX=B$

Where, $A=\left[\begin{array}{ccc}1& 1& 1\\ 2& 1& -1\\ 3& 2& k\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ and $B=\left[\begin{array}{c}2\\ 3\\ 4\end{array}\right]$

The condition for the system of linear equations has a unique solution is $|A|\ne 0$.

Therefore,

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

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

$⟹k\ne 0$

Hence, for a unique solution, k should be a non-zero real number.