If the system of linear equations $x+y+z=2, 2x+y-z=3$ and $3x+2y+kz= 4$ has a unique solution, then:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $k≠0$
Given system of linear equations,
$x+y+z=2$
$2x+y-z=3$
$3x+2y+kz= 4$
$A=\begin{bmatrix}1&1&1\\2&1&-1\\3&2&k\end{bmatrix}$
$det(A)≠0$
$\begin{vmatrix}1&1&2\\2&1&-1\\3&2&k\end{vmatrix}=1(k+2)-1(2k+3)+1(4-3)$
$=k+2-2k-3+1$
$=-k$
$⇒-k≠0$
$⇒k≠0$