For what value of $k$, the following system have a unique solution? (where R is set of real numbers)
$x + y + z = 1$
$2x + 3y+4z = 3$
$x-y+kz=5$

$k ∈R∼\{-3\}$

The correct answer is Option (2) → $k ∈R∼\{-3\}$ **

The system has a unique solution when the determinant of coefficient matrix $\neq 0$.

Coefficient matrix:

$A=\begin{bmatrix}1 & 1 & 1 \\ 2 & 3 & 4 \\ 1 & -1 & k\end{bmatrix}$

$\det(A)=1\begin{vmatrix}3 & 4 \\ -1 & k\end{vmatrix} -1\begin{vmatrix}2 & 4 \\ 1 & k\end{vmatrix} +1\begin{vmatrix}2 & 3 \\ 1 & -1\end{vmatrix}$

$= 1(3k +4) - (2k -4) + (-2 -3)$

$= 3k + 4 - 2k + 4 - 5$

$= k + 3$

Unique solution requires:

$k + 3 \neq 0$

$k \neq -3$

The system has a unique solution for all real $k$ except $k = -3$.