Question

Let S be the set of all real values of k for which the system of linear equations
$x+y+z=2$
$2x+y-z=3$
$3x+2y+kz=4$
has a unique solution. Then S is :

• A. an empty set
• B. equal to {0}
• C. equal to R
• D. equal to R-{0}

Answer 1

Answer: (D)

Writing above system of equation in matrix form
$\begin{bmatrix}1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & k\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}=\begin{bmatrix}{l}2 \\ 3 \\ 4\end{bmatrix}$
$Ax = B$
Now by Rank method, form Augmented matrix
$[ A ; B ]=\begin{bmatrix}1 & 1 & 1 & 2 \\ 2 & 1 & -1: 3 \\ 3 & 2 & k & .4\end{bmatrix}$
Applying elementary transformations to make above augment matrix in Echelon form,
$R _{3} \rightarrow R _{3}-3 R _{1} \& R _{2} $
$\rightarrow R _{2}-2 R _{1}$
$[ A ; B ]=\begin{bmatrix}1 & 1 & 1 & -2 \\ 0 & -1 & -3 & -1 \\ 0 & -1 & k -3 & -2\end{bmatrix}$
Apply, $R _{3} \rightarrow R _{3}- R _{2}$
$[ A ; B ]=\begin{bmatrix}1 & 1 & 1: 2 \\ 0 & -1 & -3 ;-1 \\ 0 & 0 & k ;-1\end{bmatrix}$
From above matrix, clearly Rank of Augmented matrix will be 3 , but Rank of A matrix depends on value of $k$, i.e if $k=0$, Rank of A matrix will be 2 but Rank of Augmented matrix is 3. Hence, in this case No solution is their
So, for vnique solution $<\neq 0$
Hence, set $S$ can have an real values except $O, S=R-\{0\}$.