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
$\left[\begin{array}{ccc}1& 1& 1\\ 2& 1& -1\\ 3& 2& k\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}l2\\ 3\\ 4\end{array}\right]$
$Ax=B$
Now by Rank method, form Augmented matrix
$[A;B]=\left[\begin{array}{cccc}1& 1& 1& 2\\ 2& 1& -1:3& \\ 3& 2& k& .4\end{array}\right]$
Applying elementary transformations to make above augment matrix in Echelon form,
$R_3\to R_3-3R_1\&R_2$
$\to R_2-2R_1$
$[A;B]=\left[\begin{array}{cccc}1& 1& 1& -2\\ 0& -1& -3& -1\\ 0& -1& k-3& -2\end{array}\right]$
Apply, $R_3\to R_3-R_2$
$[A;B]=\left[\begin{array}{ccc}1& 1& 1:2\\ 0& -1& -3;-1\\ 0& 0& k;-1\end{array}\right]$
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 $<\ne 0$
Hence, set $S$ can have an real values except $O,S=R-\{0\}$.