Question

The non zero value of K such that the system of equations,
$x+Ky+3z=0,4x+3y+Kz=0,2x+y+2z=0$
has non-trival solution is

• A. 4
• B. 2.5
• C. 3.5
• D. None

Answer 1

The correct option is D None

The set of equations is written in the form of matrix

$\left[\begin{array}{ccc}1& K& 3\\ 4& 3& K\\ 2& 1& 2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

AX = B,

$C=[A,B]=\left[\begin{array}{ccccc}1& K& 3& |& 0\\ 4& 3& K& |& 0\\ 2& 1& 2& |& 0\end{array}\right]$

On interchanging first an third rows, we have,

$\left[\begin{array}{ccccc}2& 1& 2& |& 0\\ 4& 3& K& |& 0\\ 1& K& 3& |& 0\end{array}\right]$


Applyig $R_2\to R_2-2R_{1}and{R}_3\to R_3-\frac{1}{2}{R}_1$


$\left[\begin{array}{ccccc}2& 1& 2& |& 0\\ 0& 1& K-4& |& 0\\ 0& K-\frac{1}{2}& 2& |& 0\end{array}\right]$

Applying $R_3\to R_3-(K-\frac{1}{2})R_2$

$\left[\begin{array}{ccccc}2& 1& 2& |& 0\\ 0& 1& K-4& |& 0\\ 0& 0& 2-(K-\frac{1}{2})(K-4)& |& 0\end{array}\right]$

For a nontrivial solution

$\rho \left(A\right)=\rho \left(C\right)=2$

So, $2-(K-\frac{1}{2})(K-4)=0$
$2-K^2+\frac{1}{2}K+4K-2=0$

$-K^2+\frac{9}{2}K=0$
$K(-K+\frac{9}{2})=0$

$K=\frac{9}{2}$ and 0

Hence non-zero value of K = 4.5

Alternative Solutions:

Ax = 0 has non-trivial solution
|A| = 0

$\left|\begin{array}{ccc}1& k& 3\\ 4& 3& k\\ 2& 1& 2\end{array}\right|=0$

$=1(6-k)-k(8-2k)+3(4-6)=0$
$6-k-8k+2k^2-6=0$
$2k^2-9k=0$
$k(2k-9)=0$

$k=0,k=4.5$