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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants4 Marks
Question
Find the real values of $\lambda$ for which the following system of linear equations has non-trivial solutions. Also, find the non-trivial solutions:
$2\lambda\text{x}-2\text{y}+3\text{z}=0,$
$\text{x}+\lambda\text{y}+2\text{z}=0,$
$2\text{x}+0\text{y}+\lambda\text{z}=0$
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Answer

The given system of equations can be written as,
$2\lambda\text{x}-2\text{y}+3\text{z}=0$
$\text{x}+\lambda\text{y}+2\text{z}=0$
$2\text{x}+0\text{y}+\lambda\text{z}=0$
The given system of equations will have non-trivial solutions if D = 0
 $\Rightarrow\begin{vmatrix}2\lambda&-2&3\\1&\lambda&2\\2&0&\lambda\end{vmatrix}=0$
$\Rightarrow 2\lambda(\lambda^2)+2(\lambda-4)+3(-2\lambda)=0$
$\Rightarrow 2\lambda^3 - 4\lambda - 8 = 0$
$\Rightarrow \lambda = 2$
So, the given system of equations will have non-trivial solutions if $\lambda = 2$
Now, we shall find solutions for $\lambda = 2$
Replacing z by k in the first two equations, we get
$2\lambda\text{x}-2\text{y}=-3\text{k}$
$\text{x}+\lambda\text{y}=-2\text{k}$
$\text{x}=\frac{\begin{vmatrix}-3\text{k}&-2\\-2\text{k}&\lambda\end{vmatrix}}{\begin{vmatrix}2\lambda&-2\\1&\lambda\end{vmatrix}}=\frac{-3\text{k}\lambda-4\text{k}}{2\lambda^2+2} $
$=\frac{-3\text{k}(2)-4\text{k}}{2(2)^2+2}=\frac{-6\text{k}-4\text{k}}{10}=-\text{k}$
$\text{y}=\frac{\begin{vmatrix}2\lambda&-3\text{k}\\1&-2\text{k}\end{vmatrix}}{\begin{vmatrix}2\lambda&-2\\1&\lambda\end{vmatrix}}=\frac{-4\text{k}\lambda+3\text{k}}{2\lambda^2+2}$
$=\frac{-4\text{k}(2)+3\text{k}}{2(2)^2+2}=\frac{-5\text{k}}{10}=\frac{-\text{k}}{2} $
Substituting these value of x and y in the third equation, we get
$\text{L.H.S}= 2(-\text{k})+0\Big(-\frac{\text{k}}{2}\Big)+2\text{k}$
$=0=\text{R.H.S}$
Thus,
$\lambda=2, \text{x}=-\text{k},\text{y}=-\frac{\text{k}}{2}$ and $\text{z}=\text{k}\ [\text{k}\in\text{R}]$

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