Q112 / 170 · Manipal (MET) 2020 · Mathematics
The values of $\lambda$ such that $(x, y, z) \neq(0,0,0)$ and
$(\hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) x+(3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}) y+(-4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}) z$
$=\lambda(\hat{\mathbf{i}} x+\hat{\mathbf{j}} y+\hat{\mathbf{k}} \boldsymbol{z})$ are
- A$0,1$
- B$-1,1$
- Correct answerC$-1,0$
- D$-2,0$
Solution
Given,
$\begin{aligned} &(\hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) x+(3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}) y \\ &+(-4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}) z=\lambda(\hat{\mathbf{i}} x+\hat{\mathbf{j}} y+\hat{\mathbf{k}} z)\end{aligned}$
On equating the coefficients of $\hat{\mathbf{i}}, \hat{\mathbf{j}}$ and $\hat{\mathbf{k}}$ both sides, we have
$\begin{aligned} & x+3 y-4 z=\lambda x \\ & x-3 y+5 z=\lambda y\end{aligned}$
and $\quad 3 x+y+0=\lambda z$
Above three equations can be rewritten as
$\begin{array}{r}(1-\lambda) x+3 y-4 z=0 \\ x-(3+\lambda) y+5 z=0 \\ 3 x+y-\lambda z=0\end{array}$
This is homogeneous system of equations in three variables $x, y$ and $z$.
It is consistent and have non-zero solution. i.e. $(x, y, z) \neq(0,0,0)$, if determinant of coefficient matrix is zero.
$\Rightarrow \quad\left|\begin{array}{ccc}1-\therefore & 3 & -4 \\ 1 & -(3+\lambda) & 5 \\ 3 & 1 & -\lambda\end{array}\right|=0$
On expanding along first row, we have
$\begin{aligned}(1-\lambda)[\lambda(3+\lambda)-5]-3(-\lambda-15) & \\ & -4(1+9+3 \lambda)=0\end{aligned}$
$\begin{array}{cc}\Rightarrow & (1-\lambda)\left(\lambda^2+3 \lambda-5\right)+3 \lambda+45-40-12 \lambda=0 \\ \Rightarrow & \lambda^2+3 \lambda-5-\lambda^3-3 \lambda^2+5 \lambda-9 \lambda+5=0 \\ \Rightarrow & -\lambda^3-2 \lambda^2-\lambda=0 \\ \Rightarrow & \lambda\left(\lambda^2+2 \lambda+1\right)=0\end{array}$
$\begin{aligned} \Rightarrow & \lambda(\lambda+1)^2 & =0 \\ \Rightarrow & \lambda & =0,-1\end{aligned}$
Hence, the correct answer is option: 'C'.