Question

The values of $\lambda$ such that $(x, y, z) \neq(0,0,0)$ and $(\hat{i}+\hat{j}+3 \hat{k}) x+(3 \hat{i}-3 \hat{j}+\hat{k}) y+(-4 \hat{i}+5 \hat{j}) z$
$=\lambda\left(\hat{i}{x}+ \hat{j} y + \hat{k} z\right)$ are

• A. 0, 1
• B. -1, 1
• C. -1, 0
• D. -2, 0

Answer 1

Answer: (C)

Given,
$(\hat{i}+\hat{j}+3 \hat{k})_{x+}(3 \hat{i}-3 \hat{j}+\hat{k})_{y+}(-4 \hat{i}+5 \hat{j})_{z}=\lambda(\hat{i} x+\hat{j} y+\hat{k} z)$
On equating the coefficients of $\hat{i}, \hat{j}$ and $k$ both sides, we have
$x+3 y-4 z=\lambda x$
$x-3 y+5 z=\lambda y$
and $3 x+y+0=\lambda z$
Above three equations can be rewritten as
$(1-\lambda) x+3 y-4 z=0$
$x-(3+\lambda) y+5 z=0$
$3 x +y-\lambda z=0$
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.
$\Rightarrow \begin{vmatrix}1-\lambda & 3 & -4 \\ 1 & -(3+\lambda) & 5 \\ 3 & 1 & -\lambda\end{vmatrix}=0$
On expanding along first row, we have
$(1-\lambda)[\lambda(3+\lambda)-5]-3(-\lambda-15)-4(1+9+3 \lambda)=0$
$\Rightarrow(1-\lambda)\left(\lambda^{2}+3 \lambda-5\right)+3 \lambda+45-40-12 \lambda=0$
$\Rightarrow \lambda^{2}+3 \lambda-5-\lambda^{3}-3 \lambda^{2}+5 \lambda-9 \lambda+5=0$
$\Rightarrow-\lambda^{3}-2 \lambda^{2}-\lambda=0$
$\Rightarrow \lambda\left(\lambda^{2}+2 \lambda+1\right)=0$
$\Rightarrow \lambda(\lambda+1)^{2}=0$
$\Rightarrow \lambda=0,-1$