Question

If $(x,y,z) \neq (0,0,0)$ and $(\mathbf{i} + \mathbf{j} + 3\mathbf{k})x + (3\mathbf{i} - 3\mathbf{j} + \mathbf{k})y$

$+ ( - 4\mathbf{i} + 5\mathbf{j})z = \lambda(x\mathbf{i} + y\mathbf{j} + z\mathbf{k}),$ then the value of λ will be

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

Answer 1

Answer: (D)

0, – 1

Answer 2

Comparing the coefficients of $\mathbf{i},\mathbf{j}$ and $\mathbf{k},$ the corresponding equations are

$x + 3y - 4z = \lambda x$ or $(1 - \lambda)x + 3y - 4z = 0$ ......(i)

$x - (\lambda + 3)y + 5z = 0$ ......(ii)

$3x + y - \lambda z = 0$ .....(iii)

These equations (i), (ii) and (iii) have a non-trivial solution,

if $\left| \begin{matrix} (1 - \lambda) & 3 & - 4 \\ 1 & - (\lambda + 3) & 5 \\ 3 & 1 & - \lambda \end{matrix} \right| = 0 \Rightarrow \lambda = 0, - 1.$