Question

There are n real values of $\lambda$ for which equations
(a- $\lambda$ )x+by+cz=0,
bx+(c-$\lambda$)y+az=0,
cx+ay+(b-$\lambda$)=0 are
simultaneously true, then n is equal to

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (B)

3

Given, $(a-\lambda )x+by+c=0$
$bx+(c-\lambda )y+az=0$
$cx+ay+(b-\lambda )z=0$
has solution. so,
$\left|\begin{array}{ccc}a-\lambda & b& c\\ b& c-\lambda & a\\ c& a& b-\lambda \end{array}\right|=0$
So this is a cubic in $\lambda$.
So, there are 3 real values of $\lambda$ can exist.
$\therefore n=3$