Question

$\left|\begin{array}{ccc}x+1& x+2& x+\lambda \\ x+2& x+3& x+\mu \\ x+3& x+4& x+\nu \end{array}\right|=0$, where $\lambda ,\mu ,\nu$ are in A.P. is

• A. an equation whose all roots are real
• B. an identity in x
• C. an equation with only one real root
• D. none of these

Answer 1

Answer: (B)

an identity in x

Given, $\lambda ,\mu ,\nu$ are in A.P.
$2\mu =\lambda +\nu$
Now, $\left|\begin{array}{ccc}x+1& x+2& x+\lambda \\ x+2& x+3& x+\mu \\ x+3& x+4& x+\nu \end{array}\right|=0$
$R_1\to R_1+R_3$
$\left|\begin{array}{ccc}2(x+2)& 2(x+3)& 2x+2\mu \\ x+2& x+3& x+\mu \\ x+3& x+4& x+\nu \end{array}\right|=0$
$2\left|\begin{array}{ccc}x+2& x+3& x+\mu \\ x+2& x+3& x+\mu \\ x+3& x+4& x+\nu \end{array}\right|=0$
Since, $R_1,R_2$ are identical . So the determinant is 0.
Thus, we can say that the determinant value is 0 irrespective of value of x . That means for all values of x.
Hence, it is an identity in x.