If $a^{2}+b^{2}+c^{2}=-2$ and $f(x)=\left|\begin{array}{ccc}1+a^{2} x & \left(1+b^{2}\right) x & \left(1+c^{2}\right) x \\ \left(1+a^{2}\right) x & 1+b^{2} x & \left(1+c^{2}\right) x \\ \left(1+a^{2}\right) x & \left(1+b^{2}\right) x & 1+c^{2} x\end{array}\right|$ then $f(x)$ is a polynomial of degree

JEE Main · 2005 · 08 May 2005 · Offline · Determinants

a^{2} + b^{2} + c^{2} = - 2

and

f (x) = 1 + a^{2} x (1 + a^{2}) x (1 + a^{2}) x (1 + b^{2}) x 1 + b^{2} x (1 + b^{2}) x (1 + c^{2}) x (1 + c^{2}) x 1 + c^{2} x

then

f (x)

is a polynomial of degree

Answer: (D)

Apply the column operation

C_{1} \to C_{1} + C_{2} + C_{3}

. In each row the new first entry is

1 + (a^{2} + b^{2} + c^{2} + 2) x

, and since

a^{2} + b^{2} + c^{2} = - 2

this equals

1 + 0 \cdot x = 1

, so the first column becomes

(1, 1, 1)^{T}

. Subtracting consecutive rows with

R_{1} \to R_{1} - R_{2}

and

R_{2} \to R_{2} - R_{3}

and then expanding, the determinant reduces to

(x - 1)^{2}

. Thus

f (x)

is a polynomial of degree

2

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option