Question

If a²+b²+c²=-2 and

$f(x) = \begin{vmatrix} 1+a^2x & (1+b^2)x & (1+c^2)x \\ (1+a^2)x & 1+b^2x & (1+c^2)x \\ (1+a^2)x & (1+b^2)x & 1+c^2x \end{vmatrix}$ then f(x) is a

polynomial of degree

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

1

Answer 2

We are given

$$a^2 + b^2 + c^2 = -2$$

and

$$f(x)=\begin{vmatrix} 1+a^2x & (1+b^2)x & (1+c^2)x \\ (1+a^2)x & 1+b^2x & (1+c^2)x \\ (1+a^2)x & (1+b^2)x & 1+c^2x \end{vmatrix}.$$

Step 1. Restructure the Matrix:

Write each entry of the matrix as:

• Diagonal: $1+a^2x,\, 1+b^2x,\, 1+c^2x$.
• Off-diagonal: $(1+b^2)x,\,(1+c^2)x,$ etc.

Notice that for any entry from row $i$:

$$1+\alpha^2 x = 1 + (\alpha^2 x)$$

and its off-diagonals are just $1+\beta^2x$ but without the “1” (i.e. they are effectively $(1+\beta^2x) = [1 +\beta^2x]$ when compared with the diagonal which has an extra constant 1). Define:

$$u=1+a^2x,\quad v=1+b^2x,\quad w=1+c^2x.$$

Then the matrix becomes:

$$\begin{pmatrix} u & v-1 & w-1 \\ u-1 & v & w-1 \\ u-1 & v-1 & w \end{pmatrix}.$$

Step 2. Compute the Determinant:

A direct expansion (which exploits symmetry) leads to:

$$f(x)= u(v+w-1) - (u-1)[(v-1)+(w-1)].$$

Observe:

$$(v-1)+(w-1)= v+w-2.$$

So,

$$f(x)= u(v+w-1) - (u-1)(v+w-2).$$

Substitute back $u=1+a^2x$ and note that

$$v+w = (1+b^2x)+(1+c^2x)= 2+(b^2+c^2)x.$$

Thus,

$$v+w-1 = 1+(b^2+c^2)x,\quad \text{and} \quad v+w-2 = (b^2+c^2)x.$$

Also, $u-1= a^2x$.

Now,

$$f(x)= (1+a^2x)[1+(b^2+c^2)x] - a^2x\cdot (b^2+c^2)x.$$

Expanding:

$$(1+a^2x)[1+(b^2+c^2)x] = 1 + (b^2+c^2)x + a^2x + a^2(b^2+c^2)x^2.$$

and

$$a^2x\cdot (b^2+c^2)x = a^2(b^2+c^2)x^2.$$

The quadratic terms cancel:

$$f(x)= 1 + (a^2+b^2+c^2)x.$$

Given $a^2+b^2+c^2=-2$, we have:

$$f(x)= 1-2x.$$

Thus, $f(x)$ is a linear polynomial (degree $1$).