First slide

Introduction to Determinants

Question

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

Difficult

A

0
B

1
C

2
D

3

Solution

Operating $C_{1} \to C_{1} + C_{2} + C_{3}$ , we get
$f (x) = |\begin{array}{ccc} 1 + 2 x + (a^{2} + b^{2} + c^{2}) x & (1 + b^{2}) x & (1 + c^{2}) x \\ 1 + 2 x + (a^{2} + b^{2} + c^{2}) x & 1 + b^{2} x & (1 + c^{2}) x \\ 1 + 2 x + (a^{2} + b^{2} + c^{2}) x & (1 + b^{2}) x & 1 + c^{2} x \end{array}|$
$= |\begin{array}{ccc} 1 & (1 + b^{2}) x & (1 + c^{2}) x \\ 1 & 1 + b^{2} x & (1 + c^{2}) x \\ 1 & (1 + b^{2}) x & 1 + c^{2} x \end{array}| [∵ a^{2} + b^{2} + c^{2} = - 2]$
$= |\begin{array}{ccc} 1 & (1 + b^{2}) x & (1 + c^{2}) x \\ 0 & 1 - x & 0 \\ 0 & 0 & 1 - x \end{array}| \begin{array}{r} [Operating R_{2} \to R_{2} - R_{1} \\ and R_{3} \to R_{3} - R_{1}] \end{array}$
$\begin{array}{l} = (1) [(1 - x)^{2} - 0] \\ = (1 - x)^{2} \end{array}$
which is a polynomial of degree 2.

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