First slide

Introduction to Determinants

Question

Suppose $P (x) = |\begin{array}{ccc} x & - 51 & - 71 \\ 51 & x & - 73 \\ 71 & 73 & x \end{array}|$ .Product of zeros of $P (x)$ is

Moderate

A

0
B

195
C

-195
D

–264333

Solution

Tip
Suppose $P (x) = x^{3} + a x^{2} + b x + c$ then product of zero of $P (x) is - c = - P (0)$
Now, $\begin{aligned} - c = - P (0) = - |\begin{array}{ccc} 0 & - 51 & - 71 \\ 51 & 0 & - 73 \\ 71 & 73 & 0 \end{array}| \\ = 0 \end{aligned}$
[ $∴$ Skew symmetric determinant of odd order]

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