First slide

Applications of determinant

Question

The value of x for which the matrix $A = [\begin{array}{ccc} 2 / x & - 1 & 2 \\ 1 & x & 2 x^{2} \\ 1 & 1 / x & 2 \end{array}]$ is singular is

Easy

A

± 1
B

± 2
C

± 3
D

none of these

Solution

We have
$\begin{array}{l} | A | = (\frac{2}{x}) |\begin{array}{cc} x & 2 x^{2} \\ 1 / x & 2 \end{array}| + |\begin{array}{cc} 1 & 2 x^{2} \\ 1 & 2 \end{array}| + 2 |\begin{array}{cc} 1 & x \\ 1 & 1 / x \end{array}| \\ = \frac{2}{x} (0) + 2 - 2 x^{2} + 2 (\frac{1}{x} - x) \\ = \frac{2 x (1 - x^{2}) + 2 (1 - x^{2})}{x} = \frac{2 (x + 1)^{2} (1 - x)}{x} \end{array}$
Now, $| A | = 0 \Rightarrow x = \pm 1$ .

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