If A = 1 0 0 a - 1 0 b c 1 is such that A 2 = I , then

If

A = [\begin{matrix} 1 & 0 & 0 \\ a & - 1 & 0 \\ b & c & 1 \end{matrix}]

is such that

A^{2} = I

, then

$b = \frac{a c}{2}$
$b = - \frac{a c}{2}$
$b = \frac{a + c}{2}$
$b = \sqrt{a c}$

Solution

Given:
$A = [\begin{matrix} 1 & 0 & 0 \\ a & - 1 & 0 \\ b & c & 1 \end{matrix}]$
And,
$A^{2} = I$
$\Rightarrow [\begin{matrix} 1 & 0 & 0 \\ a & - 1 & 0 \\ b & c & 1 \end{matrix}] [\begin{matrix} 1 & 0 & 0 \\ a & - 1 & 0 \\ b & c & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$
$\Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 b + a c & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$
So,
$2 b + a c = 0 \Rightarrow b = - \frac{a c}{2}$

Asked in: AP EAMCET 2022 (04 Jul Shift 2)

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