Let A=mnpq,d=|A|≠0 and |A−d(Adj A)|=0 . Then

Let A=mnpq,d=|A|≠0 and |A−d(Adj A)|=0 . Then

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Let $A = (\begin{matrix} m & n \\ p & q \end{matrix}), d = | A | \neq 0 and | A - d (Adj A) | = 0$ . Then

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$1 + d^{2} = m^{2} + q^{2}$

${(1 + d)}^{2} = {(m + q)}^{2}$

${(1 + d)}^{2} = m^{2} + q^{2}$

$1 + d^{2} = {(m + q)}^{2}$

answer is B.

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Detailed Solution

Here d=mq-pn
$\begin{array}{l} | A - d (Adj A) | = 0 \\ \Rightarrow | (\begin{matrix} m & n \\ p & q \end{matrix}) - d (\begin{matrix} q & - n \\ - p & m \end{matrix}) | = 0 \\ \Rightarrow | (\begin{array}{cl} m - dq & n + dn \\ p + dp & q - dm \end{array}) | = 0 \\ \Rightarrow (m - aq) (q - dm) - pn {(1 + d)}^{2} = 0 \\ \Rightarrow mq - {dm}^{2} - {dq}^{2} + d^{2} qm = pn + {pnd}^{2} + 2 pnd \\ \Rightarrow d + d^{3} = d (m^{2} + q^{2} + 2 (mq - d)) \\ \Rightarrow d + d^{3} = d (m^{2} + q^{2} + 2 mq) - 2 d^{2} \\ \Rightarrow {(1+d)}^{2} = {(m+q)}^{2} \end{array}$