Let A = ( m n p q ) , d = | A | 0 and | A - d ( Adj A ) | = 0 . Then [2023]

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Solution

Q.
$Let A = (\begin{matrix} m & n \\ p & q \end{matrix}), d = | A | \neq 0 and$ $| A - d (Adj A) |$ $= 0 .$ Then [2023]

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

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

3 ${(1 + d)}^{2} = m^{2} + q^{2}$

4 ${(1 + d)}^{2} = {(m + q)}^{2}$

Ans.
(4)

We have, A = $[\begin{matrix} m & n \\ p & q \end{matrix}]$ , $d =$ $| A |$ $\neq 0 .$

$and$ $| A - d (Adj A) | = 0$ $∴$ $Adj (A) =$ $[\begin{matrix} q & - n \\ - p & m \end{matrix}]$

$\Rightarrow$ $| A - d (Adj A) |$ $=$ $| [\begin{matrix} m & n \\ p & q \end{matrix}] - d [\begin{matrix} q & - n \\ - p & m \end{matrix}] |$
$=$ $| \begin{matrix} m - d q & n + d n \\ p + d p & q - d m \end{matrix} |$

$\Rightarrow (m - d q) (q - d m) - (n + d n) (p + d p) = 0$

$\Rightarrow m q - d m^{2} - d q^{2} + d^{2} m q - n p - d p n - d n p - d^{2} n p = 0$

$\Rightarrow (m q - n p) - d (m^{2} + q^{2} + 2 p n) + d^{2} (m q - n p) = 0$

$\Rightarrow (1 + d^{2}) (m q - n p) - d (m^{2} + q^{2} + 2 p n) = 0$

$\Rightarrow (1 + d^{2}) (m q - n p) - d (m^{2} + q^{2} + 2 m q - 2 m q + 2 p n) = 0$

$\Rightarrow (1 + d^{2}) (m q - n p) - d {(m + q)}^{2} + 2 d (m q - p n) = 0$

$\Rightarrow (1 + d^{2} + 2 d) (m q - p n) - d {(m + q)}^{2} = 0$

$\Rightarrow (1 + d^{2}) (m q - p n) = d {(m + q)}^{2}$

$\Rightarrow (1 + d^{2}) = {(m + q)}^{2} ∴ d = m q - p n .$

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