The number of matrices A = [ cases a & b c & d cases ] , where a,...

Given, $A =$\left[$\begin{cases}a & b \\ c & d\end{cases}$\right]$and$A = {A}^{- 1}$</p><p>So,${A}^{2} = A \cdot {A}^{- 1} = I$</p><p>$⇒ \left[$\begin{cases}a & b \\ c & d\end{cases}$\right]\left[$\begin{cases}a & b \\ c & d\end{cases}$\right] = \left[$\begin{cases}1 & 0 \\ 0 & 1\end{cases}$\right]$</p><p>$⇒ \left[$\begin{cases}{a}^{2} + bc & ab + bd \\ ac + cd & bc + {d}^{2}\end{cases}$\right] = \left[$\begin{cases}1 & 0 \\ 0 & 1\end{cases}$\right]$</p><p>On comparing both side we get,</p><p>$∴ {a}^{2} + bc = 1 ⋯ $\left(1\right)$

$ab + bd = 0 ⋯ \left(2\right)$

$ac + cd = 0 ⋯ \left(3\right)$

$bc + {d}^{2} = 1 ⋯ \left(4\right)$

Now equation \left(1\right) - \left(4\right)$gives</p><p>${a}^{2} - {d}^{2} = 0$</p><p>$⇒ \left(a + d\right) = 0$or$a - d = 0$</p><p>Case-I</p><p>$a + d = 0 ⇒ \left(a, d\right) = \left(- 1, 1\right), \left(0, 0\right), \left(1, - 1\right)$</p><p>Assuming case \left(a\right)$⇒ $\left(a, d\right) = \left(- 1, 1\right)$

Now from equation \left(1\right)$</p><p>$1 + bc = 1 ⇒ bc = 0$</p><p>When$b = 0, c = 12$possibilities</p><p>When$c = 0, b = 12$possibilities</p><p>But \left(0, 0\right)$ is repeated

$∴ 2 \times 12 = 24$

So, total case will be $24 - 1$ (repeated) $= 23$ pairs.

case \left(b\right)$⇒ \left(a, d\right) = \left(1, - 1\right) ⇒ bc = 0 \to 23$ pairs

case \left(c\right) ⇒ \left(a, d\right) = \left(0, 0\right) ⇒ bc = 1$</p><p>$⇒ \left(b, c\right) = \left(1, 1\right)$and \left(- 1, - 1\right) \to 2$ pairs

Case-II

When $a = d$

from \left(2\right)$and \left(3\right)$

$a \ne 0$ then $b = c = 0$

${a}^{2} = 1$

$⇒ a = \pm 1 = d$

\left(a, d\right) = \left(1, 1\right), \left(- 1, - 1\right) \to 2$pairs</p><p>$∴$Total$= 23 + 23 + 2 + 2$</p><p>$= 50 pairs.