Class 12th

Let A = $\left[\begin{array}{cc}\hfill 4\hfill & \hfill 5\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right]$

We write A = IA

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 4\hfill & \hfill 5\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 4-3\hfill & \hfill 5-4\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1-0\hfill & \hfill 0-1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A. (R1→R2 - R2)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 3-3(1)\hfill & \hfill 4-3(1)\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 0-3(1)\hfill & \hfill 1-3(-1)\hfill \end{array}\right]$ A. (R2→R2 - 3R1)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill -3\hfill & \hfill 4\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1-0\hfill & \hfill 1-1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1-(-3)\hfill & \hfill -1-4\hfill \\ \hfill -3\hfill & \hfill 4\hfill \end{array}\right]$ A. (R1→R1 - R2)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 4\hfill & \hfill -5\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right]$ A.

∴ A^-1 = $\left[\begin{array}{cc}\hfill 4\hfill & \hfill -5\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right]$

Let A = $\left[\begin{array}{cc}\hfill 3\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill 2\hfill \end{array}\right]$

We write A = IA.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 3\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill 2\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 2*3\hfill & \hfill 2*1\hfill \\ \hfill 5\hfill & \hfill 2\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 2*1\hfill & \hfill 2*0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A. (R1→2R1)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 6\hfill & \hfill 2\hfill \\ \hfill 5\hfill & \hfill 2\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 6-5\hfill & \hfill 2-2\hfill \\ \hfill 5\hfill & \hfill 2\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 2-0\hfill & \hfill 0-1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A. (R1→R1R2)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 5\hfill & \hfill 2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 5-5(1)\hfill & \hfill 2-5(0)\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 2\hfill & \hfill -1\hfill \\ \hfill 0-5(2)\hfill & \hfill 1-5(-1)\hfill \end{array}\right]$ A.(R2→R2 - 5R1)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 2\hfill & \hfill -1\hfill \\ \hfill -10\hfill & \hfill 6\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 2\hfill & \hfill -1\hfill \\ \hfill -5\hfill & \hfill 3\hfill \end{array}\right]$ A. $\left(R_2\to \frac{1}{2}{R}_2\right)$

∴ A^-1 = $\left[\begin{array}{cc}\hfill 2\hfill & \hfill -1\hfill \\ \hfill -5\hfill & \hfill 3\hfill \end{array}\right]$

Let A = $\left[\begin{array}{cc}\hfill 2\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 3\hfill \end{array}\right].$

We write, A = IA

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 2\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 3\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 2\hfill & \hfill 5\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$ A. (R1 $\leftrightarrow$ R2)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 2-2(1)\hfill & \hfill 5-2(3)\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1-2(0)\hfill & \hfill 0-2(1)\hfill \end{array}\right]$ A. (R2→R22R1)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill -1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -2\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]$ A. (R2→ (1)R2)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1-3(0)\hfill & \hfill 3-3(1)\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 0-3(-1)\hfill & \hfill 1-3(2)\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]$ A. (R1→R13R2)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 3\hfill & \hfill -5\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]$ A.

∴ A^-1 = $\left[\begin{array}{cc}\hfill 3\hfill & \hfill -5\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right].$

Let A = $\left[\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill 7\hfill & \hfill 4\hfill \end{array}\right].$

We write, A = IA,

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill 7\hfill & \hfill 4\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 7\hfill & \hfill 4\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$ A. (R1 $\leftrightarrow$ R2)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 7-3(2)\hfill & \hfill 4-3(1)\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 0-3(1)\hfill & \hfill 1-3(0)\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$ A.(R1→R1 3R2)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill -3\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 2-2(1)\hfill & \hfill 1-2(1)\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill -3\hfill & \hfill 1\hfill \\ \hfill 1-2(-3)\hfill & \hfill 0-2(1)\hfill \end{array}\right]$ A. (R2→R2 2R1)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill -1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill -3\hfill & \hfill 1\hfill \\ \hfill 7\hfill & \hfill -2\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill -3\hfill & \hfill 1\hfill \\ \hfill -7\hfill & \hfill 2\hfill \end{array}\right]$ A. (R2→(1)R2)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1-0\hfill & \hfill 1-1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill -3-(-7)\hfill & \hfill 1-2\hfill \\ \hfill -7\hfill & \hfill 2\hfill \end{array}\right]$ A. (R1→R1R2)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 4\hfill & \hfill -1\hfill \\ \hfill -7\hfill & \hfill 2\hfill \end{array}\right]$ A.

∴A^-1 = $\left[\begin{array}{cc}\hfill 4\hfill & \hfill -1\hfill \\ \hfill -7\hfill & \hfill 2\hfill \end{array}\right]$

Let A = $\left[\begin{array}{cc}\hfill 2\hfill & \hfill 3\hfill \\ \hfill 5\hfill & \hfill 7\hfill \end{array}\right]$

We write, A = 1A.

$\to \left[\begin{array}{cc}\hfill 2\hfill & \hfill 3\hfill \\ \hfill 5\hfill & \hfill 7\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A.

$\Rightarrow \left[\begin{array}{cc}\hfill 5\hfill & \hfill 7\hfill \\ \hfill 2\hfill & \hfill 3\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$ A((R₂ $\leftrightarrow$ R₁)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 5-2(2)\hfill & \hfill 7-2(3)\hfill \\ \hfill 2\hfill & \hfill 3\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 0-2(1)\hfill & \hfill 1-2(0)\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$ A. (R1→R1–2R2)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 3\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill -2\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 2-2(1)\hfill & \hfill 3-2(1)\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill -2\hfill & \hfill 1\hfill \\ \hfill 1-2(-2)\hfill & \hfill 0-2(1)\hfill \end{array}\right]$ A.(R2→R2–2R1)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill -2\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill -2\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1-0\hfill & \hfill 1-1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill -2-5\hfill & \hfill 1-(-2)\hfill \\ \hfill 5\hfill & \hfill -2\hfill \end{array}\right]$ A(R1→R1–R2)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill -7\hfill & \hfill 3\hfill \\ \hfill 5\hfill & \hfill -2\hfill \end{array}\right]$ A.

∴ A^-1 = $\left[\begin{array}{cc}\hfill -7\hfill & \hfill 3\hfill \\ \hfill 5\hfill & \hfill -2\hfill \end{array}\right]$

Let A = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 2\hfill & \hfill 7\hfill \end{array}\right].$

We write, A = IA.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 2\hfill & \hfill 7\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A.

$\Rightarrow \left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 2-2(1)\hfill & \hfill 7-2(3)\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0-2(1)\hfill & \hfill 1-0\hfill \end{array}\right]$ A (R₂→R₂–2R₁)

$\Rightarrow \left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill -2\hfill & \hfill 1\hfill \end{array}\right]$ A.

$\Rightarrow \left[\begin{array}{cc}\hfill 1-0\hfill & \hfill 3-3(1)\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1-3(-2)\hfill & \hfill 0-3(1)\hfill \\ \hfill -2\hfill & \hfill 1\hfill \end{array}\right]$ A(R₁→R₁–3R₂)

$\Rightarrow \left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 7\hfill & \hfill -3\hfill \\ \hfill -2\hfill & \hfill 1\hfill \end{array}\right]$ A.

∴A^-1 = $\left[\begin{array}{cc}\hfill 7\hfill & \hfill -3\hfill \\ \hfill -2\hfill & \hfill 1\hfill \end{array}\right].$

Let A = $\left[\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right].$

We write, A = IA.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 2-1\hfill & \hfill 1-1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1-0\hfill & \hfill 0-1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A (R₁→R₁–R₂)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 1-1\hfill & \hfill 1-0\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 0-1\hfill & \hfill 1-(-1)\hfill \end{array}\right]$ = A (R₂→R₂–R₁).

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]$ A.

∴ A^-1  = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right].$

Let A= $\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 3\hfill \end{array}\right].$

We write A = IA.

$\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 3\hfill \end{array}\right]={\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]}^{1}A.$

$\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 2-2(1)\hfill & \hfill 3-2(-1)\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 0-2(1)\hfill & \hfill 1-0\hfill \end{array}\right]$ (R₂ → R₂ –2R₁)

$\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 5\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill -2\hfill & \hfill 1\hfill \end{array}\right]A.$

$\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill \raisebox{1ex}{$0$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \end{array}\right]=[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \end{array}]A\left(R_2\to \frac{1}{5}{R}_2\right)$

$\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \end{array}\right].$

$\left[\begin{array}{cc}\hfill 1+0\hfill & \hfill -1+1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1-2/5\hfill & \hfill 0+1/5\hfill \\ \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \end{array}\right]A\left(R_1\to R_1+R_2\right)$

$\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=[\begin{array}{cc}\hfill \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \\ \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \end{array}]A.$

∴ A^-1  = $\left[\begin{array}{cc}\hfill \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \\ \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \end{array}\right]$

Given, A = $\left[\begin{array}{cc}\hfill cos\alpha \hfill & \hfill -sin\alpha \hfill \\ \hfill sin\alpha \hfill & \hfill cos\alpha \hfill \end{array}\right]$ . Then, A' = $\left[\begin{array}{cc}\hfill cos\alpha \hfill & \hfill sin\alpha \hfill \\ \hfill -sin\alpha \hfill & \hfill cos\alpha \hfill \end{array}\right]$

and A + A' = I.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill cos\alpha \hfill & \hfill -sin\alpha \hfill \\ \hfill sin\alpha \hfill & \hfill cos\alpha \hfill \end{array}\right]$ + $\left[\begin{array}{cc}\hfill cos\alpha \hfill & \hfill sin\alpha \hfill \\ \hfill -sin\alpha \hfill & \hfill cos\alpha \hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right].$

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 2cos\alpha +cos\alpha \hfill & \hfill -sin\alpha +sin\alpha \hfill \\ \hfill sin\alpha -sin\alpha \hfill & \hfill cos\alpha +cos\alpha \hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 2cot\alpha \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2cot\alpha \hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right].$

Equating the corresponding element of the matrix we get,

2 cos $\alpha$ = 1

$\Rightarrow$ cos $\alpha =\frac{1}{2}$

$\Rightarrow$ $\alpha$ = cos -$\frac{11}{2}$ = cos^-1$\left(cos\frac{}{3}\right)$$$

Option B is correct

Given A and B are symmetric matrices,

(E) Then, A' = A and B' = B.

Now, (AB - BA)' = (AB)'-  (BA)'

= B'A' - A'B'.

= BA - AB

(AB - BA)' = - (AB - BA)

AB - BA is a skew symmetric matrix

∴ Option A is correct.