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Definition of a Determinant

Let A =[ ...

Question

Let $A = [\begin{matrix} 0 & - tan α tan α & 0 \end{matrix}]$ ,
$B (α) = [\begin{matrix} cos α & - sin α sin α & cos α \end{matrix}]$

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Solution

$A = [\begin{matrix} 0 & - tan α tan α & 0 \end{matrix}], B = [\begin{matrix} cos α & - sin α sin α & cos α \end{matrix}]$
$I - A = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] - [\begin{matrix} 0 & - tan α tan α & 0 \end{matrix}] = [\begin{matrix} 1 & tan α - tan α & 1 \end{matrix}]$ ...(1)
${(I - A)}^{- 1} = {cos}^{2} α [\begin{matrix} 1 & - tan α tan α & 1 \end{matrix}]$ ...(2)
$I + A = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] + [\begin{matrix} 0 & - tan α tan α & 0 \end{matrix}] = [\begin{matrix} 1 & - tan α tan α & 1 \end{matrix}]$ ...(3)
${(I + A)}^{- 1} = {cos}^{2} α [\begin{matrix} 1 & tan α - tan α & 1 \end{matrix}]$ ...(4)
A) From (2) and (3)
$(I + A) {(I - A)}^{- 1} = {cos}^{2} α [\begin{matrix} 1 & - tan α tan α & 1 \end{matrix}] [\begin{matrix} 1 & - tan α tan α & 1 \end{matrix}] = {cos}^{2} α [\begin{matrix} 1 - {tan}^{2} α & - 2 tan α 2 tan α & 1 - {tan}^{2} α \end{matrix}] = [\begin{matrix} cos 2 α & - sin 2 α sin 2 α & cos 2 α \end{matrix}] = B (2 α)$
B) From (1) and (4)
$(I - A) {(I + A)}^{- 1} = {cos}^{2} α [\begin{matrix} 1 & tan α - tan α & 1 \end{matrix}] [\begin{matrix} 1 & tan α - tan α & 1 \end{matrix}] = {cos}^{2} α [\begin{matrix} 1 - {tan}^{2} α & 2 tan α - 2 tan α & 1 - {tan}^{2} α \end{matrix}] = [\begin{matrix} cos 2 α & sin 2 α - sin 2 α & cos 2 α \end{matrix}] = [\begin{matrix} cos (- 2 α) & sin (- 2 α) sin (- 2 α) & cos (- 2 α) \end{matrix}] = B (- 2 α)$
C) $B {(α)}^{2} = [\begin{matrix} cos α & - sin α sin α & cos α \end{matrix}] [\begin{matrix} cos α & - sin α sin α & cos α \end{matrix}] = [\begin{matrix} {cos}^{2} α - {sin}^{2} α & - cos α sin α - cos α sin α cos α sin α + cos α sin α & {cos}^{2} α - {sin}^{2} α \end{matrix}]$
$= [\begin{matrix} cos 2 α & - sin 2 α sin 2 α & cos 2 α \end{matrix}] = B (2 α)$
D) $B {(α)}^{- 1} = 1. [\begin{matrix} cos α & + sin α - sin α & cos α \end{matrix}]$
$\Rightarrow B {(α)}^{- 2} = [\begin{matrix} cos α & + sin α - sin α & cos α \end{matrix}] [\begin{matrix} cos α & + sin α - sin α & cos α \end{matrix}] = [\begin{matrix} {cos}^{2} α - {sin}^{2} α & cos α sin α + cos α sin α - cos α sin α - cos α sin α & {cos}^{2} α - {sin}^{2} α \end{matrix}] = [\begin{matrix} cos 2 α & sin 2 α - sin 2 α & cos 2 α \end{matrix}] = [\begin{matrix} cos (- 2 α) & sin (- 2 α) sin (- 2 α) & cos (- 2 α) \end{matrix}] = B (- 2 α)$

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