If A=0−tan⁡α/2tan⁡α/20 and I is a 2×2 unit matrix, then (I−A)cos⁡α−sin⁡αsin⁡αcos⁡α

A
$- I + A$
B
$I + A$
C
$- I - A$
D
None of these

Solution:

$Since I = [\begin{array}{ll} 1 0 \\ 0 1 \end{array}] and given A = [\begin{array}{cc} 0 & - \tan α / 2 \\ \tan α / 2 & 0 \end{array}]$
$∴ I - A = [\begin{array}{cc} 1 & \tan α / 2 \\ - \tan α / 2 & 1 \end{array}]$
$Now, (I - A) [\begin{array}{cc} \cos α & - \sin α \\ \sin α & \cos α \end{array}]$
$\begin{array}{l} = [\begin{array}{cc} 1 & \tan α / 2 \\ - \tan α / 2 & 1 \end{array}] [\begin{array}{cc} \cos α & - \sin α \\ \sin α & \cos α \end{array}] \\ = [\begin{matrix} \cos α + \sin αtan α / 2 & - \sin α + \tan α / 2 \cos α \\ - \tan α / 2 \cos α + sinα & \cos α + \sin αtan α / 2 \end{matrix}] \\ = [\begin{matrix} \frac{cosαcosα / 2 + \sin αsin α / 2}{cosα / 2} & \frac{\sin α / 2 \cos α - sinαcosα / 2}{cosα / 2} \\ \frac{- \sin α / 2 \cos α + sinαcosα / 2}{cosα / 2} & \frac{cosαcosα / 2 + \sin αsin α / 2}{cosα / 2} \end{matrix}] \\ = [\begin{matrix} \frac{\cos (α - α / 2)}{cosα / 2} & \frac{- \sin (α - α / 2)}{cosα / 2} \\ \frac{\sin (α - α / 2)}{cosα / 2} & \frac{\cos (α - α / 2)}{cosα / 2} \end{matrix}] = [\begin{array}{cc} 1 & - \tan α / 2 \\ \tan α / 2 & 1 \end{array}] \\ = I + A \end{array}$

$If A = [\begin{array}{cc} 0 & - \tan α / 2 \\ \tan α / 2 & 0 \end{array}] and I is a 2 \times 2 unit matrix, then$ $(I - A) [\begin{array}{cc} \cos α & - \sin α \\ \sin α & \cos α \end{array}]$

Solution:

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