If and I is the identity matrix of order 2, show that .

Engineering

Mathematics

Multiplication of Matrices

Symmetric and Skew Symmetric Matrices

Question

If $A = ∣ \begin{array}{cc} 0 & - \tan \frac{α}{2} \\ \tan \frac{α}{2} & 0 \end{array} ∣$ and I is the identity matrix of order 2, show that $I + A = (I - A) [\begin{array}{cc} \cos α & - \sin α \\ \sin α & \cos α \end{array}]$ .

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Given ,

A = [\begin{array}{cc} 0 & - \tan \frac{α}{2} \\ \tan \frac{α}{2} & 0 \end{array}]

LHS : I + A = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}] + [\begin{array}{cc} 0 & - \tan \frac{α}{2} \\ \tan \frac{α}{2} & 0 \end{array}] = [\begin{array}{cc} 1 & - \tan \frac{α}{2} \\ \tan \frac{α}{2} & 1 \end{array}]

RHS : (I - A) [\begin{array}{cc} \cos α & - \sin α \\ \sin α & \cos α \end{array}] = [\begin{array}{cc} 1 & \tan \frac{α}{2} \\ - \tan \frac{α}{2} & 1 \end{array}] [\begin{array}{cc} \cos α & - \sin α \\ \sin α & \cos α \end{array}]

= [\begin{array}{cc} \cos α + \sin αtan \frac{α}{2} & - \sin α + \cos αtan \frac{α}{2} \\ \tan \frac{α}{2} \cos α + \sin α & \tan \frac{α}{2} \sin αcos α \end{array}]

= [\begin{array}{cc} \frac{\cos αcos \frac{α}{2} + \sin αsin \frac{α}{2}}{\cos \frac{α}{2}} & \frac{\cos α, \sin \frac{α}{2} - \sin αcos \frac{α}{2}}{\cos \frac{α}{2}} \\ \frac{\sin αcos \frac{α}{2} - \cos αsin \frac{α}{2}}{\cos \frac{α}{2}} & \frac{\sin \frac{α}{2} \sin α + \cos αcos \frac{α}{2}}{\cos \frac{α}{2}} \end{array}]

= [\begin{array}{cc} \cos (α - \frac{α}{2}) / \cos \frac{α}{2} & - \sin (α - \frac{α}{2}) / \cos \frac{α}{2} \\ \sin (α - \frac{α}{2}) / \cos \frac{α}{2} & \cos (α - \frac{α}{2}) / \cos \frac{α}{2} \end{array}]

∵ [\begin{array}{c} \cos (A - B) = \cos Acos B + \sin Asin B \\ \sin (A - B) = \sin Asin B + \cos Acos B \end{array}]

= [\begin{array}{cc} 1 & - \tan \frac{α}{2} \\ \tan \frac{α}{2} & 1 \end{array}]

As LHS = RHS hence proved

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