Question

If we have the matrix A = \left[ {\begin{array}{*{20}{c}} 0&{ - \tan \dfrac{\alpha }{2}} \\\ {\tan \dfrac{\alpha }{2}}&0 \end{array}} \right] and $I$ the identity matrix of order $2$, show that I + A = \left( {I - A} \right)\left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{ - \sin \alpha } \\\ {\sin \alpha }&{\cos \alpha } \end{array}} \right]

Answer 1

Here we are asked to verify the given statement with the given matrix $A$. To verify any statement, we first need to find the value of the expression on the left-hand side and then find the value of the expression on the right-hand side. After finding the values we will find whether it satisfies the statement by comparing them.
Formula: Formulas that we need to know before solving this problem:
$\cos 2\theta = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$
$\sin 2\theta = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}$

Complete step-by-step solution:
It is given that the matrix A = \left[ {\begin{array}{*{20}{c}} 0&{ - \tan \dfrac{\alpha }{2}} \\\ {\tan \dfrac{\alpha }{2}}&0 \end{array}} \right]. We aim to verify the given equation I + A = \left( {I - A} \right)\left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{ - \sin \alpha } \\\ {\sin \alpha }&{\cos \alpha } \end{array}} \right] where, $I$ is the identity matrix of order two.
To verify the given equation, we first find the value of the expression on the left-hand side of the given equation. That is $I + A$.
We already have that A = \left[ {\begin{array}{*{20}{c}} 0&{ - \tan \dfrac{\alpha }{2}} \\\ {\tan \dfrac{\alpha }{2}}&0 \end{array}} \right] and $I$ is nothing but the identity matrix of order two, \left[ {\begin{array}{*{20}{c}} 1&0 \\\ 0&1 \end{array}} \right]
Now let us find the value of the left-hand side expression $I + A$.

1&0 \\\ 0&1 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 0&{ - \tan \dfrac{\alpha }{2}} \\\ {\tan \dfrac{\alpha }{2}}&0 \end{array}} \right]$$ On simplifying this we get $$I + A = \left[ {\begin{array}{*{20}{c}} 1&{ - \tan \dfrac{\alpha }{2}} \\\ {\tan \dfrac{\alpha }{2}}&1 \end{array}} \right]$$ Let us now find the value of the expression on the right-hand side. That is $$\left( {I - A} \right)\left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{ - \sin \alpha } \\\ {\sin \alpha }&{\cos \alpha } \end{array}} \right]$$ We already have that $$A = \left[ {\begin{array}{*{20}{c}} 0&{ - \tan \dfrac{\alpha }{2}} \\\ {\tan \dfrac{\alpha }{2}}&0 \end{array}} \right]$$ and $$I$$ is nothing but the identity matrix of order two, $$\left[ {\begin{array}{*{20}{c}} 1&0 \\\ 0&1 \end{array}} \right]$$ Now let us find the value of the right-hand side expression $$\left( {I - A} \right)\left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{ - \sin \alpha } \\\ {\sin \alpha }&{\cos \alpha } \end{array}} \right]$$. $$\left( {I - A} \right)\left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{ - \sin \alpha } \\\ {\sin \alpha }&{\cos \alpha } \end{array}} \right] = \left( {\left[ {\begin{array}{*{20}{c}} 1&0 \\\ 0&1 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 0&{ - \tan \dfrac{\alpha }{2}} \\\ {\tan \dfrac{\alpha }{2}}&0 \end{array}} \right]} \right) \times \left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{ - \sin \alpha } \\\ {\sin \alpha }&{\cos \alpha } \end{array}} \right]$$ On simplifying this we get $$\left( {I - A} \right)\left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{ - \sin \alpha } \\\ {\sin \alpha }&{\cos \alpha } \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&{\tan \dfrac{\alpha }{2}} \\\ { - \tan \dfrac{\alpha }{2}}&1 \end{array}} \right] \times \left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{ - \sin \alpha } \\\ {\sin \alpha }&{\cos \alpha } \end{array}} \right]$$ On multiplying the above, we get $$ = \left[ {\begin{array}{*{20}{c}} {\cos \alpha + \left( {\tan \dfrac{\alpha }{2}} \right)\left( {\sin \alpha } \right)}&{\left( { - \sin \alpha } \right) + \left( {\left( {\tan \dfrac{\alpha }{2}} \right)\left( {\cos \alpha } \right)} \right)} \\\ {\left( { - \tan \dfrac{\alpha }{2}} \right)\cos \alpha + \sin \alpha }&{\left( { - \tan \dfrac{\alpha }{2}} \right)\left( { - \sin \alpha } \right) + \cos \alpha } \end{array}} \right]$$ On further simplifying the above using the formulas $$\cos 2\theta = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$$ and $$\sin 2\theta = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}$$ by taking the angle as $$\dfrac{\alpha }{2}$$ we get $$ = \left[ {\begin{array}{*{20}{c}} {\left( {\dfrac{{1 - {{\tan }^2}\dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}} \right) + \tan \dfrac{\alpha }{2}\left( {\dfrac{{2\tan \dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}} \right)}&{\left( {\dfrac{{ - 2\tan \dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}} \right) + \tan \dfrac{\alpha }{2}\left( {\dfrac{{1 - {{\tan }^2}\dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}} \right)} \\\ { - \tan \dfrac{\alpha }{2}\left( {\dfrac{{1 - {{\tan }^2}\dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}} \right) + \left( {\dfrac{{2\tan \dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}} \right)}&{ - \tan \dfrac{\alpha }{2}\left( {\dfrac{{ - 2\tan \dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}} \right) + \left( {\dfrac{{1 - {{\tan }^2}\dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}} \right)} \end{array}} \right]$$ $$ = \left[ {\begin{array}{*{20}{c}} {\dfrac{{1 - {{\tan }^2}\dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}} + \dfrac{{2{{\tan }^2}\dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}}&{\dfrac{{ - 2\tan \dfrac{\alpha }{2} + \tan \dfrac{\alpha }{2} - {{\tan }^3}\dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}} \\\ {\dfrac{{ - \tan \dfrac{\alpha }{2}\left( {1 - {{\tan }^2}\dfrac{\alpha }{2}} \right)}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}} + \dfrac{{2\tan \dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}}&{\dfrac{{2{{\tan }^2}\dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}} + \dfrac{{1 - {{\tan }^2}\dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}} \end{array}} \right]$$ $$ = \left[ {\begin{array}{*{20}{c}} {\dfrac{{1 - {{\tan }^2}\dfrac{\alpha }{2} + 2{{\tan }^2}\dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}}&{\dfrac{{ - 2\tan \dfrac{\alpha }{2} + \tan \dfrac{\alpha }{2} - {{\tan }^3}\dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}} \\\ {\dfrac{{ - \tan \dfrac{\alpha }{2} + {{\tan }^3}\dfrac{\alpha }{2} + 2\tan \dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}}&{\dfrac{{2{{\tan }^2}\dfrac{\alpha }{2} + 1 - {{\tan }^2}\dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}} \end{array}} \right]$$ $$ = \left[ {\begin{array}{*{20}{c}} {\dfrac{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}}&{\dfrac{{ - \tan \dfrac{\alpha }{2}\left( {1 + {{\tan }^2}\dfrac{\alpha }{2}} \right)}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}} \\\ {\dfrac{{\tan \dfrac{\alpha }{2}\left( {1 + {{\tan }^2}\dfrac{\alpha }{2}} \right)}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}}&{\dfrac{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}} \end{array}} \right]$$$$\left( {I - A} \right)\left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{ - \sin \alpha } \\\ {\sin \alpha }&{\cos \alpha } \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&{ - \tan \dfrac{\alpha }{2}} \\\ {\tan \dfrac{\alpha }{2}}&1 \end{array}} \right]$$ Thus, we have found the value of the expression on the right-hand side. From both the values of the expressions we see that the left-hand side is equal to the right-hand side. That is $$I + A = \left( {I - A} \right)\left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{ - \sin \alpha } \\\ {\sin \alpha }&{\cos \alpha } \end{array}} \right]$$. Therefore, we verified the given statement. **Note:** Verifying the given expression or equation can be done in another way also that is, by finding the value of the expression on any one side, we can derive the expression of the other side in that. Addition and subtraction of matrices can be done by adding or subtracting the individual terms concerning their positions.