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# If $matrix{\text{ }}A{\text{ }} = {\text{ }}F\left( \alpha \right){\text{ }} = \left( {\begin{array}{*{20}{c}} {\cos \alpha }&{ - \sin \alpha }&0 \\ {\sin \alpha }&{\cos \alpha }&0 \\ 0&0&1 \end{array}} \right)$then find ${A^{ - 1}}$and prove that: (i) ${A^{ - 1}}A = I_3$ (ii) ${A^{ - 1}} = F( - \alpha )$(iii) $A.\left( {adjA} \right){\text{ }} = {\text{ }}\left| A \right|I{\text{ }} = {\text{ }}\left( {adj{\text{ }}A} \right).A$  Hint:- We will first find the modulus of A then we will find the minors and cofactors and then put them into the formula for ${A^{ - 1}}$.
For (i) we will multiply the matrices ${A^{ - 1}}$ and $A$ and then prove it equal to RHS.
For (ii) we will replace $\alpha$ by $- \alpha$ in the given matrix and then prove it equal to ${A^{ - 1}}$.
For (iii) we will first pre multiply $A$ with $adjA$ and prove it equal to modulus of $A$ and then we will post multiply $A$ with $adjA$ and prove it equal to modulus of $A$.

We are given:
$matrix{\text{ }}A{\text{ }} = {\text{ }}F\left( \alpha \right){\text{ }} = \left( {\begin{array}{*{20}{c}} {\cos \alpha }&{ - \sin \alpha }&0 \\ {\sin \alpha }&{\cos \alpha }&0 \\ 0&0&1 \end{array}} \right)$
Calculating the modulus of A we get:
Expanding along row 1 we get:
$|A| = \cos \alpha \left[ {\left( 1 \right)\left( {\cos \alpha } \right) - 0} \right] - \sin \alpha \left[ {\left( 1 \right)\left( { - \sin \alpha } \right) - 0} \right] + 0 \\ |A| = \cos \alpha \left[ {\cos \alpha } \right] + \sin \alpha \left[ {\sin \alpha } \right] \\ |A| = {\cos ^2}\alpha + {\sin ^2}\alpha \\$
Now since we know that:
${\cos ^2}\alpha + {\sin ^2}\alpha = 1$
Hence,
$|A| = 1$
Now we will calculate the cofactors:
$F11 = + \left| {\begin{array}{*{20}{c}} {\cos \alpha }&0 \\ 0&1 \end{array}} \right| = \cos \alpha \\ F12 = - \left| {\begin{array}{*{20}{c}} {\sin \alpha }&0 \\ 0&1 \end{array}} \right| = - \sin \alpha \\ F13 = + \left| {\begin{array}{*{20}{c}} {\sin \alpha }&{\cos \alpha } \\ 0&0 \end{array}} \right| = 0 \\ F21 = - \left| {\begin{array}{*{20}{c}} { - \sin \alpha }&0 \\ 0&1 \end{array}} \right| = \sin \alpha \\ F22 = + \left| {\begin{array}{*{20}{c}} {\cos \alpha }&0 \\ 0&1 \end{array}} \right| = \cos \alpha \\ F23 = - \left| {\begin{array}{*{20}{c}} {\cos \alpha }&{ - \sin \alpha } \\ 0&0 \end{array}} \right| = 0 \\ F31 = + \left| {\begin{array}{*{20}{c}} { - \sin \alpha }&0 \\ {\cos \alpha }&0 \end{array}} \right| = 0 \\ F32 = - \left| {\begin{array}{*{20}{c}} {\cos \alpha }&0 \\ {\sin \alpha }&0 \end{array}} \right| = 0 \\ F33 = + \left| {\begin{array}{*{20}{c}} {\cos \alpha }&{ - \sin \alpha } \\ {\sin \alpha }&{\cos \alpha } \end{array}} \right| = 1 \\$
Now we will write the matrix formed by the adjoint of A.
$B = \left( {\begin{array}{*{20}{c}} {\cos \alpha }&{ - \sin \alpha }&0 \\ {\sin \alpha }&{\cos \alpha }&0 \\ 0&0&1 \end{array}} \right)$
Therefore,
$adjA = {B^T} = {\left( {\begin{array}{*{20}{c}} {\cos \alpha }&{ - \sin \alpha }&0 \\ {\sin \alpha }&{\cos \alpha }&0 \\ 0&0&1 \end{array}} \right)^T} \\ adjA = \left( {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha }&0 \\ { - \sin \alpha }&{\cos \alpha }&0 \\ 0&0&1 \end{array}} \right) \\$
Now we know that:
${A^{ - 1}} = \dfrac{{adjA}}{{|A|}}$
Putting the values we get:
${A^{ - 1}} = \dfrac{1}{1}\left( {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha }&0 \\ { - \sin \alpha }&{\cos \alpha }&0 \\ 0&0&1 \end{array}} \right) \\ {A^{ - 1}} = \left( {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha }&0 \\ { - \sin \alpha }&{\cos \alpha }&0 \\ 0&0&1 \end{array}} \right) \\$
(i) Considering LHS we get:
$LHS = {A^{ - 1}}A$
Solving the LHS we get:
$LHS = \left( {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha }&0 \\ { - \sin \alpha }&{\cos \alpha }&0 \\ 0&0&1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} {\cos \alpha }&{ - \sin \alpha }&0 \\ {\sin \alpha }&{\cos \alpha }&0 \\ 0&0&1 \end{array}} \right) \\ LHS = \left( {\begin{array}{*{20}{c}} {{{\cos }^2}\alpha + {{\sin }^2}\alpha + 0}&{ - \cos \alpha \sin \alpha + \cos \alpha \sin \alpha + 0}&{0 + 0 + 0} \\ { - \cos \alpha \sin \alpha + \cos \alpha \sin \alpha + 0}&{{{\cos }^2}\alpha + {{\sin }^2}\alpha + 0}&{0 + 0 + 0} \\ {0 + 0 + 0}&{0 + 0 + 0}&{0 + 0 + 1} \end{array}} \right) \\ LHS = \left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right) = I_3 \\$
Hence LHS=RHS therefore verified.
(ii) The matrix is given by:
${A^{ - 1}} = \left( {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha }&0 \\ { - \sin \alpha }&{\cos \alpha }&0 \\ 0&0&1 \end{array}} \right)$
Replacing $\alpha$ by $- \alpha$in the given matrix A we get:
$F\left( { - \alpha } \right){\text{ }} = \left( {\begin{array}{*{20}{c}} {\cos \left( { - \alpha } \right)}&{ - \sin \left( { - \alpha } \right)}&0 \\ {\sin \left( { - \alpha } \right)}&{\cos \left( { - \alpha } \right)}&0 \\ 0&0&1 \end{array}} \right) \\ F\left( { - \alpha } \right){\text{ }} = \left( {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha }&0 \\ { - \sin \alpha }&{\cos \alpha }&0 \\ 0&0&1 \end{array}} \right) = {A^{ - 1}} \\$
Hence verified.
(iii) Considering the LHS we get:-
$LHS = A.adjA$
Calculating the LHS we get:
$LHS = \left( {\begin{array}{*{20}{c}} {\cos \alpha }&{ - \sin \alpha }&0 \\ {\sin \alpha }&{\cos \alpha }&0 \\ 0&0&1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha }&0 \\ { - \sin \alpha }&{\cos \alpha }&0 \\ 0&0&1 \end{array}} \right) \\ LHS = \left( {\begin{array}{*{20}{c}} {{{\cos }^2}\alpha + {{\sin }^2}\alpha + 0}&{\cos \alpha \sin \alpha - \cos \alpha \sin \alpha + 0}&{0 + 0 + 0} \\ {\cos \alpha \sin \alpha - \cos \alpha \sin \alpha + 0}&{{{\cos }^2}\alpha + {{\sin }^2}\alpha + 0}&{0 + 0 + 0} \\ {0 + 0 + 0}&{0 + 0 + 0}&{0 + 0 + 1} \end{array}} \right) \\ LHS = \left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right) = 1 \times I \\ LHS = \left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right) = |A|I....................\left( 1 \right) \\$
Now considering RHS we get:-
$RHS = adjA.A$
Calculating the RHS we get:-
$RHS = \left( {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha }&0 \\ { - \sin \alpha }&{\cos \alpha }&0 \\ 0&0&1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} {\cos \alpha }&{ - \sin \alpha }&0 \\ {\sin \alpha }&{\cos \alpha }&0 \\ 0&0&1 \end{array}} \right) \\ RHS = \left( {\begin{array}{*{20}{c}} {{{\cos }^2}\alpha + {{\sin }^2}\alpha + 0}&{ - \cos \alpha \sin \alpha + \cos \alpha \sin \alpha + 0}&{0 + 0 + 0} \\ { - \cos \alpha \sin \alpha + \cos \alpha \sin \alpha + 0}&{{{\cos }^2}\alpha + {{\sin }^2}\alpha + 0}&{0 + 0 + 0} \\ {0 + 0 + 0}&{0 + 0 + 0}&{0 + 0 + 1} \end{array}} \right) \\ RHS = \left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right) = 1 \times I \\ RHS = \left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right) = |A|I....................\left( 2 \right) \\$
Hence from equation 1 and equation2 the third statement is proved.

Note: The inverse of matrix exist if and only if $|A| \ne 0$. The inverse of a matrix A is given by: ${A^{ - 1}} = \dfrac{{adjA}}{{|A|}}$
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