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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) \[A.\left( {adjA} \right){\text{ }} = {\text{ }}\left| A \right|I{\text{ }} = {\text{ }}\left( {adj{\text{ }}A} \right).A\]

Answer Verified Verified
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 \[A\].

Complete step-by-step answer:
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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