Question

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 1

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:

{\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: $${A^{ - 1}} = \dfrac{{adjA}}{{|A|}}$$