Question

What is the inverse of the matrix
$A = \left[ {\begin{array}{*{20}{l}}
  {\cos \theta }&{\sin \theta }&0 \\
  { - \sin \theta }&{\cos \theta }&0 \\
  0&0&1
\end{array}} \right]$
A. $\left[ {\begin{array}{*{20}{l}}
  {\cos \theta }&{ - \sin \theta }&0 \\
  {\sin \theta }&{\cos \theta }&0 \\
  0&0&1
\end{array}} \right]$
B. $\left[ {\begin{array}{*{20}{l}}
  {\cos \theta }&0&{ - \sin \theta } \\
  0&1&0 \\
  {\sin \theta }&0&{\cos \theta }
\end{array}} \right]$
C. $\left[ {\begin{array}{*{20}{l}}
  1&0&0 \\
  0&{\cos \theta }&{ - \sin \theta } \\
  0&{\sin \theta }&{\cos \theta }
\end{array}} \right]$
D. $\left[ {\begin{array}{*{20}{l}}
  {\cos \theta }&{\sin \theta }&0 \\
  { - \sin \theta }&{\cos \theta }&0 \\
  0&0&1
\end{array}} \right]$

Answer 1

Hint:
For finding the inverse of the matrix first calculate the matrix of the minors and then turn it into the matrix of the cofactors the find the adjoint of the matrix and multiply by $\dfrac {1}{{{\text{determinant}}}}$ that is ${A^{ - 1}} = \dfrac {{adj(A)}}{{\left| A \right|}}$

Complete step by step solution:
Here we are given the matrix A and we need to find the value of${A^{ - 1}}$. So we need to follow the following steps.
Step 1. First we need to find the matrix of the minor for each element of the matrix, ignore the elements in that row and the columns and find the determinant of the remaining values.
For example:
${a_{11}} = \left| {\begin{array}{*{20}{l}}
  {\cos \theta }&0 \\
  0&1
\end{array}} \right| = \cos \theta $
${a_{12}} = \left| {\begin{array}{*{20}{l}}
  { - \sin \theta }&0 \\
  0&1
\end{array}} \right| = - \sin \theta $
${a_{13}} = \left| {\begin{array}{*{20}{l}}
  { - \sin \theta }&{\cos \theta } \\
  0&0
\end{array}} \right| = 0$
${a_{21}} = \left| {\begin{array}{*{20}{l}}
  {\sin \theta }&0 \\
  0&1
\end{array}} \right| = \sin \theta $
${a_{22}} = \left| {\begin{array}{*{20}{l}}
  {\cos \theta }&0 \\
  0&1
\end{array}} \right| = \cos \theta $
${a_{23}} = \left| {\begin{array}{*{20}{l}}
  {\cos \theta }&{\sin \theta } \\
  0&1
\end{array}} \right| = 0$
${a_{33}} = \left| {\begin{array}{*{20}{l}}
  {\cos \theta }&{\sin \theta } \\
  { - \sin \theta }&{\cos \theta }
\end{array}} \right| = {\cos ^2}\theta + {\sin ^2}\theta = 1$
${a_{32}} = \left| {\begin{array}{*{20}{l}}
  {\cos \theta }&0 \\
  { - \sin \theta }&0
\end{array}} \right| = 0$
${a_{31}} = \left| {\begin{array}{*{20}{l}}
  {\sin \theta }&0 \\
  {\cos \theta }&0
\end{array}} \right| = 0$
Hence we can write the matrix of minors as $\left[ {\begin{array}{*{20}{l}}
  {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
  {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right] = $$\left[ {\begin{array}{*{20}{l}}
  {\cos \theta }&{ - \sin \theta }&0 \\
  {\sin \theta }&{\cos \theta }&0 \\
  0&0&1
\end{array}} \right]$

In the second step we need to find the matrix of the cofactors. Here as we get the matrix of the minors then we have to change the alternate signs to get the matrix of the cofactors so we get that
Matrix of cofactors as$\left[ {\begin{array}{*{20}{l}}
  {\cos \theta }&{\sin \theta }&0 \\
  { - \sin \theta }&{\cos \theta }&0 \\
  0&0&1
\end{array}} \right]$
Now we need to transpose this matrix of cofactors that is the rows and columns will be interchanged. So we will get the adjoint of the matrix A
$adj(A) = \left[ {\begin{array}{*{20}{l}}
  {\cos \theta }&{ - \sin \theta }&0 \\
  {\sin \theta }&{\cos \theta }&0 \\
  0&0&1
\end{array}} \right]$
Now finding the determinant of the matrix A
$\left| A \right| = \left| {\begin{array}{*{20}{l}}
  {\cos \theta }&{\sin \theta }&0 \\
  { - \sin \theta }&{\cos \theta }&0 \\
  0&0&1
\end{array}} \right| = \cos \theta (\cos \theta - 0) - \sin \theta ( - \sin \theta - 0) + 0$$ = $${\cos ^2}\theta + {\sin ^2}\theta = 1$
So now we know the formula that
${A^{ - 1}} = \dfrac {{adj(A)}}{{\left| A \right|}}$
Putting the suitable values we get that
${A^{ - 1}} = \dfrac {1}{{\left| A \right|}}\left[ {\begin{array}{*{20}{l}}
  {\cos \theta }&{ - \sin \theta }&0 \\
  {\sin \theta }&{\cos \theta }&0 \\
  0&0&1
\end{array}} \right]$
As we know the value of the determinant is 1 so putting it we get that
${A^{ - 1}} = \dfrac {1}{1}\left[ {\begin{array}{*{20}{l}}
  {\cos \theta }&{ - \sin \theta }&0 \\
  {\sin \theta }&{\cos \theta }&0 \\
  0&0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}
  {\cos \theta }&{ - \sin \theta }&0 \\
  {\sin \theta }&{\cos \theta }&0 \\
  0&0&1
\end{array}} \right]$

Note:
We all know that for any matrix $A$ we can write that $A.{A^{ - 1}} = I$ where $I$ is the identity matrix whose determinant is equal to one and ${A^{ - 1}}$ is the inverse of the matrix $A$.