Question

If \[A = \left[ {\begin{array}{*{20}{c}}{\cos \theta }&{ - \sin \theta }\\{\sin \theta }&{\cos \theta }\end{array}} \right]\], then which of the following statement is not correct?
A. A is an orthogonal matrix
B. \[A'\] is orthogonal matrix
C. Determinant A = 1
D. A is not invertible

Answer 1

Hint: To check whether the matrix is orthogonal or not, we will find whether the product of A and its transpose is an identity matrix. Then we find the determinant value of the given matrix to check whether the given matrix is invertible or not.

Formula used:
If A is an orthogonal matrix, then \[AA' = I\] where \[A'\] is a transpose matrix of A and I is an identity matrix.

Complete step by step solution:
Given matrix is \[A = \left[ {\begin{array}{*{20}{c}}{\cos \theta }&{ - \sin \theta }\\{\sin \theta }&{\cos \theta }\end{array}} \right]\].
To transpose a matrix, we will interchange rows into columns.
The transpose of A is
\[A' = \left[ {\begin{array}{*{20}{c}}{\cos \theta }&{\sin \theta }\\{ - \sin \theta }&{\cos \theta }\end{array}} \right]\]
Now we will compute the product of \[A\] and \[A'\]
\[AA' = \left[ {\begin{array}{*{20}{c}}{\cos \theta }&{ - \sin \theta }\\{\sin \theta }&{\cos \theta }\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\cos \theta }&{\sin \theta }\\{ - \sin \theta }&{\cos \theta }\end{array}} \right]\]
\[ \Rightarrow AA' = \left[ {\begin{array}{*{20}{c}}{{{\cos }^2}\theta + {{\sin }^2}\theta }&{\cos \theta \sin \theta - \cos \theta \sin \theta }\\{\cos \theta \sin \theta - \cos \theta \sin \theta }&{{{\sin }^2}\theta + {{\cos }^2}\theta }\end{array}} \right]\]
\[ \Rightarrow AA' = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\]
Thus, A is an orthogonal matrix.
We will check whether \[A'\] is an orthogonal matrix.
If \[A'{\left( {A'} \right)^\prime } = I\], then \[A'\] is an orthogonal matrix.
We know that \[{\left( {A'} \right)^\prime } = A\].
Putting \[{\left( {A'} \right)^\prime } = A\] in \[A'{\left( {A'} \right)^\prime }\]
\[A'{\left( {A'} \right)^\prime }\]
\[ = A'A\]
\[ = \left[ {\begin{array}{*{20}{c}}{\cos \theta }&{\sin \theta }\\{ - \sin \theta }&{\cos \theta }\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\cos \theta }&{ - \sin \theta }\\{\sin \theta }&{\cos \theta }\end{array}} \right]\]
\[ = \left[ {\begin{array}{*{20}{c}}{{{\cos }^2}\theta + {{\sin }^2}\theta }&{ - \cos \theta \sin \theta + \cos \theta \sin \theta }\\{ - \cos \theta \sin \theta + \cos \theta \sin \theta }&{{{\sin }^2}\theta + {{\cos }^2}\theta }\end{array}} \right]\]
\[ = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\]
\[ = I\]
Thus \[A'\] is an orthogonal matrix.
Now we calculate the determinant of A.
\[\left| A \right| = \left| {\begin{array}{*{20}{c}}{\cos \theta }&{ - \sin \theta }\\{\sin \theta }&{\cos \theta }\end{array}} \right|\]
\[ \Rightarrow \left| A \right| = {\cos ^2}\theta + {\sin ^2}\theta \]
\[ \Rightarrow \left| A \right| = 1 \ne 0\]
Since \[\left| A \right| \ne 0\], the matrix is an invertible matrix.
Hence option D is the correct option.

Note: Students often confused with invertible and non-invertible matrix. If the determinant value of a matrix is not equal to zero, then the matrix is invertible. If the determinant value of a matrix is equal to zero, then the matrix is noninvertible.