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 matrixB. $$A'$$ is orthogonal matrixC. Determinant A = 1D. A is not invertible

Question

Vedantu Content Team · Accepted Answer

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.

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 matrixB. \[A'\] is orthogonal matrixC. Determinant A = 1D. A is not invertible

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