Answer
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Hint: In this problem, we have to prove the given condition with the given matrix. We have to find the transpose of the given matrix to multiply the given matrix and the transpose of that matrix to get the identity matrix, which will be equal to 1. We know that we can find the transpose of a matrix, by interchanging its rows and columns. We can multiply two matrices by matrix multiplication to prove that the multiplication of the matrix and its transpose equals 1.
Complete step by step answer:
We know that the given matrix is,
\[A=\left( \begin{matrix}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha \\
\end{matrix} \right)\]
Now we can find the transpose of the above matrix.
We know that we can find the transpose of a matrix, by interchanging its rows and columns.
\[A'=\left( \begin{matrix}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha \\
\end{matrix} \right)\]
We have to prove \[A'\times A=1\].
We can multiply the matrix A and A’, we get
\[\Rightarrow A\times A'=\left( \begin{matrix}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha \\
\end{matrix} \right)\left( \begin{matrix}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha \\
\end{matrix} \right)\]
Now we can multiply the above step using matrix multiplication, we get
\[\Rightarrow A\times A'=\left( \begin{matrix}
{{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha & \cos \alpha \sin \alpha -\cos \alpha \sin \alpha \\
\cos \alpha \sin \alpha -\cos \alpha \sin \alpha & {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \\
\end{matrix} \right)\]
We know that \[{{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha =1\], we can now substitute this in the above step, we get
\[\Rightarrow A\times A'=\left( \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right)=1\]
Therefore, \[A'-A=1\]is proved.
Note: Students make mistakes while multiplying two matrices, we should multiply the first row of the first matrix and the first column of the second matrix and add that, similarly we can do multiplication. We should also know that we can find the transpose of a matrix, by interchanging its rows and columns. We should also know some trigonometric properties and formulas to solve or prove these types of problems.
Complete step by step answer:
We know that the given matrix is,
\[A=\left( \begin{matrix}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha \\
\end{matrix} \right)\]
Now we can find the transpose of the above matrix.
We know that we can find the transpose of a matrix, by interchanging its rows and columns.
\[A'=\left( \begin{matrix}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha \\
\end{matrix} \right)\]
We have to prove \[A'\times A=1\].
We can multiply the matrix A and A’, we get
\[\Rightarrow A\times A'=\left( \begin{matrix}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha \\
\end{matrix} \right)\left( \begin{matrix}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha \\
\end{matrix} \right)\]
Now we can multiply the above step using matrix multiplication, we get
\[\Rightarrow A\times A'=\left( \begin{matrix}
{{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha & \cos \alpha \sin \alpha -\cos \alpha \sin \alpha \\
\cos \alpha \sin \alpha -\cos \alpha \sin \alpha & {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \\
\end{matrix} \right)\]
We know that \[{{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha =1\], we can now substitute this in the above step, we get
\[\Rightarrow A\times A'=\left( \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right)=1\]
Therefore, \[A'-A=1\]is proved.
Note: Students make mistakes while multiplying two matrices, we should multiply the first row of the first matrix and the first column of the second matrix and add that, similarly we can do multiplication. We should also know that we can find the transpose of a matrix, by interchanging its rows and columns. We should also know some trigonometric properties and formulas to solve or prove these types of problems.
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