If \[A = \left[ {\begin{array}{*{20}{c}}{\cos x}&{\sin x}\\{ - \sin x}&{\cos x}\end{array}} \right]\], then find the value of \[A\left( {adj A} \right)\].
A. \[\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\]
B. \[\left[ {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right]\]
C. \[\left[ {\begin{array}{*{20}{c}}1&1\\0&0\end{array}} \right]\]
D. \[\left[ {\begin{array}{*{20}{c}}{ - 2}&0\\0&{ - 2}\end{array}} \right]\]
Answer
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Hint: First, find the adjoint matrix of the given matrix \[A\]. Substitute the values in the given required equation. Solve the equation by u\sing the matrix multiplication method and get the required answer.
Formula used:
The adjoint matrix of a \[2 \times 2\] matrix \[A = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\] is: \[adj A = \left[ {\begin{array}{*{20}{c}}d&{ - b}\\{ - c}&a\end{array}} \right]\]
\[\sin^{2}x + \cos^{2}x = 1\]
Complete step by step solution:
The given matrix is \[A = \left[ {\begin{array}{*{20}{c}}{\cos x}&{\sin x}\\{ - \sin x}&{\cos x}\end{array}} \right]\].
Let’s calculate the adjoint matrix of the given \[2 \times 2\] matrix \[A\].
Apply the rule for the adjoint matrix of a \[2 \times 2\] matrix.
We get,
\[adj A = \left[ {\begin{array}{*{20}{c}}{\cos x}&{ - \sin x}\\{\sin x}&{\cos x}\end{array}} \right]\]
Now substitute the above value in the expression \[A\left( {adj A} \right)\].
\[A\left( {adj A} \right) = \left[ {\begin{array}{*{20}{c}}{\cos x}&{\sin x}\\{ - \sin x}&{\cos x}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\cos x}&{ - \sin x}\\{\sin x}&{\cos x}\end{array}} \right]\]
Solve the right-hand side by u\sing the matrix multiplication method.
\[A\left( {adj A} \right) = \left[ {\begin{array}{*{20}{c}}{\cos^{2}x + \sin^{2}x}&{ - \sin x \cos x + \cos x \sin x}\\{ - \sin x \cos x + \cos x \sin x}&{\sin^{2}x + \cos^{2}x}\end{array}} \right]\]
Apply the standard trigonometric formula \[\sin^{2}x + \cos^{2}x = 1\].
\[A\left( {adj A} \right) = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\]
Hence the correct option is A.
Note: Students should keep in mind that the product of two matrices is defined if the number of columns of the first matrix is equal to the number of rows of the second matrix. And the number of rows of the resulting matrix is equal to the number of rows of the first matrix and the number of columns is equal to the number of columns of the second matrix.
Formula used:
The adjoint matrix of a \[2 \times 2\] matrix \[A = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\] is: \[adj A = \left[ {\begin{array}{*{20}{c}}d&{ - b}\\{ - c}&a\end{array}} \right]\]
\[\sin^{2}x + \cos^{2}x = 1\]
Complete step by step solution:
The given matrix is \[A = \left[ {\begin{array}{*{20}{c}}{\cos x}&{\sin x}\\{ - \sin x}&{\cos x}\end{array}} \right]\].
Let’s calculate the adjoint matrix of the given \[2 \times 2\] matrix \[A\].
Apply the rule for the adjoint matrix of a \[2 \times 2\] matrix.
We get,
\[adj A = \left[ {\begin{array}{*{20}{c}}{\cos x}&{ - \sin x}\\{\sin x}&{\cos x}\end{array}} \right]\]
Now substitute the above value in the expression \[A\left( {adj A} \right)\].
\[A\left( {adj A} \right) = \left[ {\begin{array}{*{20}{c}}{\cos x}&{\sin x}\\{ - \sin x}&{\cos x}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\cos x}&{ - \sin x}\\{\sin x}&{\cos x}\end{array}} \right]\]
Solve the right-hand side by u\sing the matrix multiplication method.
\[A\left( {adj A} \right) = \left[ {\begin{array}{*{20}{c}}{\cos^{2}x + \sin^{2}x}&{ - \sin x \cos x + \cos x \sin x}\\{ - \sin x \cos x + \cos x \sin x}&{\sin^{2}x + \cos^{2}x}\end{array}} \right]\]
Apply the standard trigonometric formula \[\sin^{2}x + \cos^{2}x = 1\].
\[A\left( {adj A} \right) = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\]
Hence the correct option is A.
Note: Students should keep in mind that the product of two matrices is defined if the number of columns of the first matrix is equal to the number of rows of the second matrix. And the number of rows of the resulting matrix is equal to the number of rows of the first matrix and the number of columns is equal to the number of columns of the second matrix.
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