If (i) A=\(\begin{bmatrix} \cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha \end{bmatrix}\),then verify that A'A=I
(ii) A= \(\begin{bmatrix} \sin\alpha & \cos\alpha\\ -\cos \alpha & \sin\alpha \end{bmatrix}\),then verify that A'A=I
(ii) A= \(\begin{bmatrix} \sin\alpha & \cos\alpha\\ -\cos \alpha & \sin\alpha \end{bmatrix}\),then verify that A'A=I
Solution and Explanation
(i) \(A=\) \(\begin{bmatrix} \cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha \end{bmatrix}\)
\(\therefore A'=\) \(\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}\)
A'A= \(\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}\) \(\begin{bmatrix} \cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha \end{bmatrix}\)
= \(\begin{bmatrix} (\cos\alpha) (cos\alpha) + (- \sin\alpha)( -\sin\alpha) & (\cos\alpha)(\sin\alpha)+(-\sin\alpha)(\cos\alpha)\\ (\sin\alpha)(\cos\alpha)+(\cos\alpha)(-\sin\alpha) & (\sin\alpha)(\sin\alpha)+(\cos\alpha)(\cos\alpha) \end{bmatrix}\)
= \(\begin{bmatrix} \cos^2α+\sin^2α & \sinα\cosα-\sinα\cosα\\ \ sinα\cosα-\sinα\cosα & \sin^2α+\cos^2α \end{bmatrix}\)
= \(\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}= I\)
Hence we verified that: A'A=I
(ii) \(\begin{bmatrix} \sin\alpha & \cos\alpha\\ -\cos\alpha & \sin\alpha \end{bmatrix}\)
so A'= \(\begin{bmatrix} \sin\alpha & -\cos\alpha\\ \cos\alpha & \sin\alpha \end{bmatrix}\)
A'A= \(\begin{bmatrix} \sin\alpha & -\cos\alpha\\ \cos\alpha & \sin\alpha \end{bmatrix}\)\(\begin{bmatrix} \sin\alpha & \cos\alpha\\ -\cos\alpha & \sin\alpha \end{bmatrix}\)
= \(\begin{bmatrix} (\sin\alpha)(\sin\alpha)+(-\cos\alpha)(-\cos\alpha) & (\sin\alpha)(\cos\alpha)+(-\cos\alpha)(\sin\alpha)\\ (\cos\alpha)(\sin\alpha)+(\sin\alpha)(-\cos\alpha) & (\cos\alpha)(\cos\alpha)+(\sin\alpha)(\sin\alpha) \end{bmatrix}\)
= \(\begin{bmatrix} \sin^2α\cos^2α & \sinα\cosα-\sin\alpha\cos\alpha & \\ \sinα\cosα-\sinα\cosα & \cos^2α+\sin^2α \end{bmatrix}\)
= \(\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}= I\)
Hence we verified that: \(A'A=I\)
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If ai,j denotes the (i, j)th entry of M-1 , then a4,1 equals __________ (rounded off to two decimal places). - If \( A = \begin{bmatrix} 5 & 1 \\ -2 & 0 \end{bmatrix} \) and \( B^T = \begin{bmatrix} 1 & 10 \\ -2 & -1 \end{bmatrix} \), then the matrix \( AB \) is:
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Concepts Used:
Transpose of a Matrix
The matrix acquired by interchanging the rows and columns of the parent matrix is called the Transpose matrix. The transpose matrix is also defined as - “A Matrix which is formed by transposing all the rows of a given matrix into columns and vice-versa.”
The transpose matrix of A is represented by A’. It can be better understood by the given example:
Now, in Matrix A, the number of rows was 4 and the number of columns was 3 but, on taking the transpose of A we acquired A’ having 3 rows and 4 columns. Consequently, the vertical Matrix gets converted into Horizontal Matrix.
Hence, we can say if the matrix before transposing was a vertical matrix, it will be transposed to a horizontal matrix and vice-versa.
Read More: Transpose of a Matrix
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