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Q6(i):
If (i) $A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$, then verify that $A' A = I$

Solution :

Step 1: Identifying the Transpose of Matrix $A$

Given the matrix $A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$, the transpose matrix $A'$ is obtained by interchanging the rows and columns [Definition of Transpose].

$A' = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$

Step 2: Setting up the Product $A'A$

We perform the matrix multiplication $A'A$ using the row-by-column method [Matrix Multiplication Rule: $(AB)_{ij} = \sum_{k=1}^{n} a_{ik}b_{kj}$].

$A'A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$

Step 3: Executing Matrix Multiplication

We compute each element of the resulting $2 \times 2$ matrix:

  • Element $(1,1) = (\cos \alpha)(\cos \alpha) + (-\sin \alpha)(-\sin \alpha) = \cos^2 \alpha + \sin^2 \alpha$
  • Element $(1,2) = (\cos \alpha)(\sin \alpha) + (-\sin \alpha)(\cos \alpha) = \cos \alpha \sin \alpha - \sin \alpha \cos \alpha$
  • Element $(2,1) = (\sin \alpha)(\cos \alpha) + (\cos \alpha)(-\sin \alpha) = \sin \alpha \cos \alpha - \cos \alpha \sin \alpha$
  • Element $(2,2) = (\sin \alpha)(\sin \alpha) + (\cos \alpha)(\cos \alpha) = \sin^2 \alpha + \cos^2 \alpha$

Step 4: Applying Trigonometric Identities

Utilizing the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ and the property of additive inverses:

$\cos^2 \alpha + \sin^2 \alpha = 1$ $\cos \alpha \sin \alpha - \sin \alpha \cos \alpha = 0$
$\sin \alpha \cos \alpha - \cos \alpha \sin \alpha = 0$ $\sin^2 \alpha + \cos^2 \alpha = 1$

Step 5: Final Matrix Verification

Substituting the computed values back into the matrix structure:

$A'A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Since $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ is the Identity matrix of order 2, denoted as $I$ [Definition of Identity Matrix].

A'A = 1 0 0 1 = I

Final Solution: $A'A = I$ is verified.


More Questions from Class 12 Mathematics Matrices EXERCISE 3.3


CBSE Solutions for Class 12 Mathematics Matrices


Chapters in CBSE - Class 12 Mathematics


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