Q.If

Concept: Orthogonal Matrix Verification — showing that a matrix multiplied by its transpose yields the identity.

(i) For $A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$, compute $A'$ (the transpose):

Now multiply $A'A$:

The $(1,1)$ entry: $\cos^2 \alpha + \sin^2 \alpha = 1$.

The $(1,2)$ entry: $\cos\alpha \sin\alpha - \sin\alpha \cos\alpha = 0$.

The $(2,1)$ entry: $\sin\alpha \cos\alpha - \cos\alpha \sin\alpha = 0$.

The $(2,2)$ entry: $\sin^2 \alpha + \cos^2 \alpha = 1$.

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

(ii) For $A = \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}$, the transpose is:

Then: …

Why this works: Orthogonal Matrix Verification

A matrix $A$ is called orthogonal if $A'A = I$ (or equivalently $AA' = I$). Geometrically, orthogonal matrices preserve lengths and angles — they represent rotations or reflections. The key condition is that the rows (or columns) of $A$ must be unit vectors that are perpendicular to each other.

For a $2 \times 2$ matrix, this means:

• Each row has length $1$: the sum of squares of its entries equals $1$.
• The dot product of the two rows equals $0$.

Both matrices given are classic rotation matrices (the first rotates by $-\alpha$, the second by $\alpha$ with a sign twist). So we expect $A'A = I$ to hold for all $\alpha$.

Part (i): $A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$

Step 1: Write down $A'$ (the transpose).

Transpose swaps rows and columns:

Step 2: Multiply $A'A$.

We compute the product:

Multiply entry by entry:

• Top-left: $(\cos \alpha)(\cos \alpha) + (-\sin \alpha)(-\sin \alpha) = \cos^2\alpha + \sin^2\alpha = 1$
• Top-right: $(\cos \alpha)(\sin \alpha) + (-\sin \alpha)(\cos \alpha) = \cos\alpha\sin\alpha - \sin\alpha\cos\alpha = 0$
• Bottom-left: $(\sin \alpha)(\cos \alpha) + (\cos \alpha)(-\sin \alpha) = \sin\alpha\cos\alpha - \cos\alpha\sin\alpha = 0$
• Bottom-right: $(\sin \alpha)(\sin \alpha) + (\cos \alpha)(\cos \alpha) = \sin^2\alpha + \cos^2\alpha = 1$

So:

Part (ii): $A = \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}$

Step 1: Write $A'$.

Step 2: Multiply $A'A$. …

Method: Verifying a matrix is orthogonal ($A'A=I$)

To check $A'A=I$, write the transpose $A'$, multiply $A'A$, and simplify each entry — for trigonometric (rotation) matrices the identity $\sin^2\alpha+\cos^2\alpha=1$ does the work.

Steps

Step 1: Write $A'$

Swap the rows and columns of $A$.

Step 2: Multiply $A'A$ entry-by-entry …

Common Mistakes

Mistake 1: Computing $AA'$ when the problem asks for $A'A$

Why it's wrong: the stated product is $A'A$; although both equal $I$ for an orthogonal matrix, following the wrong order invites sign errors mid-computation. Correct approach: multiply exactly in the order asked.

Mistake 2: Writing the transpose incorrectly …