Q.If $A = \begin{bmatrix}\cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha\end{bmatrix}$, and $A^{-1} = A'$, find value of $\alpha$.

The key idea is that $A$ is a rotation matrix, and for a rotation matrix, the inverse equals the transpose.

Step 1: For $A = \begin{bmatrix}\cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha\end{bmatrix}$, the transpose is

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

Step 2: The inverse of a $2\times2$ matrix $\begin{bmatrix}a & b\\ c & d\end{bmatrix}$ is $\frac{1}{ad-bc}\begin{bmatrix}d & -b\\ -c & a\end{bmatrix}$.

Here $\det(A) = \cos^2\alpha + \sin^2\alpha = 1$, so

$A^{-1} = \begin{bmatrix}\cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha\end{bmatrix}$.

Step 3: Given $A^{-1} = A'$, we have …

Step 1 — Write $A$ and its transpose.

Step 2 — Compute $A^{-1}$. The determinant is

so, using $\begin{bmatrix} a & b \\c & d \end{bmatrix}^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix} d & -b \\-c & a \end{bmatrix}$,

$A^{-1}=\begin{bmatrix} \cos\alpha & -\sin\alpha \\\sin\alpha & \cos\alpha \end{bmatrix}.$ …

Method: Using the orthogonality of a rotation matrix ($A^{-1}=A'$)

A rotation matrix $\begin{bmatrix}\cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha\end{bmatrix}$ is orthogonal, meaning its inverse equals its transpose. Questions built on the condition $A^{-1}=A'$ are really testing this property.

Steps

Step 1: Write down the transpose $A'$ by swapping rows and columns.

Step 2: Compute the inverse $A^{-1}$.

For a $2\times2$ matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix}$, $A^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$. For a rotation matrix $\det=\cos^2\alpha+\sin^2\alpha=1$, which simplifies this.

Step 3: Compare $A^{-1}$ and $A'$. …

Common Mistakes

Mistake 1: Expecting a single specific value of $\alpha$.

Why it's wrong: because every rotation matrix is orthogonal, $A^{-1}=A'$ holds identically; the correct answer is "all real $\alpha$", not one number. Correct approach: compute both sides and observe they always agree.

Mistake 2: Forgetting $\det A=1$ and dividing by the wrong quantity when finding $A^{-1}$.

Why it's wrong: an incorrect determinant scales the inverse wrongly. Correct approach: use $\cos^2\alpha+\sin^2\alpha=1$ so the $\tfrac{1}{\det}$ factor is just $1$. …