Question

If $A = \begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}$, then $\text{adj}\ A =$

• A. $\begin{bmatrix} -\cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}$
• B. $\begin{bmatrix} \cos\theta & \sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}$
• C. $\begin{bmatrix} \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}$
• D. $\begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Hint:

Recognize that the upper-left $2 \times 2$ block of this matrix is a standard rotation matrix. Rotation matrices are orthogonal, meaning their inverse and adjoint are equal to their transpose.
To transpose a rotation matrix, keep the diagonal intact and simply flip the signs of the off-diagonal symmetric elements!

Answer 1

The Correct Option is C

Step 1: Look at the matrix.
$A$ is a rotation matrix built from $\cos\theta$ and $\sin\theta$. We need its adjoint, $\text{adj}\,A$.
Step 2: Recall a handy link.
For any square matrix, $A^{-1} = \dfrac{\text{adj}\,A}{|A|}$, so $\text{adj}\,A = |A|\,A^{-1}$. If we know $|A|$ and $A^{-1}$, we are done.
Step 3: Find the determinant.
Expanding using the bottom row, only the $1$ in the corner contributes: \[ |A| = 1\cdot(\cos^2\theta + \sin^2\theta) = 1 \]
Step 4: Use the rotation property.
This $A$ is an orthogonal matrix, so its inverse is its transpose: $A^{-1} = A^T$.
Step 5: Combine the facts.
Since $|A| = 1$, we get $\text{adj}\,A = 1 \cdot A^{-1} = A^T$.
Step 6: Write the transpose.
Swap rows and columns of $A$: \[ A^T = \begin{bmatrix} \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \] This is the adjoint. \[ \boxed{\begin{bmatrix} \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}} \]