Matrices & Determinants
Matrices and Determinants
Allen Star Batch
Grade 12
Question:
If $A(\theta) = \begin{pmatrix} \sin \theta & i \cos \theta \\ i \cos \theta & \sin \theta \end{pmatrix}$, then which of the following is not true
$A(\theta)^{-1} = A(\pi - \theta)$
$A(\theta) + A(\pi + \theta)$ is a null matrix
$A(\theta)$ is invertible for all $\theta \in \mathbb{R}$
$A(\theta)^{-1} = A(-\theta)$
Step-by-Step Solution
Key Concept: Use the relationship between matrix transformations under angle shifts and inverse formulas for 2×2 matrices.
Since $|A(0)| = 1$, matrix $A(0)$ is invertible. Computing $A(\pi + \theta)$ and observing that it equals $-A(0)$, we find the adjugate is $\text{adj}A(0) = \begin{pmatrix}\sin 0 & -i\cos 0 \\ -i\cos 0 & \sin 0\end{pmatrix}$. Therefore $A(0)^{-1} = \text{adj}A(0) = A(\pi - 0)$.
Correct Answer: 1,2,3