Question

If $A = \begin{bmatrix} \cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix}$, where $A_{21}, A_{22}, A_{23}$ are cofactors of $a_{21}, a_{22}, a_{23}$ respectively, then the value of $a_{21}\text{A}_{21} + \text{a}_{22}\text{A}_{22} + \text{a}_{23}\text{A}_{23} =$

• A. 1
• B. -1
• C. 0
• D. 2

Hint:

Row $\times$ cofactors = determinant.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The given expression $a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23}$ is mathematically identical to the expansion formula for the determinant of matrix $A$ along its second row. Therefore, solving this problem simply requires us to calculate the determinant of the entire matrix $A$ by any convenient method. Step 2: Key Formula or Approach:
Definition of determinant by cofactor expansion along row $i$: $|A| = \sum_{j} a_{ij} A_{ij}$. Here, with $i=2$, it's exactly the determinant $|A|$. Calculate $|A|$ by expanding along the most convenient row or column, which is the 3rd row (or 3rd column) since it contains two zeros. Step 3: Detailed Explanation:
The expression is defined as the determinant of $A$: \[ \text{Value} = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23} = |A| \] Let's find the determinant of matrix $A$: \[ |A| = \begin{vmatrix} \cos \theta & \sin \theta & 0
-\sin \theta & \cos \theta & 0
0 & 0 & 1 \end{vmatrix} \] Expanding along the third row is the most efficient method due to the zeros: \[ |A| = 0 \cdot (\dots) - 0 \cdot (\dots) + 1 \cdot \begin{vmatrix} \cos \theta & \sin \theta
-\sin \theta & \cos \theta \end{vmatrix} \] Calculate the $2\times2$ determinant: \[ |A| = 1 \cdot [(\cos \theta)(\cos \theta) - (\sin \theta)(-\sin \theta)] \] \[ |A| = \cos^2 \theta - (-\sin^2 \theta) \] \[ |A| = \cos^2 \theta + \sin^2 \theta \] Applying the fundamental trigonometric identity: \[ |A| = 1 \] Step 4: Final Answer:
The value of the expression is 1.