Question

If \[ A = \begin{bmatrix} \cos\theta & \sin\theta & 0
-\sin\theta & \cos\theta & 0
0 & 0 & 1 \end{bmatrix} \] and \( A_{11}, A_{12}, A_{13} \) are the cofactors of \( a_{11}, a_{12}, a_{13} \) respectively, then the value of \( a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13} \) is:

• A. \( -1 \)
• B. \( 1 \)
• C. \( 0 \)
• D. \( 2 \)

Hint:

The sum \( \sum a_{ij} A_{ij} \) along any row \( i \) of a matrix is equal to the determinant of the matrix. This is a result of cofactor (Laplace) expansion. Especially useful when dealing with orthogonal or rotation matrices.

Answer 1

The Correct Option is B

We are given a \( 3 \times 3 \) orthogonal matrix: \[ A = \begin{bmatrix} \cos\theta & \sin\theta & 0
-\sin\theta & \cos\theta & 0
0 & 0 & 1 \end{bmatrix} \] We are asked to evaluate the expression: \[ a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} \] This is the dot product of the first row of the matrix with the first row of cofactors, which is equivalent to the first row of \( A \cdot \text{adj}(A)^T \). However, more directly: \[ \sum_{j=1}^{3} a_{1j} A_{1j} = \text{cofactor expansion of determinant along the first row} \] Hence, \[ a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} = \det(A) \] Now, \( A \) is a rotation matrix (since the upper-left \( 2 \times 2 \) submatrix is a 2D rotation matrix), and its determinant is: \[ \det(A) = \det\left( \begin{bmatrix} \cos\theta & \sin\theta & 0
-\sin\theta & \cos\theta & 0
0 & 0 & 1 \end{bmatrix} \right) = \det\left( \begin{bmatrix} \cos\theta & \sin\theta
-\sin\theta & \cos\theta \end{bmatrix} \right) \cdot \det(1) = (\cos^2\theta + \sin^2\theta)(1) = 1 \] Therefore, \[ a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} = 1 \]