Matrices & Determinants
Matrices and Determinants
Allen Star Batch
Grade 12
Question:
For any real values of $X, Y, Z, L, M, N$ value of $\begin{vmatrix} \cos(X - L) & \cos(X - M) & \cos(X - N) \\ \cos(Y - L) & \cos(Y - M) & \cos(Y - N) \\ \cos(Z - L) & \cos(Z - M) & \cos(Z - N) \end{vmatrix} =$
0
1
$\cos X \cos Y \cos Z + \cos L \cos M \cos N$
$(\cos X - \cos Y)(\cos Y - \cos Z)(\cos Z - \cos X)(\cos L - \cos M)(\cos M - \cos N)(\cos N - \cos M)$
Step-by-Step Solution
Key Concept: The determinant can be evaluated using the product-to-sum formula: cos(A-B) = cos A cos B + sin A sin B. Substituting this expansion transforms the matrix into a form where the determinant becomes the difference of two identical column structures, yielding zero.
The determinant $\begin{vmatrix} \cos X & \sin X & 0 \\ \cos Y & \sin Y & 0 \\ \cos Z & \sin Z & 0 \end{vmatrix}$ equals zero because the third column is entirely zero, making all rows linearly dependent.
Correct Answer: 1