Matrices & Determinants
Matrices And Determinants
nta_abhyas_2025
Grade 12
Question:
If $A$, $B$, $C$ are the angles of a triangle and $\begin{vmatrix} 1+\sin A & 1+\sin B & 1+\sin C \\ \sin A + \sin^2 A & \sin B + \sin^2 B & \sin C + \sin^2 C \\ 1+\sin A & 1+\sin B & 1+\sin C \end{vmatrix} = 0$, then triangle $ABC$ is
(A) right angled isosceles
None
(C) equilateral
(D) scalene
Step-by-Step Solution
Key Concept: Using row operations on determinants and the sine rule relating angles to sides in a triangle.
Expanding the determinant $R_1 \to R_1 - R_3$ and simplifying the third row. The determinant becomes $(\sin A - \sin B) \times (\sin B - \sin C) \times (\sin C - \sin A) = 0$. Therefore, either $\sin A = \sin B$, $\sin B = \sin C$, or $\sin C = \sin A$. By the sine rule, this implies the corresponding sides are equal, making the triangle isosceles.
Correct Answer: 1