If $A,B$ and $C$ are the angles of a triangle and $\left| {\begin{array}{*{20}{c}} 1&1&1\\ {1 + \sin A}&{1 + \sin B}&{\sin C}\\ {\sin A + {{\sin }^2}A}&{\sin B + {{\sin }^2}B}&{\sin C + {{\sin }^2}C} \end{array}} \right| = 0$ then the triangle must be

JEE MATHEMATICS MCQ: MATRICES

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Analyzing +4 | -1

A,B

and

C

are the angles of a triangle and

\left| {\begin{array}{cccccccccccccccccccc} 1&1&1\\ {1 + \sin A}&{1 + \sin B}&{\sin C}\\ {\sin A + {{\sin }^2}A}&{\sin B + {{\sin }^2}B}&{\sin C + {{\sin }^2}C} \end{array}} \right| = 0

then the triangle must be

View answer and explanation

Correct option: B

Isosceles

The triangle must be isosceles because the determinant simplifies to a product of differences of sines of angles, which equals zero only when at least two angles have equal sines.

More Questions from MATRICES

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\left|\begin{array}{ccc}a & b-c & b+c \\ c+a & b & c-a \\ a-b & a+b & c\end{array}\right|=0

and

a, b, c

are real numbers then the line

a x+b y+c=0

passes through the fixed point is The number of solutions of the system of equations

{x_2} - {x_3} = 1,\; - {x_1} + 2{x_3} = - 2,\;{x_1} - 2,\;{x_1} - 2{x_2} = 3

is If the equations

2x + 3y + 1 = 0,3x + y - 2 = 0

and

ax + 2y - b = 0

are consistent, then The value of the determinant

\left|\begin{array}{ccc}1 & \left(\alpha^{2 x}-\alpha^{-2 x}\right)^2 & \left(\alpha^{2 x}+\alpha^{-2 x}\right)^2 \\ 1 & \left(\beta^{2 x}-\beta^{-2 x}\right)^2 & \left(\beta^{2 x}+\beta^{-2 x}\right)^2 \\ 1 & \left(\gamma^{2 x}-\gamma^{-2 x}\right)^2 & \left(\gamma^{2 x}+\gamma^{-2 x}\right)^2\end{array}\right|

, is If

\omega

is the cube root of unity, then

\left| {\begin{array}{cccccccccccccccccccc} 1&\omega &{{\omega ^2}}\\ \omega &{{\omega ^2}}&1\\ {{\omega ^2}}&1&\omega \end{array}} \right|

is equal to For any matrix

A

of order

2 \times 2

, if

A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]

, then

|A|

is equal to