Class 12Mathematics3 and 4 .Determinants and MatricesExpansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line
Easy$\left| \begin{matrix} 0 & a & -b \\ -a & 0 & c \\ b & -c & 0 \end{matrix} \right| = $
- A$ -2abc $
- B$ abc $
- C$ 0 $
- D$ a^2 + b^2 + c^2 $
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Similar Questions
Find the equation of the line joining $(1, 2)$ and $(3, 6)$ using determinants.
Easy
View Solution$\Delta = \left| \begin{array}{ccc} a & a+b & a+b+c \\ 3a & 4a+3b & 5a+4b+3c \\ 6a & 9a+6b & 11a+9b+6c \end{array} \right|$ where $a = i, b = \omega, c = \omega^2$,then $\Delta$ is equal to
Medium
View SolutionEvaluate the determinant: $\left|\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{array}\right|$
Easy
View Solution$\left|\begin{array}{rr}\sin 35^{\circ} & -\cos 35^{\circ} \\ \sin 55^{\circ} & \cos 55^{\circ}\end{array}\right|=$ . . . . . .
The number of real values of $t$ such that the system of homogeneous equations
$\begin{aligned}
t x+(t+1) y+(t-1) z &=0 \\
(t+1) x+t y+(t+2) z &=0 \\
(t-1) x+(t+2) y+t z &=0
\end{aligned}$
has non-trivial solutions is
$\begin{aligned}
t x+(t+1) y+(t-1) z &=0 \\
(t+1) x+t y+(t+2) z &=0 \\
(t-1) x+(t+2) y+t z &=0
\end{aligned}$
has non-trivial solutions is
DifficultTS EAMCET 2013
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