Class 12Mathematics3-4. Matrices and DeterminantsExpansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line
EasyIf $f(x) = \left| \begin{array}{ccc} x & x+1 & x+3 \\ x+2 & x+4 & x+7 \\ x+6 & x+9 & x+13 \end{array} \right|$, then $f(5) =$
- A-$15$
- B$10$
- C-$2$
- D$0$
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Similar Questions
If $\left| {\begin{array}{*{20}{c}}{ - {a^2}}&{ab}&{ac}\\{ab}&{ - {b^2}}&{bc}\\{ac}&{bc}&{ - {c^2}}\end{array}} \right| = K{a^2}{b^2}{c^2},$ then $K = $
Medium
View SolutionIf $\Delta=\left|\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right|=K(a-b)(b-c)(c-a)$, then $K=$
If $\left| \begin{array}{ccc} \cos 2x & \sin^2 x & \cos 4x \\ \sin^2 x & \cos 2x & \cos^2 x \\ \cos 4x & \cos^2 x & \cos 2x \end{array} \right| = a_0 + a_1 \sin x + a_2 \sin^2 x + \dots$,then $a_0$ is equal to:
The constant term in the expansion of $ \left|\begin{array}{ccc} 3x+1 & 2x-1 & x+2 \\ 5x-1 & 3x+2 & x+1 \\ 7x-2 & 3x+1 & 4x-1 \end{array}\right| $ is
DifficultKCET 2019
View SolutionThe area of a triangle with vertices $(K, 0), (4, 0), (0, 2)$ is $4$ square units. Find the value of $K$.
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