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
EasyThe area of a triangle whose vertices are $(2, -6)$,$(5, 4)$,and $(k, 4)$ is $35$ sq. units. Then,the value of $k$ is . . . . . . .
- A$12, -2$
- B$-2$
- C$-12, -2$
- D$12$
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If $f(x) = \left|\begin{array}{ccc} 1 & x & x+1 \\ 2x & x(x-1) & x(x+1) \\ 3x(x-1) & x(x-1)(x-2) & (x-1)x(x+1) \end{array}\right|$, then $f(2012)$ is equal to:
If $A = \begin{bmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{bmatrix}$,then for all $\theta \in \left( \frac{3\pi}{4}, \frac{5\pi}{4} \right)$,$\det(A)$ lies in the interval:
DifficultJEE MAIN 2019
View SolutionIf $\left| {\begin{array}{*{20}{c}}{{x^2} + x}&{x + 1}&{x - 2}\\ {2{x^2} + 3x - 1}&{3x}&{3x - 3}\\ {{x^2} + 2x + 3}&{2x - 1}&{2x - 1}\end{array}} \right| = Ax - 12$,then the value of $A$ is
MediumIIT 1982
View SolutionIf $A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$, then $\det(A)$ is equal to
If ${a^{ - 1}} + {b^{ - 1}} + {c^{ - 1}} = 0$ such that $\left| {\begin{array}{*{20}{c}}{1 + a}&1&1\\1&{1 + b}&1\\1&1&{1 + c}\end{array}} \right| = \lambda $,then the value of $\lambda $ is
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