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
EasyIf $\left|\begin{array}{cc}x & 2 \\ 18 & x\end{array}\right|=\left|\begin{array}{cc}6 & 2 \\ 18 & 6\end{array}\right|,$ then $x$ is equal to
- A$6$
- B$\pm 6$
- C$-6$
- D$0$
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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 SolutionFind the values of $x$ for which $\left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|$.
Easy
View SolutionIf $x^4+y^4+z^4=0$ then,$\left|\begin{array}{ccc}1 & xy & yz \\ zx & 1 & xy \\ yz & zx & 1\end{array}\right|=$ . . . . . . . $(\because x, y, z \in \mathbb{R})$
If matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$,then $|A|^{-1}$ is equal to
If $t_1, t_2$ and $t_3$ are distinct,then the points $(t_1, 2at_1 + at_1^3)$,$(t_2, 2at_2 + at_2^3)$ and $(t_3, 2at_3 + at_3^3)$ are collinear if:
Difficult
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