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
EasyLet $A = \begin{bmatrix} x+2 & 3x \\ 3 & x+2 \end{bmatrix}$ and $B = \begin{bmatrix} x & 0 \\ 5 & x+2 \end{bmatrix}$. Then all solutions of the equation $\det(AB) = 0$ are:
- A$1, -1, 0, 2$
- B$1, 4, 0, -2$
- C$1, -1, 4, 3$
- D$-1, 4, 0, 3$
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If $\Delta = \begin{vmatrix} 1 & \cos \theta & 1 \\ -\cos \theta & 1 & \cos \theta \\ -1 & -\cos \theta & 1 \end{vmatrix}$,then $\Delta$ lies in the interval
If matrix $A = \begin{bmatrix} \sin \theta & \csc \theta & 1 \\ \csc \theta & 1 & \sin \theta \\ 1 & \sin \theta & \csc \theta \end{bmatrix}$ is a non-invertible matrix,then the possible value of $\theta$ is $(n \in \mathbb{Z})$
If $\left| \begin{array}{ccc} x+1 & 3 & 5 \\ 2 & x+2 & 5 \\ 2 & 3 & x+4 \end{array} \right| = 0$,then $x =$
Medium
View SolutionAt what value of $x$ will $\left| \begin{array}{ccc} x + \omega^2 & \omega & 1 \\ \omega & \omega^2 & 1 + x \\ 1 & x + \omega & \omega^2 \end{array} \right| = 0$?
Easy
View SolutionIf $x, y, z$ are all different and not equal to zero and $\left|\begin{array}{ccc}1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z\end{array}\right|=0$,then the value of $x^{-1}+y^{-1}+z^{-1}$ is equal to
DifficultKCET 2016
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