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 $A = \begin{bmatrix} 5 & 5x & x \\ 0 & x & 5x \\ 0 & 0 & 5 \end{bmatrix}$ and $|A^2| = 25$, then $|x|$ is equal to
- A$\frac{1}{5}$
- B$5$
- C$25$
- D$1$
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If the determinant of the matrix $A = \begin{bmatrix} 0 & a & b \\ -a & 0 & \beta \\ -b & -\alpha & 0 \end{bmatrix}$ is zero for all $a, b$, then $\alpha + \beta =$
The solution of the equation $\left| \begin{array}{ccc} \cos \theta & \sin \theta & \cos \theta \\ -\sin \theta & \cos \theta & \sin \theta \\ -\cos \theta & -\sin \theta & \cos \theta \end{array} \right| = 0$ is:
Medium
View SolutionIf $\left| \begin{array}{ccc} a & b & c \\ b & c & a \\ c & a & b \end{array} \right| = k(a + b + c)(a^2 + b^2 + c^2 - bc - ca - ab)$,then $k =$
Medium
View Solution$A$ determinant is chosen at random from the set of all determinants of order $2 \times 2$ with elements $0$ or $1$ only. The probability that the determinant chosen is non-zero is .........
The least value of the product $xyz$ for which the determinant $\left| \begin{array}{ccc} x & 1 & 1 \\ 1 & y & 1 \\ 1 & 1 & z \end{array} \right|$ is non-negative,is
DifficultJEE MAIN 2015
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