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
Easy$\left| \begin{matrix} 0 & a & -b \\ -a & 0 & c \\ b & -c & 0 \end{matrix} \right| = $
- A$ -2abc $
- B$ abc $
- C$ 0 $
- D$ a^2 + b^2 + c^2 $
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Similar Questions
For all values of $A, B, C$ and $P, Q, R$,the value of $\left| \begin{array}{ccc} \cos(A-P) & \cos(A-Q) & \cos(A-R) \\ \cos(B-P) & \cos(B-Q) & \cos(B-R) \\ \cos(C-P) & \cos(C-Q) & \cos(C-R) \end{array} \right|$ is
DifficultIIT 1994
View SolutionFor how many value$(s)$ of $x$ in the closed interval $[-4, -1]$ is the matrix $\begin{bmatrix} 3 & -1+x & 2 \\ 3 & -1 & x+2 \\ x+3 & -1 & 2 \end{bmatrix}$ singular?
Medium
View SolutionLet $A = \begin{bmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{bmatrix}$. If $|A|^2 = 25$,then $|\alpha|$ is equal to:
If $A = \begin{bmatrix} 5 & 5x & x \\ 0 & x & 5x \\ 0 & 0 & 5 \end{bmatrix}$ and $|A^2| = 25$,then $|x|$ is equal to
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 $\alpha + \beta =$
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