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
MediumThe solutions of the equation $\left| \begin{array}{ccc} x & 2 & -1 \\ 2 & 5 & x \\ -1 & 2 & x \end{array} \right| = 0$ are
- A$3, -1$
- B$-3, 1$
- C$3, 1$
- D$-3, -1$
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The value of $\left| \begin{array}{ccc} 1 & 1 & 1 \\ bc & ca & ab \\ b+c & c+a & a+b \end{array} \right|$ is
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View SolutionIf $\alpha \neq a, \beta \neq b, \gamma \neq c$ and $\left|\begin{array}{lll}\alpha & b & c \\ a & \beta & c \\ a & b & \gamma\end{array}\right|=0$,then $\frac{a}{\alpha-a}+\frac{b}{\beta-b}+\frac{\gamma}{\gamma-c}$ is equal to :
DifficultJEE MAIN 2024
View SolutionIf $A = \begin{vmatrix} \sin(\theta + \alpha) & \cos(\theta + \alpha) & 1 \\ \sin(\theta + \beta) & \cos(\theta + \beta) & 1 \\ \sin(\theta + \gamma) & \cos(\theta + \gamma) & 1 \end{vmatrix}$,then
Medium
View SolutionIf $P, Q$ and $R$ are $3 \times 3$ matrices such that $\begin{bmatrix} 3x^2+x+3 & 2x^2-x+4 & 7x^2+8x+5 \\ 5x^2+3x+2 & 4x^2-2x-1 & 7x^2+5x+8 \\ 3x^2+2x+5 & 4x^2-x-2 & 3x^2+8x+7 \end{bmatrix} = Px^2+Qx+R$,then $\det R = $
DifficultAP EAMCET 2023
View SolutionThe matrix $\left[ {\begin{array}{*{20}{c}}2&\lambda &{ - 4}\\{ - 1}&3&4\\1&{ - 2}&{ - 3}\end{array}} \right]$ is non-singular,if
Easy
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