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
EasyThe roots of the equation $\left| \begin{matrix} 0 & x & 16 \\ x & 5 & 7 \\ 0 & 9 & x \end{matrix} \right| = 0$ are
- A$0, 12, 12$
- B$0, 12, -12$
- C$0, 12, 16$
- D$0, 9, 16$
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Similar Questions
If $1, \omega, \omega^2$ are the cube roots of unity,then $\Delta = \begin{vmatrix} 1 & \omega^n & \omega^{2n} \\ \omega^n & \omega^{2n} & 1 \\ \omega^{2n} & 1 & \omega^n \end{vmatrix} = $
MediumAIEEE 2003
View SolutionThe sum of the values of $x$ such that the matrix $\begin{bmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{bmatrix} - x \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ is singular,is
The number of distinct real roots of $\left| {\begin{array}{*{20}{c}}{\sin x}&{\cos x}&{\cos x}\\{\cos x}&{\sin x}&{\cos x}\\{\cos x}&{\cos x}&{\sin x}\end{array}} \right| = 0$ in the interval $-\frac{\pi}{4} \le x \le \frac{\pi}{4}$ is
MediumIIT 2001
View SolutionIf the system of equations
$(k+1)^3 x + (k+2)^3 y = (k+3)^3$
$(k+1) x + (k+2) y = k+3$
$x + y = 1$
is consistent,then the value of $k$ is
$(k+1)^3 x + (k+2)^3 y = (k+3)^3$
$(k+1) x + (k+2) y = k+3$
$x + y = 1$
is consistent,then the value of $k$ is
DifficultAP EAMCET 2010
View SolutionThe set of all values of $\lambda$ for which the system of linear equations $2x_1 - 2x_2 + x_3 = \lambda x_1$,$2x_1 - 3x_2 + 2x_3 = \lambda x_2$,and $-x_1 + 2x_2 = \lambda x_3$ has a non-trivial solution:
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