Let A = begin{bmatrix} -2 & x & 1 x & 1 & 1 | vedclass

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(C) Given the matrix $A = \begin{bmatrix} -2 & x & 1 \ x & 1 & 1 \ 2 & 3 & -1 \end{bmatrix}$.The determinant of $A$ is given by $\operatorname{det}(

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$19$

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(C) Given the matrix $A = \begin{bmatrix} -2 & x & 1 \ x & 1 & 1 \ 2 & 3 & -1 \end{bmatrix}$.The determinant of $A$ is given by $\operatorname{det}(A) = -2(-1 - 3) - x(-x - 2) + 1(3x - 2) = 0$.Simplifying the expression: $-2(-4) - x(-x - 2) + 3x - 2 = 0$.$8 + x^2 + 2x + 3x - 2 = 0$.$x^2 + 5x + 6 = 0$.Factoring the quadratic equation: $(x + 2)(x + 3) = 0$.The roots are $l = -2$ and $m = -3$ (or vice versa).We need to find $l^3 - m^3$.If $l = -2$ and $m = -3$,then $l^3 - m^3 = (-2)^3 - (-3)^3 = -8 - (-27) = -8 + 27 = 19$.If $l = -3$ and $m = -2$,then $l^3 - m^3 = (-3)^3 - (-2)^3 = -27 - (-8) = -27 + 8 = -19$.Since $19$ is an option,the correct value is $19$.

Let $A = \begin{bmatrix} -2 & x & 1 \\ x & 1 & 1 \\ 2 & 3 & -1 \end{bmatrix}$. If the roots of the equation $\operatorname{det}(A) = 0$ are $l$ and $m$,then find the value of $l^3 - m^3$.

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