Sum of the roots of the equation left|begin{a | vedclass

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(C) Let the given equation be $D_1 - D_2 = 0$. Expanding the first determinant $D_1$ along the second row: $D_1 = -1 	imes \left|\begin{array}{ccc} x

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(C) Let the given equation be $D_1 - D_2 = 0$. Expanding the first determinant $D_1$ along the second row: $D_1 = -1 	imes \left|\begin{array}{ccc} x & 0 & 0 \ 0 & x & 0 \ 2 & 0 & x-1 \end{array}ight| = -1 	imes [x(x(x-1) - 0)] = -x^2(x-1) = -x^3 + x^2$. Expanding the second determinant $D_2$ along the first row: $D_2 = -x 	imes \left|\begin{array}{cc} 0 & x-1 \ 2 & 0 \end{array}ight| = -x(0 - 2(x-1)) = -x(-2x + 2) = 2x^2 - 2x$. Substituting these into the equation: $(-x^3 + x^2) - (2x^2 - 2x) = 0$ $-x^3 - x^2 + 2x = 0$ $x^3 + x^2 - 2x = 0$ $x(x^2 + x - 2) = 0$ $x(x+2)(x-1) = 0$ The roots are $x = 0, -2, 1$. Sum of the roots $= 0 + (-2) + 1 = -1$.

Sum of the roots of the equation $\left|\begin{array}{cccc} x & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & x & 0 & 0 \\ 2 & 0 & x-1 & 0 \end{array}\right| - \left|\begin{array}{ccc} 0 & x & 0 \\ 0 & 0 & x-1 \\ 2 & 2 & 0 \end{array}\right| = 0$ is

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