The solution(s) of the equation left| begin{m | vedclass

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(A) Given the determinant $\Delta = \left| \begin{matrix} x & a & b \ a & x & a \ b & b & x \end{matrix} ight| = 0$. Applying $C_1 ightarrow C_1

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$x = -(a + b)$

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(A) Given the determinant $\Delta = \left| \begin{matrix} x & a & b \ a & x & a \ b & b & x \end{matrix} ight| = 0$. Applying $C_1 ightarrow C_1 + C_2 + C_3$,we get: $\left| \begin{matrix} x+a+b & a & b \ x+a+b & x & a \ x+a+b & b & x \end{matrix} ight| = 0$. Taking $(x+a+b)$ common from $C_1$: $(x+a+b) \left| \begin{matrix} 1 & a & b \ 1 & x & a \ 1 & b & x \end{matrix} ight| = 0$. Applying $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$: $(x+a+b) \left| \begin{matrix} 1 & a & b \ 0 & x-a & a-b \ 0 & b-a & x-b \end{matrix} ight| = 0$. Expanding along $C_1$: $(x+a+b) [(x-a)(x-b) - (a-b)(b-a)] = 0$. $(x+a+b) [x^2 - bx - ax + ab + (a-b)^2] = 0$. $(x+a+b) [x^2 - (a+b)x + ab + a^2 - 2ab + b^2] = 0$. $(x+a+b) [x^2 - (a+b)x + a^2 - ab + b^2] = 0$. Thus,$x = -(a+b)$ is a solution. The quadratic part $x^2 - (a+b)x + a^2 - ab + b^2 = 0$ yields complex roots unless $a=b=x$. If $a=b$,the determinant becomes $\left| \begin{matrix} x & a & a \ a & x & a \ a & a & x \end{matrix} ight| = (x-a)^2(x+2a) = 0$,giving $x=a$ or $x=-2a$. Since the question implies general solutions,$x=-(a+b)$ is the primary root.

The solution$(s)$ of the equation $\left| \begin{matrix} x & a & b \\ a & x & a \\ b & b & x \end{matrix} \right| = 0$ is/are:

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