Solve the equation left|begin{array}{ccc}x+a | vedclass

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(A) Given the determinant equation: $\left|\begin{array}{ccc}x+a & x & x \ x & x+a & x \ x & x & x+a\end{array}ight|=0$ Applying the row operation

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$x=-\frac{a}{3}$

Vedclass · Answer

(A) Given the determinant equation: $\left|\begin{array}{ccc}x+a & x & x \ x & x+a & x \ x & x & x+a\end{array}ight|=0$ Applying the row operation $R_{1} ightarrow R_{1}+R_{2}+R_{3}$,we get: $\left|\begin{array}{ccc}3x+a & 3x+a & 3x+a \ x & x+a & x \ x & x & x+a\end{array}ight|=0$ Taking $(3x+a)$ as a common factor from $R_{1}$: $(3x+a) \left|\begin{array}{ccc}1 & 1 & 1 \ x & x+a & x \ x & x & x+a\end{array}ight|=0$ Applying column operations $C_{2} ightarrow C_{2}-C_{1}$ and $C_{3} ightarrow C_{3}-C_{1}$: $(3x+a) \left|\begin{array}{ccc}1 & 0 & 0 \ x & a & 0 \ x & 0 & a\end{array}ight|=0$ Expanding along $R_{1}$: $(3x+a) [1(a 	imes a - 0 	imes 0)] = 0$ $(3x+a) a^{2} = 0$ Since it is given that $a 
eq 0$,then $a^{2} 
eq 0$. Therefore,we must have: $3x+a = 0$ $x = -\frac{a}{3}$

Solve the equation $\left|\begin{array}{ccc}x+a & x & x \\ x & x+a & x \\ x & x & x+a\end{array}\right|=0$,where $a \neq 0$.

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