If a neq 1, b neq -1, c neq -1 and the system | vedclass

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(B) The given system of equations is:$x - ay - az = 0$$-bx + y - bz = 0$$-cx - cy + z = 0$For a non-trivial solution,the determinant of the coefficien

Vedclass · Accepted Answer

$\frac{a}{a+1} + \frac{b}{b+1} + \frac{c}{c+1} = 1$

Vedclass · Answer

(B) The given system of equations is:$x - ay - az = 0$$-bx + y - bz = 0$$-cx - cy + z = 0$For a non-trivial solution,the determinant of the coefficient matrix must be zero:$\Delta = \begin{vmatrix} 1 & -a & -a \ -b & 1 & -b \ -c & -c & 1 \end{vmatrix} = 0$Expanding the determinant:$1(1 - bc) + a(-b - bc) - a(bc + c) = 0$$1 - bc - ab - abc - abc - ac = 0$$1 - (ab + bc + ca) - 2abc = 0$Divide by $(a+1)(b+1)(c+1)$:Note that $(a+1)(b+1)(c+1) = abc + ab + bc + ca + a + b + c + 1$.Alternatively,divide the determinant equation by $(a+1)(b+1)(c+1)$ or manipulate the system:From $x = a(y+z)$,we get $x = a(x+y+z) - ax$,so $x(1+a) = a(x+y+z)$.Thus,$\frac{x}{x+y+z} = \frac{a}{a+1}$.Similarly,$\frac{y}{x+y+z} = \frac{b}{b+1}$ and $\frac{z}{x+y+z} = \frac{c}{c+1}$.Adding these three equations:$\frac{x+y+z}{x+y+z} = \frac{a}{a+1} + \frac{b}{b+1} + \frac{c}{c+1}$.Since the solution is non-trivial,$x+y+z 
eq 0$,so:$1 = \frac{a}{a+1} + \frac{b}{b+1} + \frac{c}{c+1}$.

If $a \neq 1, b \neq -1, c \neq -1$ and the system of equations $x = a(y+z), y = b(z+x), z = c(x+y)$ has a non-trivial solution,then:

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