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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

If $A = \begin{bmatrix} \alpha^2 & 5 \\ 5 & -\alpha \end{bmatrix}$ and $\det(A^{10}) = 1024$,then $\alpha = $

If $a, b, c$ are respectively the $5^{\text{th}}, 8^{\text{th}}, 13^{\text{th}}$ terms of an arithmetic progression,then $\left|\begin{array}{ccc}a & 5 & 1 \\ b & 8 & 1 \\ c & 13 & 1\end{array}\right|=$

$\left| {\begin{array}{ccc} 19 & 17 & 15 \\ 9 & 8 & 7 \\ 1 & 1 & 1 \end{array}} \right| = $

If $x=k$ satisfies the equation $\left|\begin{array}{ccc}x-2 & 3x-3 & 5x-5 \\ x-4 & 3x-9 & 5x-25 \\ x-8 & 3x-27 & 5x-125\end{array}\right|=0$, then $x=k$ also satisfies the equation

The value of $\left| \begin{array}{ccc} 1 & 1 & 1 \\ bc & ca & ab \\ b+c & c+a & a+b \end{array} \right|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module