If the system of equations ax + y + z = 0,x + | vedclass

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

Question

(C) Since the system of equations has a non-trivial solution,the determinant of the coefficient matrix must be zero:$\begin{vmatrix} a & 1 & 1 \ 1 &

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Since the system of equations has a non-trivial solution,the determinant of the coefficient matrix must be zero:$\begin{vmatrix} a & 1 & 1 \ 1 & b & 1 \ 1 & 1 & c \end{vmatrix} = 0$Applying row operations $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$:$\begin{vmatrix} a & 1 & 1 \ 1-a & b-1 & 0 \ 1-a & 0 & c-1 \end{vmatrix} = 0$Expanding along the first row:$a(b-1)(c-1) - 1(1-a)(c-1) + 1(0 - (1-a)(b-1)) = 0$$a(b-1)(c-1) - (1-a)(c-1) - (1-a)(b-1) = 0$Divide the entire equation by $(1-a)(1-b)(1-c)$ (noting $a, b, c 
eq 1$):$\frac{a(b-1)(c-1)}{(1-a)(1-b)(1-c)} - \frac{(1-a)(c-1)}{(1-a)(1-b)(1-c)} - \frac{(1-a)(b-1)}{(1-a)(1-b)(1-c)} = 0$$\frac{-a}{(1-a)} - \frac{1}{(1-b)} - \frac{1}{(1-c)} = 0$$\frac{a}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 0$Since $\frac{a}{1-a} = \frac{a-1+1}{1-a} = -1 + \frac{1}{1-a}$,we substitute this into the equation:$-1 + \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 0$$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$.

If the system of equations $ax + y + z = 0$,$x + by + z = 0$ and $x + y + cz = 0$,where $a, b, c \neq 1$,has a non-trivial solution,then the value of $\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}$ is

Similar Questions

Let $\left|\begin{array}{ccc}x^2+x+1 & x+1 & 2x-3 \\ 3x^2-1 & x+2 & x-1 \\ x^2+5x+1 & 2x+3 & x+4\end{array}\right| = ax^4+bx^3+cx^2+dx+e$ be an identity in $x$. If $a, b, c, d$ are known, then the value of $e$ is

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

If $a_i^2 + b_i^2 + c_i^2 = 1$ for $i = 1, 2, 3$ and $a_ia_j + b_ib_j + c_ic_j = 0$ for $i \ne j$ where $i, j = 1, 2, 3$,then the value of the determinant $\left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right|$ is:

If $a, b, c$ are in $A.P.$,find the value of $\left|\begin{array}{ccc} 2y+4 & 5y+7 & 8y+a \\ 3y+5 & 6y+8 & 9y+b \\ 4y+6 & 7y+9 & 10y+c \end{array}\right|$.

The area of the triangle with vertices $(3, 8)$, $(-4, 2)$ and $(5, 1)$ is $\frac{P}{4}$. The value of $P$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module