If a, b, c are non-zero real numbers and if t | vedclass

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

Question

(C) The given system of equations is:$(a-1)x - y - z = 0$$-x + (b-1)y - z = 0$$-x - y + (c-1)z = 0$For a non-trivial solution,the determinant of the c

Vedclass · Accepted Answer

$abc$

Vedclass · Answer

(C) The given system of equations is:$(a-1)x - y - z = 0$$-x + (b-1)y - z = 0$$-x - y + (c-1)z = 0$For a non-trivial solution,the determinant of the coefficient matrix must be zero:$\begin{vmatrix} a-1 & -1 & -1 \ -1 & b-1 & -1 \ -1 & -1 & c-1 \end{vmatrix} = 0$Adding $R_2$ and $R_3$ to $R_1$:$\begin{vmatrix} a-3 & b-3 & c-3 \ -1 & b-1 & -1 \ -1 & -1 & c-1 \end{vmatrix} = 0$Alternatively,expanding the determinant:$(a-1)((b-1)(c-1) - 1) + 1(-(c-1) - 1) - 1(1 + (b-1)) = 0$$(a-1)(bc - b - c + 1 - 1) + (-c + 1 - 1) - (1 + b - 1) = 0$$(a-1)(bc - b - c) - c - b = 0$$abc - ab - ac - bc + b + c - c - b = 0$$abc - ab - ac - bc = 0$Therefore,$ab + bc + ca = abc$.

If $a, b, c$ are non-zero real numbers and if the equations $(a-1) x=y+z, (b-1) y=z+x, (c-1) z=x+y$ have a non-trivial solution,then $ab+bc+ca=$

Similar Questions

Let $A = \begin{bmatrix} 2 & 0 & 3 \\ 4 & 7 & 11 \\ 5 & 4 & 8 \end{bmatrix}$. Then

$f(x)$ is an $n^{\text{th}}$ degree polynomial satisfying $f(x) = \frac{1}{2} \begin{vmatrix} f(x) & f(\frac{1}{x}) - f(x) \\ 1 & f(\frac{1}{x}) \end{vmatrix}$. If $f(2) = 33$,then the value of $f(3)$ is

If $2x + 3y - 5z = 7$,$x + y + z = 6$,and $3x - 4y + 2z = 1$,then $x =$

Evaluate the determinant: $\left| {\begin{array}{*{20}{c}}{b + c}& a& a\\b& {c + a}& b\\c& c& {a + b}\end{array}} \right|$

$\left|\begin{array}{rr}\sin 35^{\circ} & -\cos 35^{\circ} \\ \sin 55^{\circ} & \cos 55^{\circ}\end{array}\right|=$ . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module