If a + b + c = 0,then the solution of the equ | vedclass

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

Question

(C) Given the determinant equation: $\left| \begin{array}{ccc} a - x & c & b \ c & b - x & a \ b & a & c - x \end{array} ight| = 0$.Apply the oper

Vedclass · Accepted Answer

$0, \pm \sqrt{\frac{3}{2}(a^2 + b^2 + c^2)}$

Vedclass · Answer

(C) Given the determinant equation: $\left| \begin{array}{ccc} a - x & c & b \ c & b - x & a \ b & a & c - x \end{array} ight| = 0$.Apply the operation $C_1 	o C_1 + C_2 + C_3$:$\left| \begin{array}{ccc} a + b + c - x & c & b \ a + b + c - x & b - x & a \ a + b + c - x & a & c - x \end{array} ight| = 0$.Since $a + b + c = 0$,the first column becomes $-x$ in every entry:$\left| \begin{array}{ccc} -x & c & b \ -x & b - x & a \ -x & a & c - x \end{array} ight| = 0$.Taking $-x$ common from the first column:$-x \left| \begin{array}{ccc} 1 & c & b \ 1 & b - x & a \ 1 & a & c - x \end{array} ight| = 0$.This gives $x = 0$ as one solution.For the remaining part,expand the determinant:$1((b-x)(c-x) - a^2) - c(c-x-a) + b(a-b+x) = 0$.Simplifying this leads to $x^2 - (a^2 + b^2 + c^2) + (ab + bc + ca) = 0$.Since $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca) = 0$,we have $ab+bc+ca = -\frac{1}{2}(a^2+b^2+c^2)$.Substituting this: $x^2 - (a^2+b^2+c^2) - \frac{1}{2}(a^2+b^2+c^2) = 0$.$x^2 = \frac{3}{2}(a^2+b^2+c^2) \Rightarrow x = \pm \sqrt{\frac{3}{2}(a^2+b^2+c^2)}$.Thus,the solutions are $x = 0, \pm \sqrt{\frac{3}{2}(a^2+b^2+c^2)}$.

If $a + b + c = 0$,then the solution of the equation $\left| \begin{array}{ccc} a - x & c & b \\ c & b - x & a \\ b & a & c - x \end{array} \right| = 0$ is

Similar Questions

If $a$ is a non-real complex number for which the system of equations $ax - a^2y + a^3z = 0$,$-a^2x + a^3y + az = 0$,and $a^3x + ay - a^2z = 0$ has non-trivial solutions,then $|a|$ is:

If $\omega$ is the cube root of unity,then $\left| \begin{array}{ccc} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{array} \right| = $

If the area of a triangle with vertices $(2, 6)$,$(5, 4)$,and $(k, 4)$ is $35$ square units,then $k = \text{ . . . . . . }$.

If $a+x=b+y=c+z+1,$ where $a, b, c, x, y, z$ are non-zero distinct real numbers,then $\left|\begin{array}{lll}x & a+y & x+a \\ y & b+y & y+b \\ z & c+y & z+c\end{array}\right|$ is equal to

The area of a triangle with vertices $(-3, 0)$,$(3, 0)$,and $(0, k)$ is $9$ sq units. Find the value of $k$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module