If x^2+y^2+z^2 neq 0, quad x=c y+b z, quad y= | vedclass

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

Question

(A) The given system of linear equations is:$x - cy - bz = 0$$-cx + y - az = 0$$-bx - ay + z = 0$Since $x^2 + y^2 + z^2 
eq 0$,the system has a non-t

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) The given system of linear equations is:$x - cy - bz = 0$$-cx + y - az = 0$$-bx - ay + z = 0$Since $x^2 + y^2 + z^2 
eq 0$,the system has a non-trivial solution. Therefore,the determinant of the coefficient matrix must be zero:$\begin{vmatrix} 1 & -c & -b \ -c & 1 & -a \ -b & -a & 1 \end{vmatrix} = 0$Expanding the determinant along the first row:$1(1 - a^2) - (-c)(-c - ab) + (-b)(ca + b) = 0$$1 - a^2 - c^2 - abc - abc - b^2 = 0$$1 - a^2 - b^2 - c^2 - 2abc = 0$Rearranging the terms,we get:$a^2 + b^2 + c^2 + 2abc = 1$

If $x^2+y^2+z^2 \neq 0, \quad x=c y+b z, \quad y=a z+c x$ and $z=b x+a y$,then $a^2+b^2+c^2+2 a b c$ is equal to

Similar Questions

What is a phycobiont?

Which of the following pairs of transition metal ions have the same calculated values of magnetic moment?

$P_4O_{10}$ has short and long $P-O$ bonds. The number of short $P=O$ bonds in this compound is

If $\cos ^{-1}\left(\frac{5}{13}\right)+\cos ^{-1}\left(\frac{3}{5}\right)=\cos ^{-1} x$,then $x$ is equal to

The figure below shows two paths that may be taken by a gas to go from a state $A$ to a state $C$. In process $AB$,$400 \, J$ of heat is added to the system and in process $BC$,$100 \, J$ of heat is added to the system. The heat absorbed by the system in the process $AC$ will be ............. $J$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module