Question

Suppose that there are real numbers x,y,z not all zero such that x=cy+bz,y=az+cx and z=bx+ay, then a2+b2+c2+2abc is equal to. ​

Answer 1

System of equations

x – cy – bz = 0

cx – y + az = 0

bx + ay – z = 0

$\begin{pmatrix}1&-c&-b\\ c&-1&a\\ b&a&-1\end{pmatrix}$

⇒1(1−a  

2

)+c(−c−ab)−b(a+b)=0

⇒a  

2

+b  

2

+c  

2

+2abc=1

has non trivial solution if the determinant of coefficient matrix is zero

or

a,b,c real numbers

x,y,z real numbers not all zero

x=cy+bz→x−cy−bz=0−−−(1)

y=az+cx→cx+y−az=0−−−(2)

z=bx+cy→−bx−ay+z=0−−−(3)

$\begin{pmatrix}1&-c&-b\\ c&-1&a\\ b&a&-1\end{pmatrix}$

Answer 2

$\huge\underline{\overline{\mid{\bold{\pink{ANSWER-}}\mid}}}$

The given equations are

–x + cy + bz = 0

cx –y + az = 0

bx + ay – z = 0

∵ x, y, z are not all zero

∴ The above system should not have unique (zero) solution

⇒ –1(1– a2) – c(– c – ab) + b(ac + b) = 0

⇒–1 + a2 + b2 + c2 + 2abc = 0

⇒ a2 + b2 + c2 + 2abc = 1