Matrices and Determinants 4 Question 32
33. Given, $x=c y+b z, y=a z+c x, z=b x+a y$, where $x, y$, $z$ are not all zero, prove that $a^{2}+b^{2}+c^{2}+2 a b=1$.
$(1978,2 \mathrm{M})$
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Solution:
- Given systems of equations can be rewritten as $-x+c y+b y=0, c x-y+a z=0$ and $b x+a y-z=0$
Above system of equations are homogeneous equation. Since, $x, y$ and $z$ are not all zero, so it has non-trivial solution.
Therefore, the coefficient of determinant must be zero.
$\therefore \quad\left|\begin{array}{ccc}-1 & c & b \\ c & -1 & a \\ b & a & -1\end{array}\right|=0$
$\Rightarrow-1\left(1-a^{2}\right)-c(-c-a b)+b(c a+b)=0$
$\Rightarrow \quad a^{2}+b^{2}+c^{2}+2 a b c-1=0$
$\Rightarrow \quad a^{2}+b^{2}+c^{2}+2 a b c=1$
