If $x = c y + b z, y = a z + c x, z = b x + a y$ where $x, y, z$ are not all zero, then $a^{2} + b^{2} + c^{2} + 2 a b c = k$

Open in App

Solution

We have $x - c y - b z = 0 c x - y + a z = 0 b x + a y - z = 0$
Since $x, y, z$ are not all zero, so the sysmtem will have a non-trivial solution. Therefore,
$∣ ∣ ∣ ∣ \begin{matrix} 1 & - c & - b c & - 1 & a b & a & - 1 \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow 1. (1 - a^{2}) + c (- c - a b) - b (a c + b) = 0 \Rightarrow a^{2} + b^{2} + c^{2} + 2 a b c = 1$

Suggest Corrections

Similar questions

x = c y + b z, y = a z + c x, z = b x + a y

x, y, z

a^{2} + b^{2} + c^{2} + 2 a b c

x = c y + b z, y = a z + c x, z = b x + a y

x, y, z

a^{2} + b^{2} + c^{2} + 2 a b c + 1 = 0

x = c y + b z, y = a z + c x, z = b x + a y

x, y, z

a^{2} + b^{2} + c^{2} + 2 a b c

x = c y + b z, y = a z + c x, z = b x + a y

a^{2} + b^{2} + c^{2} + 2 a b c

Join BYJU'S Learning Program