in Set Theory & Algebra edited by
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If $x=cy+bz, \: y=az+cx, \: z=bx+ay,$ where $x,y,z$ are not all zero, then $a^2+b^2+c^2=$

  1. $1+2abc$
  2. $1-2abc$
  3. $1+abc$
  4. $abc-1$
in Set Theory & Algebra edited by
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2 Answers

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Rearranging the system of equations, we get: x – cy – bz = 0, cx – y + az = 0 and bx + ay – z = 0. 
Put in Determinant and expand.

\begin{vmatrix} 1 &-c &-b \\ c&-1 &a \\ b& a &-1 \end{vmatrix}=0 So a^2+b^2+c^2+2abc=1 Ans= B
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Why do we take the determinant value and why is it equal to zero?
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Why determinant taken equal to 0

@Sukhbir Singh

@Ruturaj Mohanty

 

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Because you are finding the solution of the determinant in terms of the given values.

Think of this as compared with the quadratic equation.

How do you find the roots?

Simply by equating it to $\mathbf 0$, right?

So, the same is true here.
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Write the equations in matrix form and use the theorem - "A square matrix M from the equation Mx = 0 is invertible if and only if this homoegenous system of equations has no non-trivial solution".

In this case, we want a non-trivial solution (as indicated by "x, y and z are not all zero"), so we set the determinant to zero.
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