Matrix

Algebra Level 3

The system of linear equations

X + γ Y Z = 0 γ X Y Z = 0 X + Y γ Z = 0 \begin{aligned} X + \gamma Y - Z &=0 \\ \gamma X - Y - Z &=0 \\ X + Y - \gamma Z &=0 \end{aligned}

has a non-trivial solution for __________ . \text{\_\_\_\_\_\_\_\_\_\_}.

exactly one values of γ \gamma exactly two values of γ \gamma exactly three values of γ \gamma infinitely many values of γ \gamma insufficient information

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1 solution

Laurent Shorts
Apr 7, 2016

If the matrix ( 1 γ 1 γ 1 1 1 1 γ ) \begin{pmatrix}1&\gamma&-1\\\gamma&-1&-1\\1&1&-\gamma\end{pmatrix} is invertible, there is a unique solution, but it would be the trivial one. For another solution to exist, the matrix must not be invertible. This append when the determinant of the matrix is 0.

1 γ 1 γ 1 1 1 1 γ = ( γ 1 ) γ ( γ + 1 ) = 0 \begin{vmatrix}1&\gamma&-1\\\gamma&-1&-1\\1&1&-\gamma\end{vmatrix}=(\gamma-1)\gamma(\gamma+1)=0 . There are three values of γ \gamma .

What does “trival solution” mean here?

Tom Dunlap - 11 months, 2 weeks ago

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maybe non-zero solution

Luke Smith - 10 months, 1 week ago

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