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show that determinant $$\left|\matrix{ x^2+L & xy & xz \\ xy & y^2+L & yz \\ xz & yz & z^2+L \\ }\right| = L^2(x^2+y^2+z^2+L)$$

without expanding by using the appropriate properties of determinant.

All i can do is LHS

$$x^2y^2z^2\left|\matrix{ 1+L/x^2 & 1 & 1 \\ 1 & 1+L/y^2 & 1 \\ 1 & 1 & 1+L/z^2 \\ }\right|$$

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  • $\begingroup$ What part are you having problems with? $\endgroup$ – shade4159 Nov 23 '13 at 17:19
  • $\begingroup$ I have no idea to start. $\endgroup$ – haohao Nov 23 '13 at 17:23
  • $\begingroup$ All i can do was just extracting x^2 y^2 z^2 from each row and column. $\endgroup$ – haohao Nov 23 '13 at 17:24
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Look at this form of your matrix $$\pmatrix{L & 0 & 0 \\ 0 & L & 0 \\ 0 & 0 & L\\} + \pmatrix{x & 0 & 0 \\ 0 & y & 0\\ 0 & 0 & z}\pmatrix{ x & y & z \\ x & y & z \\ x & y & z \\}$$

and consider your right eigenvector $\pmatrix{x \\ y \\ z}$ with eignevalue $ x^2 + y^2 + z^2 + L$

I suspect you may not yet be familiar with eigenvalues, so if that is true, maybe this is some good motivation to learn.

Eigenvector review:
A right eigenvector $\mathbf{v}$ with eigenvalue $\lambda$ is defined for a matrix $\mathbf{A}$ as $$\mathbf{A}\mathbf{v} = \mathbf{v}\lambda$$ Here we see that $$\pmatrix{x^2 + L & xy & xz \\ xy & y^2 + L & yz \\ xz & yz & z^2 + L}\pmatrix{x \\ y \\ z} = \pmatrix{x \\ y \\ z}(x^2 + y^2 + z^2 + L)$$ In this case all other eigenvectors come from the two dimensional subspace defined as the space of vectors orthogonal to the first eigenvector, and they have eignevalue $L$ (this can be seen more easily from the first form I stated for you matrix).

The unit hypercube of hypervolume one with edges defined by those eigenvectors then is transformed by your matrix to have volume $(x^2 + y^2 + z^2 + L)L^2$ and that is the determinant for your matrix.

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  • $\begingroup$ Thank you, but i am stuck with "right eigenvector with eignevalue x^2+y^2+z^2+L" $\endgroup$ – haohao Nov 23 '13 at 17:56
  • $\begingroup$ How do i relate the eigenvalue into finding determinant? $\endgroup$ – haohao Nov 23 '13 at 17:59
  • $\begingroup$ I say hypercube, but this case it is just a three dimensional cube. $\endgroup$ – adam W Nov 23 '13 at 18:10

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