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This question already has an answer here:

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

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

Is it a must to relate to eigenvalue problem?

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marked as duplicate by Git Gud, Johannes Kloos, Davide Giraudo, azimut, Dominic Michaelis Nov 24 '13 at 10:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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$$\left|\matrix{ x^2+L & xy & xz \\ xy & y^2+L & yz \\ xz & yz & z^2+L \\ }\right| = $$ $$xyz\left|\matrix{ x+L/x & y & z \\ x & y+L/y & z \\ x & y & z+L/z \\ }\right| = $$ $$=\left|\matrix{ x^2+L & y^2 & z^2 \\ x^2 & y^2+L & z^2 \\ x^2 & y^2 & z^2+L \\ }\right| = $$ $$=\left|\matrix{ x^2+y^2+z^2+L & y^2 & z^2 \\ x^2+y^2+z^2+L & y^2+L & z^2 \\ x^2+y^2+z^2+L & y^2 & z^2+L \\ }\right| = $$ $$=(x^2+y^2+z^2+L)\left|\matrix{ 1 & y^2 & z^2 \\ 0 & L & 0\\ 0 & 0 & L \\ }\right|=L^2(x^2+y^2+z^2+L)$$

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  • $\begingroup$ Thank you very much i understand after working it out using your way. $\endgroup$ – haohao Nov 24 '13 at 7:00
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HINT: do reduced row reduction, and reduce it to an upper triangular matrix, so the determinant of the matrix is just the product of the 3 diagonal elements.

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