# Prove that LHS=RHS using ONLY properties of Det.

• Feb 20th 2010, 08:58 PM
romit
Prove that LHS=RHS using ONLY properties of Det.
Q) Without directly expanding the det., but using only the well-known properties, prove that:
http://www.mathhelpforum.com/math-he...1&d=1266731073
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Tried solving many times but couldn't find a proper method/approach. Keep getting stuck every time...

• Feb 21st 2010, 12:31 PM
Opalg
Let A denote the matrix $\displaystyle \begin{bmatrix}-bc&b^2+bc&c^2+bc\\ a^2+ac&-ac&c^2+ac\\ a^2+ab&b^2+ab&-ab\end{bmatrix}$. The eigenvalues of A are the roots of the equation $\displaystyle \begin{vmatrix}-bc-\lambda&b^2+bc&c^2+bc\\ a^2+ac&-ac-\lambda&c^2+ac\\ a^2+ab&b^2+ab&-ab-\lambda\end{vmatrix} = 0$. If we try putting $\displaystyle \lambda = -(ab+bc+ca)$, then that equation becomes $\displaystyle \begin{vmatrix}a(b+c)&b(b+c)&c(b+c)\\ a(a+c)&b(a+c)&c(a+c)\\ a(a+b)&b(a+b)&c(a+b)\end{vmatrix} = 0$. The vector (x,y,z) will be an eigenvector for $\displaystyle \lambda$ if $\displaystyle \begin{bmatrix}a(b+c)&b(b+c)&c(b+c)\\ a(a+c)&b(a+c)&c(a+c)\\ a(a+b)&b(a+b)&c(a+b)\end{bmatrix}\begin{bmatrix}x\ \y\\z\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}$. But that system of equations reduces to $\displaystyle ax+by+cz=0$, which has two linearly independent solutions. Therefore $\displaystyle \lambda = -(ab+bc+ca)$ is indeed an eigenvalue of A, with multiplicity at least 2. But the sum of the three eigenvalues of A is equal to the trace of A, which is $\displaystyle -(ab+bc+ca)$. Therefore the third eigenvalue must be $\displaystyle ab+bc+ca$. Finally, the determinant of A is the product of its eigenvalues, namely $\displaystyle (ab+bc+ca)^3$.

I don't know if that counts as a proof "using ONLY properties of Det.", but at least it didn't involve expanding the determinant.
• Feb 22nd 2010, 01:54 AM
romit
Thanks for reply Opalg, much appreciated!... Using eigenvalues will work to(Nod). Main thing, was not to use expansion(Angry).

BTW "properties of determinants" mean manipulation of the det. by basic math functions (+-/*), changing of rows/columns/etc...

Yeah! got it, after a fresh start...

$\displaystyle L.H.S.= \begin{bmatrix} -bc&b^2+bc&c^2+bc\\ a^2+ac&-ac&c^2+ac\\ a^2+ab&b^2+ab&-ab\\ \end{bmatrix}$

$\displaystyle =1/abc \begin{bmatrix} -abc&ab^2+abc&ac^2+abc\\ a^2b+abc&-abc&bc^2+abc\\ a^2c+abc&b^2c+abc&-abc\\ \end{bmatrix}$

$\displaystyle =abc/abc \begin{bmatrix} -bc&ab+ac&ac+ab\\ ab+bc&-ac&bc+ab\\ ac+bc&bc+ac&-ab\\ \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -bc+ab+bc+ac+bc&ab+ac-ac+bc+ac&ac+ab+bc+ab-ab\\ ab+bc&-ac&bc+ab\\ ac+bc&bc+ac&-ab\\ \end{bmatrix}$

$\displaystyle = \begin{bmatrix} ab+bc+ca&ab+bc+ca&ab+bc+ca\\ ab+bc&-ac&bc+ab\\ ac+bc&bc+ac&-ab\\ \end{bmatrix}$

$\displaystyle = (ab+bc+ca) \begin{bmatrix} 1&1&1\\ ab+bc&-ac&bc+ab\\ ac+bc&bc+ac&-ab\\ \end{bmatrix}$

$\displaystyle = (ab+bc+ca) \begin{bmatrix} 0&1&1\\ 0&-ac&bc+ab\\ ac+bc+ca&bc+ac&-ab\\ \end{bmatrix}$

$\displaystyle = (ab+bc+ca)^2 \begin{bmatrix} 1&1\\ -ac&bc+ab\\ \end{bmatrix}$

$\displaystyle = (ab+bc+ca)^2 \begin{bmatrix} 1&0\\ -ac&ab+bc+ca\\ \end{bmatrix}$

$\displaystyle = (ab+bc+ca)^3$

$\displaystyle =R.H.S. [Proved]$
• Feb 22nd 2010, 03:54 AM
Opalg
Quote:

Originally Posted by romit
Yeah! got it, after a fresh start...

Nice proof! (Happy)