Prove that LHS=RHS using ONLY properties of Det.

**romit** · February 20th 2010, 08:58 PM

Q) Without directly expanding the det., but using only the well-known properties, prove that:

__________________________________________________ ________

Tried solving many times but couldn't find a proper method/approach. Keep getting stuck every time...

Please help guys... Thanks in advance!

**Opalg** · February 21st 2010, 12:31 PM

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

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.

**romit** · February 22nd 2010, 01:54 AM

Thanks for reply Opalg, much appreciated!... Using eigenvalues will work to

. Main thing, was not to use expansion

.

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...

$ 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} $

$ =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} $

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

$ = \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} $

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

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

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

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

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

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

$ =R.H.S. [Proved] $

**Opalg** · February 22nd 2010, 03:54 AM

Originally Posted by romit

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

Nice proof!

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