# A proof of determinant.?

Prove the following:

[ -bc , bc+b² , ba+c²]

[ ca+a² , -ca , ca+c²] = (ab+bc+ca)³

[ ab+a² , ab+b² , -ab ]

Oh.. yeah. . . I typed a mistake, It should be:

[-bc , bc+b² , bc+c²]

[ca+a² , -ca , ca+c²]

[ab+a² , ab+b², -ab ]

Sorry. . .

### 3 Answers

- 1 decade agoFavorite Answer
The statement is not true, but I suspect that you made a typo, because the statement *is* true if you change entry (1,3) from ba+c² to bc+c².

The matrix would look like this:

[-bc , bc+b² , bc+c²]

[ca+a² , -ca , ca+c²]

[ab+a² , ab+b², -ab ]

EDIT:

Well, in that case, here's how you could do it (it's probably not the only way):

I take it you know the determinant formula for a 3x3 matrix:

|[a, b, c]|

|[d, e, f ]| = aei+bfg+cdh-ceg-bdi-afh

|[g, h, i ]|

Use it on your matrix. Fully expand it and you'll get something like this:

a³b³ + 3a³b²c + 3a²b³c + 3a³bc² + 6a²b²c² + 3ab³c² + a³c³ + 3a²bc³ + 3ab²c³ + b³c³

Now fully expand (ab+bc+ca)³ and you'll get something like this:

a³b³ + 3a³b²c + 3a²b³c + 3a³bc² + 6a²b²c² + 3ab³c² + a³c³ + 3a²bc³ + 3ab²c³ + b³c³

It's the same!!! Yay!! Q.E.D. :)

- intc_escapeeLv 71 decade ago
Prove the following:

[ -bc , bc+b² , ba+c²]

[ ca+a² , -ca , ca+c²] = (ab+bc+ca)³

[ ab+a² , ab+b² , -ab ]

It's a fool's errand. Let c = 0 and it's obvious.

[ 0 , b² , ba]

[ a² , 0, 0]

[ ab+a² , ab+b² , -ab ]

The determinant of that matrix is easily calculated by Laplace expansion on row 2 as a²(2ab³ + a²b²)

a²(2ab³ + a²b²) ≠ (ab)³

Answer: see above

- Facts MatterLv 71 decade ago
I could solve this by brute force expansion, but so, I hope, could you.

There may be some elegant way of subtracting rows or columns, which of course leads the value of the determinant unchanged, and the form of the answer even suggests that if you are clever enough at doing this, you can diagonalise and be left with (ab + bc + ca) on the diagonals, but I'm not smart enough to do this.