Anonymous

# 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 ]

Update:

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. . .

Relevance

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. :)

• 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