Factoring Determinant of 3x3 Matrix - Step by Step Solution

takercena
Messages
25
Reaction score
0

Homework Statement


Code:
| x       x^2   1  |
| x^2      x    1  |
| x^3    x^3   1  |

The answer is = x^2(1-x)^2(1+2x)

Homework Equations



none

The Attempt at a Solution


By factoring the matrix step by step.

Thanks :)
 
Physics news on Phys.org
What's the problem here? Just use co-factor expansion along any row or column.
 
Take advantage of the fact that you have a column of ones in the third column...

\begin{vmatrix}<br /> x &amp; x^2 &amp; 1 \\<br /> x^2 &amp; x &amp; 1 \\<br /> x^3 &amp; x^3 &amp; 1<br /> \end{vmatrix}<br /> = <br /> \begin{vmatrix}<br /> x^2 &amp; x \\<br /> x^3 &amp; x^3 <br /> \end{vmatrix} -<br /> \begin{vmatrix}<br /> x &amp; x^2 \\<br /> x^3 &amp; x^3 <br /> \end{vmatrix} +<br /> \begin{vmatrix}<br /> x &amp; x^2 \\<br /> x^2 &amp; x <br /> \end{vmatrix}<br />

So what does this end up being?
 
Sorry for the question. I was confused by the question a moment ago because of language barrier.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top