Cofactor Expansion: Find Determinant of 4x3 Matrix

  • Thread starter Thread starter fk378
  • Start date Start date
  • Tags Tags
    Expansion
fk378
Messages
366
Reaction score
0

Homework Statement


Find the determinant of
4...3...-5
5...2...-3
0..-1...2



The Attempt at a Solution


I've tried to get the answer using each of the 3 rows and each time I get a different answer. For the first row I get 14, the second row I get -31, and the third row I get -1. However, I thought that the determinant would be the same value for whichever row (or column) you choose to expand.
 
Physics news on Phys.org
fk378 said:

Homework Statement


Find the determinant of
4...3...-5
5...2...-3
0..-1...2



The Attempt at a Solution


I've tried to get the answer using each of the 3 rows and each time I get a different answer. For the first row I get 14, the second row I get -31, and the third row I get -1. However, I thought that the determinant would be the same value for whichever row (or column) you choose to expand.
Yes, it certainly should be!

Expanding by the first column (since it has only 2 non-zero entries):
\left|\begin{array}{ccc}4 & 3 & -5 \\ 6 & 2 & -3 \\ 0 & -1 & 2 \end{array}\right|= 4\left|\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\left|- 5\right|\begin{array}{cc} 3 & -5 \\ -1 & 2\end{array}\right|
4(4- 3)- 5(6- 5)= 4(1)- 5(1)= -1
The determinant is -1. I notice that is what you got using the third row which also has only two non-zero entries. Perhaps it is that third entry that is confusing you.
 
But if you try to work out the determinant using the other rows that do not contain the 0 entry, it does not come out to -1! Why?
 
Ah, I just realized I wasn't multiplying one of the terms with the cofactor expansion! I feel silly...thanks for your help though!
 
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