5
$\begingroup$

Photograph of worked example in textbook

I'm currently learning about finding determinant using row operations. This method requires the values below the main diagonal to all be zero. I'm looking at this example and I don't understand the last matrix. They have done r4 = 1/2r3 + r4. However, what if I had done r4 = 2r4 + r3 ? I would still get my desired zeros below the diagonals but my last value on my diagonal becomes -13 instead of -13/2. This changes my determinant result. Why is this happening? Can I not do r4 = 2r4 + r3?

$\endgroup$
3
  • $\begingroup$ When doing row operations, you're allowed to add multiples of one row to another. But that's not what you'd be doing in your proposal; instead, you'd be doubling the fourth row and then adding the third row to it. The second step in there doesn't change the determinant, but the first one does. (Replacing a row by a multiple of itself increases the determinant by that factor as well). Hence it's not permissible. $\endgroup$ – Semiclassical Jul 10 '17 at 4:11
  • $\begingroup$ @Semiclassical so in other words, when finding determinants using row operations, assuming I'm changing row Rx, I can't add a multiple of Rx to an other row to modify Rx itself? This is surprising to me because I use methodology like the one I proposed all the time when just solving matrices in Gaussian elimination methodology and I'd still get the correct values required. But for finding determinant, I can't use my proposition? $\endgroup$ – Compsci Jul 10 '17 at 4:18
  • $\begingroup$ Well, consider a simple case: Suppose the last row is 0002. If I'm doing Gaussian elimination, this corresponds to the equation $2x_4=0$. This is equivalent to $x_4=0$, which would correspond to the row 0001. But this plainly changes the determinant. So the operations which are permissible under Gaussian elimination won't be the same as which keep the determinant unchanged. (A rescaling operation, however, will never turn a nonzero determinant into a zero determinant.) $\endgroup$ – Semiclassical Jul 10 '17 at 4:25
10
$\begingroup$

There is a result that performing

$$r_{\color{blue}{i}}=r_{\color{blue}{i}}+kr_j$$

has no effect on the determinant.

This is different from performing

$$r_j = r_i + kr_j$$

If you perform

$$r_3 = r_3+ 2r_4$$

you do not change the determinant.

If you perform $$r_4 = 2r_4+r_3$$

What you are doing is actually first multiplying the $4$-th row by $2$ and then $r_4 = r_4+r_3$. Hence that is why your answer differs by a multiplication factor of $2$ from the correct answer.

$\endgroup$
5
$\begingroup$

Multiplying a row by a number other than $1$ changes the determinant. What you can do (and is done in your notes) is $R_4=R_4+\tfrac12\,R_3$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.