Use row operations to find the determinant:

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Can someone give me a full answer please?

Also can anyone tell me if the sign of the determinant matters ?

Row operations :

Det ( e(A) ) = -det(A) ; if e is Ri interchanged with Rj

Det ( e(A) ) = `Cdet(A) ; if e is CRi where C not equal to 0

Det( e(A) ) = det(A) ; if e is cRi + Rj

Where e i s a row operation.

  • $\begingroup$ The sign of the determinant doesn't matter to me --- nothing about this determinant matters to me --- but it might matter to someone else. It depends on what you have planned for the determinant, that is, on why you are interested in the determinant in the first place. So, why are you interested in this determinant? $\endgroup$ – Gerry Myerson Nov 22 '14 at 5:35
  • $\begingroup$ @Amzoti I am stuck on deciding how to subtract the rows. In my answer I said R1 - R2, then I interchanged R2 and R3. So it left me with (-1) |3 1 2 ; 0 1 4 ; 0 0 2| and the determinant of this gives -6. However my lecturer did it as : -R1 + R2, then interchanged R2 and R3 and was left with (-1)|3 1 2 ; 0 1 4 ; 0 0 -2| which gave the determinant as 6. Is it that we have to subtract rows only by one method ? $\endgroup$ – Izzy92 Nov 22 '14 at 6:20
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    $\begingroup$ Consider the row operation R1-R2. If you replace R1 by R1-R2, the sign of the determinant does not change, because you did not change the sign of R1. But, what you did was to replace R2 by R1-R2, which changed the sign of the determinant. In effect, you multiplied R2 by negative one, and then added another row to it. Your lecturer actually replaced R2 with R2-R1, which did not change the sign of the determinant. Then she/he swapped two rows, which did change the sign of the determinant. $\endgroup$ – LouisB Nov 22 '14 at 10:41
  • $\begingroup$ Thanks louis, though I had to read your response a second time, your'e right. $\endgroup$ – Izzy92 Nov 22 '14 at 15:46

Perform row operations to get your matrix into upper triangular form. Then the determinant is the product of the main diagonal.

  • $\begingroup$ If you do any row swapping you have to multiply by $-1$ for each row swap. $\endgroup$ – Laars Helenius Nov 22 '14 at 15:52

When you perform $\mathrm R_1-\mathrm R_2$, the row $\mathrm R_1$ changes, the other two rows remain the same. $$\begin{vmatrix}3&1&2\\3&1&0\\0&1&4\end{vmatrix}$$ $\mathrm R_1-\mathrm R_2$ gives $$\begin{vmatrix}0&0&2\\3&1&0\\0&1&4\end{vmatrix}$$ When you interchange rows $\mathrm R_2$ and $\mathrm R_3$, the sign of the determinant changes. $$(-1)\begin{vmatrix}0&0&2\\0&1&4\\3&1&0\end{vmatrix}$$ For the row operations method, since you need to have an upper triangular matrix, you can exchange columns $\mathrm C_1$ and $\mathrm C_3$. In this step, just like the case of rows, the sign of the determinant would change again.

Interchanging columns $\mathrm C_1$ and $\mathrm C_2$ gives $$(+1)\begin{vmatrix}2&0&0\\4&1&0\\0&1&3\end{vmatrix}$$

And now, the product of the diagonal elements gives $6$

  • $\begingroup$ Isn't the determinant the product of all the diagonal elements of a n by n triangular matrix i.e (3)(1)(2) and recall we have that (-1) outside therefore the determinant will be (-1)*(3)(1)(2) = -6 ? $\endgroup$ – Izzy92 Nov 22 '14 at 15:12
  • $\begingroup$ I'm not aware of finding the determinant by that method. How I would open the original determinant of your question is: $3[(1)(4)-(0)(1)]-1[(3)(4)-(0)(0)]+2[(3)(1)-(1)(0)]=3(4)-1(12)+2(3)=6$ $\endgroup$ – Tejas Nov 22 '14 at 15:19
  • $\begingroup$ that's the laplace method to find the determinant. I was looking for the row operation method. You kinda started of the way i was looking for by saying when you interchanged you will get a (-1) in front of the determinant. Also yea, the multiplication of the triangular elements should give you the determinant. I would have referred you to some literature on that method but I am pressed for time at the moment. $\endgroup$ – Izzy92 Nov 22 '14 at 15:39
  • $\begingroup$ Hmm.. Maybe if you could include a brief explanation of the row operation method in your question, I might be able to edit my answer to suit the method. :) $\endgroup$ – Tejas Nov 22 '14 at 15:43
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    $\begingroup$ I edited it.. :) $\endgroup$ – Tejas Nov 23 '14 at 14:11

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