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I understand that you can find the determinant of a matrix along it's diagonal if it is in triangular form. For a matrix such as this: $$ \begin{pmatrix} 1 & 5 & 0\\ 2 & 4 & -1\\ 0 &-2 & 0 \end{pmatrix} $$ When put into triangular form I get: $$ \begin{pmatrix} 1 & 5 & 0\\ 0 & 1 & 1/6\\ 0 & 0 & 1/3 \end{pmatrix} $$ Since I multiplied row two by -1/6 during the row reduction I would expect the determinant to be $$ 1\cdot 1\cdot 1/3\cdot (-1/6),$$ but the answer for the determinant of the original matrix is -2. Where exactly am I going wrong?

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  • $\begingroup$ Write your row operations as a matrix product and use the multiplication rule for determinants. Then you will minimize your risk of mistakes. $\endgroup$ – mathreadler Dec 12 '17 at 16:30
  • $\begingroup$ You should multiply by $-6$ rather than $-1/6$ $\endgroup$ – eepperly16 Dec 12 '17 at 16:31
  • $\begingroup$ @mathreadler I understand the multiplication rule, but was just wondering why this scenario did not work when trying to solve using a triangular matrix $\endgroup$ – Pulse Dec 12 '17 at 16:32
  • $\begingroup$ it will work if you do it right. there is a larger chance you will do it right if you write it down step by step with matrix products $\endgroup$ – mathreadler Dec 12 '17 at 16:33
  • $\begingroup$ @eepperly16 I multiplied by -1/6 because during the row operation, row 2 I multiplied by -1/6 to make the 6 a 1 in that row. $\endgroup$ – Pulse Dec 12 '17 at 16:33
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You can add a multiple of any row to another row without changing the determinant. So I'll do these kinds of operations to get to the triangular form.

Start with

$$\begin{matrix} 1 & 5 & 0 \\ 2 & 4 & -1 \\ 0 &-2 &0 \end{matrix}$$

Subtract twice the first row from the second:

$$\begin{matrix} 1 & 5 & 0 \\ 0 & -6 & -1 \\ 0 &-2 &0 \end{matrix}$$

Subtract one-third the new second row from the third:

$$\begin{matrix} 1 & 5 & 0 \\ 0 & -6 & -1 \\ 0 & 0 & \frac{1}{3} \end{matrix}$$

The product along the diagonal is $-2$.

Pulling out a factor from one of the rows, however, does change the determinant by that factor. You pulled out $-1/6$ from the second row, so the calculated determinant from the diagonal in your question statement was low by this factor.

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  • $\begingroup$ This really does not help with the understanding as you have not shown that the operations you perform don't change the determinant. $\endgroup$ – mathreadler Dec 12 '17 at 16:38
  • $\begingroup$ @mathreadler Added a couple of sentences at the beginning to address your concern. $\endgroup$ – John Dec 12 '17 at 16:41
  • $\begingroup$ @mathreadler Was there a problem with my approach to putting it in triangular form? I took the approach of how I would usually go about putting a matrix into reduced echelon form. Is there some property that I may be missing? $\endgroup$ – Pulse Dec 12 '17 at 16:43
  • $\begingroup$ The matrix you ended up with is the same as mine, except your second row is multiplied by $-1/6$. "Pulling out" that constant changes the determinant by the same factor: $-2 \times -1/6 = 1/3$, which was what you got. $\endgroup$ – John Dec 12 '17 at 16:46
  • $\begingroup$ You need to keep track of how the operations you apply may change the determinant. $\endgroup$ – mathreadler Dec 12 '17 at 16:48

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