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I have a matrix and I'm supposed to find the determinant. I chose to use the method of row reduction into echelon form and then multiplication across the diagonal. I row reduce the matrix but the answer I get is not the same as what my calculator says. I've gone over this 5 times now, and I can't find where I'm making a mistake. I feel like I'm missing some property of elementary operations but I dont see it.

The matrix:

$\left[\begin{matrix}-1&-3&0&1\\3&5&8&-3\\-2&-6&3&2\\0&-1&2&1\end{matrix}\right]$

After doing these row operations, I'm left with the matrix:

$3R_1+R_2$ to replace row 2.

$-2R_1+R_3$ to replace row 3.

$-4R_4+R_2$ to replace row 4.

$\left[\begin{matrix}-1&-3&0&1\\0&-4&8&0\\0&0&3&0\\0&0&0&-4\end{matrix}\right]$

If you multiply down the diagonal you get -48 but the answer is supposed to be 12 according to my calculator. What am I doing wrong?

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  • $\begingroup$ Are we to assume that the row $3R_1+R_2$ replaces $R_2$, and so on? $\endgroup$ – Rory Daulton Mar 8 '15 at 22:52
  • $\begingroup$ Yes, sorry I will edit. $\endgroup$ – Sabien Mar 8 '15 at 22:53
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Your last operation "$-4R_4+R_2$ to replace row 4" is wrong. When you add a multiple of a row to another row, that must replace the other row. You should replace row 2 in that operation for it to be valid, not row 4. Of course, that does not give you a triangular matrix, so the correct operation is

$-\frac 14R_2+R_4$ to replace $R_4$.

For more details on how row operations affect the determinant, you can read this.

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  • $\begingroup$ So when using row replacement, you can only multiply a row that you are not replacing and add it to the row you're replacing? Am I interpreting that correctly? $\endgroup$ – Sabien Mar 8 '15 at 23:00
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    $\begingroup$ @Sabien: Yes, that is correct. That is what you did in your first two row operations. $\endgroup$ – Rory Daulton Mar 8 '15 at 23:00
  • $\begingroup$ Thanks, that explains why I've been getting a lot of other determinants wrong. I get it right sometimes and wrong others, and could never figure it out. I was missing that detail. Will mark as answer when timer allows. $\endgroup$ – Sabien Mar 8 '15 at 23:02
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    $\begingroup$ @Sabien It may help to know that you can always verify your row reductions as seen here. $\endgroup$ – Daniel W. Farlow Mar 8 '15 at 23:05
  • $\begingroup$ @crash: That method seems to give only reduced row echelon forms, which always have a determinant of one or zero. Is there a way to get an upper triangular matrix with the same determinant at WolframAlpha? $\endgroup$ – Rory Daulton Mar 8 '15 at 23:07
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The operations contribute to the determinant as well.

The following operations correspond to multiplying on the left by the following matrices:

  • $3R_1 + R_2$ to replace row 2: $M_1 = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 3 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix}$, which has determinant 1
  • $ -2R_1 + R_3$ to replace row 3: $M_2 = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -2 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix}$, which has determinant 1
  • $ -4R_4 + R_2$ to replace row 4: $M_3 = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & -4\end{pmatrix}$, which has determinant -4

If $A$ was your original matrix and $T$ is the upper-triangular matrix you obtained at the end, then

$$ T = M_3 M_2 M_1 A$$

so \begin{align} \det T &= \det M_3 \cdot \det M_2 \cdot \det M_1 \cdot \det A \\ &= (-4)\cdot(1)\cdot(1)\cdot\det A, \end{align}

thus $\det A = \frac{\det T}{-4} = 12$.

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