[Math] Finding determinant of 4×4 by using echelong form and multiplying across diagonal

linear algebramatrices

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?

Best Answer

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.