# How to find the determinant using elementary row or column operations

I have the matrix:

$$\begin{vmatrix}4&-7&9&1\\6&2&7&0\\3&6&-3&3\\0&7&4&-1\end{vmatrix}$$

Does anyone see an easy move to eliminate for a diagonal? I tried factoring 3 out of row 3 and then solving via elementary row operations but I end up with fractions that make it really difficult to properly calculate.

• Don't take $3$ out of the third row. Rather add the first row to the last one and subtract $3$ times the first row from the third row to get a very nice $4$th column. Sep 24, 2018 at 23:26
• $$\begin{vmatrix}4&-7&9&1\\6&2&7&0\\-9&27&-30&0\\4&0&13&0\\\end{vmatrix}$$ Do I swap column 4 and 1 and begin normal row operations from there? Sep 24, 2018 at 23:36
• You don't have to swap whatever: you can expand the determinant along the 4th column staring with a $-$ sign). Sep 24, 2018 at 23:45
• I thought the problem was asking me to not use cofactor expansion so I was trying to avoid expanding Sep 24, 2018 at 23:48

Given $$\begin{vmatrix} 4 & -7 & 9 & 1 \\ 6 & 2 & 7 & 0 \\ 3 & 6 & -3 & 3 \\ 0 & 7 & 4 & -1 \end{vmatrix}_{R_1<->R_2}$$ $$\begin{vmatrix} 6 & 2 & 7 & 0 \\ 4 & -7 & 9 & 1 \\ 3 & 6 & -3 & 3 \\ 0 & 7 & 4 & -1 \end{vmatrix}_{R_2\rightarrow R_2-\frac23 R_1}$$ $$\begin{vmatrix} 6 & 2 & 7 & 0 \\ 0 & -\frac { 25 }{ 3 } & \frac { 13 }{ 3 } & 1 \\ 3 & 6 & -3 & 3 \\ 0 & 7 & 4 & -1 \end{vmatrix}_{R_3\rightarrow R_3-\frac12R_1}$$ $$\begin{vmatrix} 6 & 2 & 7 & 0 \\ 0 & -\frac { 25 }{ 3 } & \frac { 13 }{ 3 } & 1 \\ 0 & 5 & -\frac { 13 }{ 2 } & 3 \\ 0 & 7 & 4 & -1 \end{vmatrix}_{R_3\rightarrow R_3+\frac35R_2}$$ $$\begin{vmatrix} 6 & 2 & 7 & 0 \\ 0 & -\frac { 25 }{ 3 } & \frac { 13 }{ 3 } & 1 \\ 0 & 0 & -\frac { 39 }{ 10 } & \frac { 18 }{ 5 } \\ 0 & 7 & 4 & -1 \end{vmatrix}$$ and so on.... till you get $$0's$$ in the first $$3$$ columns of the last row and then you will get the value of the determinant