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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.

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  • $\begingroup$ 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. $\endgroup$
    – Math Lover
    Sep 24, 2018 at 23:26
  • $\begingroup$ $$\begin{vmatrix}4&-7&9&1\\6&2&7&0\\-9&27&-30&0\\4&0&13&0\\\end{vmatrix}$$ Do I swap column 4 and 1 and begin normal row operations from there? $\endgroup$
    – Charlatan
    Sep 24, 2018 at 23:36
  • $\begingroup$ You don't have to swap whatever: you can expand the determinant along the 4th column staring with a $-$ sign). $\endgroup$
    – Bernard
    Sep 24, 2018 at 23:45
  • $\begingroup$ I thought the problem was asking me to not use cofactor expansion so I was trying to avoid expanding $\endgroup$
    – Charlatan
    Sep 24, 2018 at 23:48

1 Answer 1

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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

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  • $\begingroup$ was there a specific reason for you to swap R1 and R2? I did something similar but without swapping and got really messy numbers; i.e. what about R2 stuck out that you decided to swap it for R1? $\endgroup$
    – Charlatan
    Sep 25, 2018 at 0:43
  • $\begingroup$ was it because the last element of the column was 0 and it simplified adding rows? that was about my best guess why after completing the row operations $\endgroup$
    – Charlatan
    Sep 25, 2018 at 1:01
  • $\begingroup$ @Charlatan Yes that was one of the reasons and you can change the rows and columns as you like to make the calculations easier $\endgroup$
    – Key Flex
    Sep 25, 2018 at 1:24

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