Why can't I solve the determinant this way?

[flyingpig]

Homework Statement

Solve this determinant

\begin{vmatrix}<br /> 2 &amp; 5 &amp; 4 &amp; 1\\ <br /> 4&amp; 7 &amp; 6 &amp;2 \\ <br /> 6&amp; -2&amp; -4 &amp; 0\\ <br /> -6&amp; 7&amp; 7&amp;0 <br /> \end{vmatrix}

The Attempt at a Solution

I decided to get rid of the 1 at the corner of the first row by doing a row operation of

R₁ -> R₁ - (1/2)R₂

\begin{vmatrix}<br /> 0 &amp; 5 - \frac{7}{2} &amp; 1 &amp; 0\\ <br /> 4&amp; 7 &amp; 6 &amp;2 \\ <br /> 6&amp; -2&amp; -4 &amp; 0\\ <br /> -6&amp; 7&amp; 7&amp;0 <br /> \end{vmatrix}

So then I crossed the last column and second row and carried the +2 and I got
2\begin{vmatrix}<br /> 0 &amp; 5-\frac{7}{2} &amp; 1 \\ <br /> <br /> 6&amp; -2&amp; -4 \\ <br /> -6&amp; 7&amp; 7 <br /> \end{vmatrix}

So I row-reduced the second and third row

R₂ -> R₂ + R₃
R₃ -> R₃ + R₂

2\begin{vmatrix}<br /> 0 &amp; 5-\frac{7}{2} &amp; 1 \\ <br /> <br /> 0&amp; 5&amp; 3 \\ <br /> 0&amp; 5&amp; 3 <br /> \end{vmatrix}

Now I am stuck because my determinant is 0...

 

[micromass]

R₂ -> R₂ + R₃
R₃ -> R₃ + R₂

This is something you cannot do. In each step, you can only change one row (or column). You are changing two. Of course, when you do this, you change both row 2 and row 3 to thesame row, so you'll get linear dependent rows.

 

[LCKurtz]

2\begin{vmatrix}<br /> 0 &amp; 5-\frac{7}{2} &amp; 1 \\ <br /> <br /> 6&amp; -2&amp; -4 \\ <br /> -6&amp; 7&amp; 7 <br /> \end{vmatrix}

So I row-reduced the second and third row

R₂ -> R₂ + R₃
R₃ -> R₃ + R₂

2\begin{vmatrix}<br /> 0 &amp; 5-\frac{7}{2} &amp; 1 \\ <br /> <br /> 0&amp; 5&amp; 3 \\ <br /> 0&amp; 5&amp; 3 <br /> \end{vmatrix}

Now I am stuck because my determinant is 0...

You can't reduce the same two rows at the same time like that. First change row 2 to get a 0 and write your new determinant. Once you have done that you will see you can't get 0 in row three using row 2 anymore. One step at a time.

By the way, it is quite possible for a determinant to come out 0, but you haven't shown this one does.

 

[flyingpig]

No I checked the Key and it isn't 0 that's how I know lol

What do you think I can't row reduce like that? Why can't I keep zeroing?

 

[LCKurtz]

No I checked the Key and it isn't 0 that's how I know lol

What do you think I can't row reduce like that? Why can't I keep zeroing?

Did you try what I suggested? Get your zero in row 2 and write that determinant down. Then you will see why you can't get a zero in row 3. What you have done by simultaneously using two rows to change each other is not one of the row operations.

 

[flyingpig]

2\begin{vmatrix}<br /> 0 &amp; 5-\frac{7}{2} &amp; 1 \\ <br /> <br /> 6&amp; -2&amp; -4 \\ <br /> 0&amp; 5&amp; 3 <br /> \end{vmatrix}

Then I cross out the first column and the second row and I get

2(-6)\begin{vmatrix}<br /> 5-\frac{7}{2} &amp; 1 \\ <br /> 5&amp; 3\\ <br /> \end{vmatrix}

Which is still wrong...very wrong

Also, if I am not doing a row operation, then what is it? I am confused, am I not just applying the row reduction algorithm

 

[LCKurtz]

...
Which is still wrong...very wrong

Also, if I am not doing a row operation, then what is it? I am confused, am I not just applying the row reduction algorithm

Re-check your arithmetic; I did not check all of your steps. I would only comment that at your very first step, the best choice would have been to subtract 2*row 1 from row 2, thereby avoiding the need to introduce fractions.

To answer your last question, a row operation is one of the operations listed in your book that leaves the value of the determinant unchanged. The calculation you did is not one of them, and "what it is" is an error that changes the problem.