How Does Changing Rows Affect a 4x4 Matrix Determinant?

[snoggerT]

If a 4X4 matrix A with rows v1, v2, v3, and v4 has determinant det A = -8

then det
|(4v1+3v4)|
|___v2___|
|___v3___| =?
|(8v1+9v4)|

- I put the underscores for spacing.

The Attempt at a Solution

- I haven't attempting anything on this problem. There are no examples even remotely like this in the text and we haven't been shown anything like this in class, so I really have no clue how to even start. Please help.

 

[HallsofIvy]

Use "row reduction". Subtract 2 times the first row from the second row. Do you know what effect row reduction of a matrix has on a determinant.

 

[snoggerT]

use row reduction on the 1X4 determinant given?

 

[HallsofIvy]

There is no such thing as a "one by four" determinant! Every determinant, by definition, must be "square". There exist "one by four", or other dimension, matrices but only square matrices have determinants. You said yourself "a 4X4 matrix A with rows v1, v2, v3, and v4". The "rows" v1, v2, v3, and v4 must be 4 dimensional vectors. Subtracting twice the first row from the second row gives you a matrix with rows 8v1+ 3v4, v2, v3, 3v4. Subtracting the new third row from the first row gives a matrix with rows 8v1, v2, v3, 3v4. Dividing the first row by 8 gives a matrix with rows v1, v2, v3, 3v4. Finally, dividing the fourth rwo by 3 gives a matrix with rows v1, v2, v3, v4. How did those row operations change the value of the determinant of that matrix? If the final result is -8, what must the original determinant have been?

 

[snoggerT]

Maybe I'm reading the way they have the determinant setup wrong. I was seeing it as 4v1+3v4 as row 1 and v2 as row 2 and v3 as row3 and and 8v1+9v4 as the 4th row (because I was viewing the determinant as a matrix I suppose). What is the correct way to read it? That might help a lot in me understanding the problem.

 

[snoggerT]

This is where I'm confused. I can use row reduction on the new matrix and it is the same as the original (v1,v2,v3,v4), but I know the determinant isn't the same. How does that relate them?

 

[HallsofIvy]

Yes, that is exactly right. And since v1, v2, v3, v4 have 4 members, this is a 4 by 4 determinant. For example, if v1= <1, 2, 0, 1> and v4= <-1, 0, 2, 1> are the top and bottom rows of the original matrix, then 4v1+ 3v4= <4, 8, 0, 4>+ <-3, 0, 6, 3>= <1, 8, 6, 7> and 8v1+ 9v4= <8, 16, 0, 8>+ <-9, 0, 18, 9>= <-1, 15, 18, 17> are the top and bottom rows of the new matrix, and of its determinant.

Here's one way to "cheat". Since the problem clearly expects a single answer, just set up a simple matrix, A, that has determinant -8.
A= \left(\begin{array}{cccc} 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; -2 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 4 &amp; 0\\ 0 &amp; 0 &amp; 0 &amp; 1\end{array}\right)
where v1= <1, 0, 0, 0>, v2= <0, -2, 0, 0>, v3= <0, 0, 4, 0>, and v4= < 0, 0, 0, 1> will do nicely.
Now, 4v1+ 3v4= <4, 0, 0, 3> and 8v1+ 9v4= <8, 0, 0, 9>.

What is the determinant of
A= \left|\begin{array}{cccc} 4 &amp; 0 &amp; 0 &amp; 3 \\ 0 &amp; -2 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 4 &amp; 0\\ 8 &amp; 0 &amp; 0 &amp; 9\end{array}\right|?

That's "cheating" because it assumes there is one single answer to this question and uses an example to find that one answer.

 

[HallsofIvy]

You stuck in a message while I was responding to the previous one.

That's probably precisely what the problem is testing: your knowledge of how row operations affect the determinant of a matrix.

If you swap two rows in a matrix, you multiply the determinant by -1.

If you subtract a multiple of one row from another the determinant is not changed!

If you multiply an entire row by a number, the determinant is multiplied by that number.

 

[snoggerT]

the determinant is -96 using the "cheating" method. Can you explain to me what row operations are being performed in this example? I'm trying to understand this before I move on to the vector space problems. Obviously rows are being multiplied by a number, but also being added to each other, so I'm not sure which rules that would apply to.