Determinant of 4 x 4 matrix

[__SB]

Hi all, I was hoping somebody could help me find the determinantof a 4 x 4 matrix.
I tried to make the numbers below the main diagonal all 0’sso the determinant is just the product of the main diagonal. However my answeris way off.
On a side note (if I can ever get this method to work), canyou do e.g. 2R1 + 3R2 or can you only multiply one row by a scalar?
Thanks in advance.

 

[__SB]

Apologies that my photo is horizontal - if anybody know why it rotates it after importing please shed light.

 

[ksdhart]

The first thing I note is that your work is extremely difficult to follow as you never state what transformations you make in between steps. Here's what I think the first few steps you took are... for brevity, I'll use R# to refer to Row number (whatever). You started with:

\(\displaystyle \begin{pmatrix}3&0&1&2\\6&1&-1&2\\-1&1&0&0\\5&2&0&3\end{pmatrix}\)

First, you subtracted 2*R1 from R2. Then I think you multiplied R3 by 3, and then added R1, leaving:

\(\displaystyle \begin{pmatrix}3&0&1&2\\0&1&-3&-2\\0&3&1&2\\5&2&0&3\end{pmatrix}\)

But after this I'm not sure what you did to get from the above matrix to the next one you wrote on the paper. Can you please tell us, in words, what row operations you performed?

\(\displaystyle \begin{pmatrix}3&0&1&2\\0&1&-3&-2\\0&3&1&2\\0&7&0&3\end{pmatrix}\)

 

[Ishuda]

...On a side note (if I can ever get this method to work), canyou do e.g. 2R1 + 3R2 or can you only multiply one row by a scalar? ...

Review the 'rules' here
http://www.utdallas.edu/dept/abp/PDF_Files/LinearAlgebra_Folder/Determinants.pdf

If you were to replace R1 by 2 R1 + 3 R2, the would be like multiplying R1 by 2 [so you need to divide the final answer by 2] and then adding a multiple of a row to the new R1, i.e. add 3 R2 to the new R1. That would not change the determinate.

 

[__SB]

The first thing I note is that your work is extremely difficult to follow as you never state what transformations you make in between steps. Here's what I think the first few steps you took are... for brevity, I'll use R# to refer to Row number (whatever). You started with:

\(\displaystyle \begin{pmatrix}3&0&1&2\\6&1&-1&2\\-1&1&0&0\\5&2&0&3\end{pmatrix}\)

First, you subtracted 2*R1 from R2. Then I think you multiplied R3 by 3, and then added R1, leaving:

\(\displaystyle \begin{pmatrix}3&0&1&2\\0&1&-3&-2\\0&3&1&2\\5&2&0&3\end{pmatrix}\)

But after this I'm not sure what you did to get from the above matrix to the next one you wrote on the paper. Can you please tell us, in words, what row operations you performed?

\(\displaystyle \begin{pmatrix}3&0&1&2\\0&1&-3&-2\\0&3&1&2\\0&7&0&3\end{pmatrix}\)

Thanks for the response.
From the matrix you have shown, I did:
Replace Row 3 with : -3R2 + R3
Replace Row 4 with: -7R2 + R4

 

[ksdhart]

Right, okay, moving forward from the last matrix I posted, I agree that those would be good steps. However, what I was asking about was how did you get from the second matrix in my post to the third? Specifically, you appear to only be modifying the first and second entries in the row, but the last two do not change, which is why I'm confused as to what row operations you performed. I cannot think of any row operations that would produce the desired transformation. I apologize if I was unclear about what I was asking before.

 

[__SB]

Right, okay, moving forward from the last matrix I posted, I agree that those would be good steps. However, what I was asking about was how did you get from the second matrix in my post to the third? Specifically, you appear to only be modifying the first and second entries in the row, but the last two do not change, which is why I'm confused as to what row operations you performed. I cannot think of any row operations that would produce the desired transformation. I apologize if I was unclear about what I was asking before.

Apologies; I’ll start again to make it clear. I reallyappreciate your patience here.
My Row Operations:
1)Replace R2 with: R2 – 2R1
2)Replace R4 with: R4 + 5R3
3)Replace R3 with: R1 + 3R3
4)Replace R3 with: R3 – 3R2
5)Replace R4 with: R4 – 7R2

 

[ksdhart]

Alright, now I understand what you did. Thank you. Every thing you've done checks out. After your last step, you're left with this matrix:

\(\displaystyle \begin{pmatrix}3&0&1&2\\0&1&-3&-2\\0&0&10&8\\0&0&21&17\end{pmatrix}\)

From here, I would let R4 = 10*R4 - 21R3, leaving:

\(\displaystyle \begin{pmatrix}3&0&1&2\\0&1&-3&-2\\0&0&10&8\\0&0&0&2\end{pmatrix}\)

Now, this matrix has determinant equal to it's main diagonal multiplied together, or 60. And as you note, this doesn't connect with the expected result of the determinant being 2. However, you've forgotten one thing, which Ishuda has already pointed out. In your workings, you say:

"3)Replace R3 with: R1 + 3R3"

In this step, you've multiplied R3 by 3, which will multiply the determinant by 3. Therefore, to get the determinant of the original matrix, you must divide by 3. And similarly, in my final step above, I multiplied R4 by 10, which multiplies the determinant by 10. Thus, we take the final determinant of our new matrix (60), and divide by 3 then divide by 10. That leaves us with 2, exactly as expected.

 

[__SB]

Alright, now I understand what you did. Thank you. Every thing you've done checks out. After your last step, you're left with this matrix:

\(\displaystyle \begin{pmatrix}3&0&1&2\\0&1&-3&-2\\0&0&10&8\\0&0&21&17\end{pmatrix}\)

From here, I would let R4 = 10*R4 - 21R3, leaving:

\(\displaystyle \begin{pmatrix}3&0&1&2\\0&1&-3&-2\\0&0&10&8\\0&0&0&2\end{pmatrix}\)

Now, this matrix has determinant equal to it's main diagonal multiplied together, or 60. And as you note, this doesn't connect with the expected result of the determinant being 2. However, you've forgotten one thing, which Ishuda has already pointed out. In your workings, you say:

"3)Replace R3 with: R1 + 3R3"

In this step, you've multiplied R3 by 3, which will multiply the determinant by 3. Therefore, to get the determinant of the original matrix, you must divide by 3. And similarly, in my final step above, I multiplied R4 by 10, which multiplies the determinant by 10. Thus, we take the final determinant of our new matrix (60), and divide by 3 then divide by 10. That leaves us with 2, exactly as expected.

I get it! Thank you so much for your help. I never realised you had to divide; we were taught to use row reduction but that methods rather cumbersome leaving you with 3 3x3 to do so I thought this method looked better.

Thanks again - help is much appreciated