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4x4 matrix determinant 
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#1
Nov1306, 08:06 AM

P: 43

Use row and/or column operations to simplify the determinant of the following matrix A, by reduction to upper triangular form, then evaluate.
[tex]A = \left(\begin{array}{cccc} 2 & 3 & 4 & 5\\ 0 & 1 & 2 & 1\\ 0 & 0 & 2 & 4\\ 0 & 3 & 6 & 0 \end{array} \right)[/tex] Is there an simpler way to find the determinant so that I don't have to expand cofactors etc? Because it would be: 2(3x3 matrix)  3(3x3 matrix + 4(3x3 matrix)  5(3x3 matrix) and then I have to find the determinants of each 3x3 matrix... 


#2
Nov1306, 08:15 AM

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P: 3,224

Yes, there is a simpler way, and the text of the problem itself suggests which way it is. Investigate row and column operations and try to solve the problem.



#3
Nov1306, 08:21 AM

P: 43

I have reduced to this:
[tex]A = \left(\begin{array}{cccc} 1 & 1.5 & 2 & 2.5\\ 0 & 1 & 2 & 1\\ 0 & 0 & 1 & 2\\ 0 & 0 & 0 & 1 \end{array} \right)[/tex] How would I determine the determinant using this? 


#4
Nov1306, 08:25 AM

HW Helper
P: 3,224

4x4 matrix determinant
Good work. You can evaluate the determinant by using the fourth row now.



#5
Nov1306, 08:48 AM

P: 43

The determinant would be 1.
However, the answer is listed as 12, and the second part is to check my answer by expanding the third row of the determinant. I'm at a loss as to what to do now. 


#6
Nov1306, 08:56 AM

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#7
Nov1306, 07:23 PM

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PF Gold
P: 39,363

You had better have kept track of your divisors while row reducing. Dividing one row by a number divides the determinant by that number.
Swapping two rows multiplies the determinant by 1. Add a multiple of one row to another does not change the determinant. Starting with [tex]\left(\begin{array}{cccc}2 & 3 & 4 & 5\\0 & 1 & 2 & 1\\0 & 0 & 2 & 4\\0 & 3 & 6 & 0\end{array}\right)[/tex] since the first column is already 0's (except for the first row), start on the second column: add 3 times the second row to the fourth row to get [tex]\left(\begin{array}{cccc}2 & 3 & 4 & 5\\0 & 1 & 2 & 1\\0 & 0 & 2 & 4\\0 & 0 & 0& 3\end{array}\right)[/tex] Now we have an upper triangular matrix: it's determinant is just the product of the numbers on the main diagonal:2(1)(2)(3)= 12. Since the only row operation used was "add a multiple of one row to another", that is the determinant of the original matrix. 


#8
Nov1306, 10:03 PM

P: 43

I realised that I reduced to upper triangular row echelon form rather than just upper triangular form.
The next part asks me to check this determinant by expanding the third row. How do I go about expanding it? Would it be using cofactors etc.? 


#9
Nov1406, 06:06 AM

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PF Gold
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Yes, that's what "expanding" a determinant means. Using the third row is particularly easy since it has only 2 nonzero numbers.



#10
Jul2408, 11:25 PM

P: 1

I thought I'd mention
that the 12 seems to be the right answer, because I am getting 48 + 48 12 + many zero terms, so 12 seems right. I am using a pictoral technique from post#7 from this forum: http://www.mathisfunforum.com/viewto...d=95302#p95302 (I made that red/black diagram, so I was checking my work with this example) 


#11
Jul2508, 04:19 AM

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PF Gold
P: 39,363

Jack, this thread is two years old! And it was said two years ago that 12 is the determinant.



#12
Oct1710, 06:09 PM

P: 1

this way you end up with 2(3x3 matrix)  0(3x3 matrix + 0(3x3 matrix)  0(3x3 matrix) so now the det is 2 times the determinant of the remaining matrix 1, 2, 1 0, 2, 4 0, 0, 3 Which is 6, and 6*2= 12. 


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