# Linear Algebra: Determinants

• Sep 24th 2007, 11:54 PM
googoogaga
Linear Algebra: Determinants
Hello, could I please receive some assistance in the solving of the following homework problem. My answer does not math the book's. Thanks! Here it is...

Evaluate the determinant of the following 4 by 4 matrix

2 1 2 1
3 0 1 1
-1 2 -2 1
-3 2 3 1

Book's answer: 20 (How?)

• Sep 25th 2007, 12:16 AM
Puubes
[quote=googoogaga;72360]Hello, could I please receive some assistance in the solving of the following homework problem. My answer does not math the book's. Thanks! Here it is...

Evaluate the determinant of the following 4 by 4 matrix

2 1 2 1
3 0 1 1
-1 2 -2 1
-3 2 3 1

Book's answer: 20 (How?)

2 1 2 1
3 0 1 1 <-- i chose this row to cross out and then cross out each column
-1 2 -2 1 with each corresponding entry in that row and take det times that
-3 2 3 1 entry

3 * det[] + 0 ( zero times det of 3X3 = 0) + det[] + det[]
1 2 1 2 1 1 2 1 2
2 -2 1 2 -2 1 -1 2 1 match the colors
2 3 1 -3 2 1 -3 2 1

then 3 * the det of that 3X3 matrix by choosing a row to do the same as before then you are left with two by twos. then add the det of the second and then the third.

3*{det[]+2*det[]+det[]} + {2*det[]+det[]+det[]} + {2*det[]+det[]+2*det[]}

-2 1 2 1 2 -2 2 1 -1 1 -1 2 2 -2 -1 -2 -1 2
3 1 2 1 2 3 2 1 -3 1 -3 2 2 3 -3 3 -3 2

det of 2X2 upper left times lower right minus lower left times upper right (-2*1)-(3*1)
3(-5+0+2) + (0+2+8) + (20+-9+8)
-9 + 10 + 19 = 20

or if you have a ti-82 or above just go to 2nd matrix then edit and insert the 4X4 matrix then quit then 2nd matrix again and go down to det( enter and then 2nd matrix again and select A B C or whatever slot you entered that matrix in and voila.

hope the colors helped can't get the numbers to line up like a want it would be easier to show that way
• Sep 25th 2007, 01:05 AM
googoogaga
Thank you much for the help but I would like to know how it is solved using the determinant equations because the colors are really confusing me. you know like A_11= .... or a_11 A_11= A_11...
or A_ij= (-1) ^(i+j) [M_11] something like that..
• Sep 25th 2007, 05:19 AM
topsquark
Quote:

Originally Posted by googoogaga
Hello, could I please receive some assistance in the solving of the following homework problem. My answer does not math the book's. Thanks! Here it is...

Evaluate the determinant of the following 4 by 4 matrix

2 1 2 1
3 0 1 1
-1 2 -2 1
-3 2 3 1

Book's answer: 20 (How?)

This is called expansion by minors. (You can use the Leibniz formula to make a nice compact formula, but I've found that to be difficult to put into use unless I'm writing a computer program to do determinants.)

Uck. This is going to be awful to type!
$\left | \begin{matrix} 2 & 1 & 2 & 1 \\ 3& 0 & 1 & 1 \\ -1 & 2 & -2 & 1 \\ -3 & 2 & 3 & 1 \end{matrix} \right | = 2 \left | \begin{matrix} 0 & 1 & 1 \\ 2 & -2 & 1 \\ 2 & 3 & 1 \end{matrix} \right | - 1 \left | \begin{matrix} 3 & 1 & 1 \\ -1 & -2 & 1 \\ -3 & 3 & 1 \end{matrix} \right |$ $+ 2 \left | \begin{matrix} 3 & 0 & 1 \\ -1 & 2 & 1 \\ -3 & 2 & 1 \end{matrix} \right | - 1 \left | \begin{matrix} 3 & 0 & 1 \\ -1 & 2 & -2 \\ -3 & 2 & 3 \end{matrix} \right |$

I presume you know how to do 3x3 determinants. If not, do an expansion by minors on each of the 4 3x3 determinants listed here.

-Dan
• Sep 25th 2007, 08:18 AM
Jhevon
Quote:

Originally Posted by topsquark
This is called expansion by minors. (You can use the Leibniz formula to make a nice compact formula, but I've found that to be difficult to put into use unless I'm writing a computer program to do determinants.)

Uck. This is going to be awful to type!
$\left | \begin{matrix} 2 & 1 & 2 & 1 \\ 3& 0 & 1 & 1 \\ -1 & 2 & -2 & 1 \\ -3 & 2 & 3 & 1 \end{matrix} \right | = 2 \left | \begin{matrix} 0 & 1 & 1 \\ 2 & -2 & 1 \\ 2 & 3 & 1 \end{matrix} \right | - 1 \left | \begin{matrix} 3 & 1 & 1 \\ -1 & -2 & 1 \\ -3 & 3 & 1 \end{matrix} \right |$ $+ 2 \left | \begin{matrix} 3 & 0 & 1 \\ -1 & 2 & 1 \\ -3 & 2 & 1 \end{matrix} \right | - 1 \left | \begin{matrix} 3 & 0 & 1 \\ -1 & 2 & -2 \\ -3 & 2 & 3 \end{matrix} \right |$

I presume you know how to do 3x3 determinants. If not, do an expansion by minors on each of the 4 3x3 determinants listed here.

-Dan

Hmm, i was taught the name was "the cofactor expansion method"...or am i mixing that with something else. i always have trouble remembering names and formulas.

anyway, the nice thing about this is, you can actually expand along ANY row or even ANY column. for this problem, i would expand along the second row (or maybe the second column depending on my mood), that way, the 0 in that row (or column) would wipe out one of the determinants i have to find, so i have to do less work, which is good.

so, with that in mind, we could say:

(expanding along the second row)

$\left | \begin{matrix} 2 & 1 & 2 & 1 \\ 3& 0 & 1 & 1 \\ -1 & 2 & -2 & 1 \\ -3 & 2 & 3 & 1 \end{matrix} \right | = -3 \left | \begin{matrix} 1 & 2 & 1 \\ 2 & -2 & 1 \\ 2 & 3 & 1 \end{matrix} \right | - 1 \left | \begin{matrix} 2 & 1 & 1 \\ -1 & 2 & 1 \\ -3 & 2 & 1 \end{matrix} \right |$ $+ 1 \left | \begin{matrix} 2 & 1 & 2 \\ -1 & 2 & -2 \\ -3 & 2 & 3 \end{matrix} \right |$

or

(expanding along the second column)

$\left | \begin{matrix} 2 & 1 & 2 & 1 \\ 3& 0 & 1 & 1 \\ -1 & 2 & -2 & 1 \\ -3 & 2 & 3 & 1 \end{matrix} \right | = -1 \left | \begin{matrix} 3 & 1 & 1 \\ -1 & -2 & 1 \\ -3 & 3 & 1 \end{matrix} \right | - 2 \left | \begin{matrix} 2 & 2 & 1 \\ 3 & 1 & 1 \\ -3 & 3 & 1 \end{matrix} \right |$ $+ 2 \left | \begin{matrix} 2 & 2 & 1 \\ 3 & 1 & 1 \\ -1 & -2 & 1 \end{matrix} \right |$

now, we only have to find three determinants. and in the first one, two of the determinants are multiplied by 1, so that's even better!
• Sep 25th 2007, 08:20 AM
Jhevon
Quote:

Originally Posted by googoogaga
Thank you much for the help but I would like to know how it is solved using the determinant equations because the colors are really confusing me. you know like A_11= .... or a_11 A_11= A_11...
or A_ij= (-1) ^(i+j) [M_11] something like that..

i think Puubes' post deserves a thanks anyway. he seems like he worked hard on it, and he did end up with the right answer so it's a solution
• Sep 25th 2007, 09:32 AM
googoogaga
You are absolutel right Jhevon. I did thank him.(Smile)
• Sep 25th 2007, 09:35 AM
Jhevon
Quote:

Originally Posted by googoogaga
You are absolutel right Jhevon. I did thank him.(Smile)

do you understand the method topsquark and i used?

this is my 38:):)th post!!!!!
• Sep 26th 2007, 05:19 AM
topsquark
Quote:

Originally Posted by Jhevon
Hmm, i was taught the name was "the cofactor expansion method"

Like practically every other method I know of in Math it probably goes by several names. ;)

-Dan