SOLUTION: Help evaluate the determinate? Please explain everything so I can comprehend this...please. |1 -3 2 0| |-3 -1 0 -2| |2 1 3 1| |2 0 -2 0

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

Pick a row of column, with the most 0's in it
that you can find, and use row and/or column 
operations to get all but one 0 in that row 
or column.

The red row below already has two 0's,

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

and all we have to do to get a 0 where that 
red -2 is on the bottom row, is to add the 
numbers in the first column to the corresponding
numbers in the third column.  We get this:

| 1  -3   2+1   0|
|-3  -1   0-3  -2|
| 2   1   3+2   1|
| 2   0  -2+2   0|

or

| 1  -3   3   0|
|-3  -1  -3  -2|
| 2   1   5   1|
| 2   0   0   0|

Now all the numbers on the bottom row are 0's
all except for one, the 2 in the lower left 
corner.

So let's cross out all the other elements in the 
same row and column that that 2 is in:

| 1  -3   3   0|
|-3  -1  -3  -2|
| 2   1   5   1|
| 2   0   0   0|  

Now multiply that 2 by the 3x3 determinant formed
by the 9 numbers in the upper right:

     |-3   3   0|
( )2×|-1  -3  -2|
     | 1   5   1|

But we must now check the sign scheme to see whether
we keep the positive sign of the 2 or whether we must
change its sign to get its opposite, -2. 

This is the sign scheme for the 4x4 determinant.



Notice that since the 2 is in the bottom left hand
corner and there is a - in the bottom left-corner in
the sign scheme, we must change the sign of the 2, to
a -2. (If the 2 had been in a position where there is a 
+ sign in the sign scheme, we would have have just kept
the sign and used 2.)

So we have:

    |-3   3   0|
 -2×|-1  -3  -2|
    | 1   5   1|

Now we have to expand that determinant.  So we
pick a row with the most 0's in it, and use row 
and/or column operations to get all but one 0 
in that row or column.

The top (red) row below already has one 0 already,

    |-3   3   0|
 -2×|-1  -3  -2|
    | 1   5   1|

and all we have to do to get a 0 where that 
upper left 3 in on the top row, is to add the 
numbers in the second column to the corresponding
numbers in the first column. 

    |-3   3   0|
 -2×|-1  -3  -2|
    | 1   5   1|

    |-3+3   3   0|
 -2×|-1-3  -3  -2|
    | 1+5   5   1|

    | 0   3   0|
 -2×|-4  -3  -2|
    | 6   5   1|

So let's cross out all the other elements in the 
same row and column that that 3 is in:

    | 0    3  0|
 -2×|-4  -3  -2|
    | 6   5   1|

Now multiply that 3 by the 2x2 determinant formed
by the 4 numbers that haven't been crossed out,
so we have:

  -2*( )3*

But we must now check the sign scheme to see whether
we keep the positive sign of the 3 or whether we must
change its sign to get its opposite, -3. 

This is the 3x3 sign scheme.



Notice that since the 3 is in the middle of the top row,
and there is a + in the middle of the top row in the sign
scheme, we change the sign of the 3 to -3. 

So we have

  

or

   =

   

Edwin