Question 174415
<pre><b>
what is the value of this determinant?
{{{abs(matrix(6,6,
1,   1,   1,   1,   1,   1,
0,   1,   0,  -1,   0,  -1,
1,   2,   0,   3,   2,   0,
1,   0,   3,   2,   0,   2,
1,   2,   0,   4,   1,   5,
1,   0,   0,   2,   3,   4))}}}
                             thank you!

The third column has the most 0's, so
we'll expand about it after we make it
have all 0's but one.

Let's make the 3 in the 3rd column a 0
by adding -3 times the 1st row to 1
times the 4th row:

{{{matrix(6,1,-3,"","",1,"","")}}}{{{abs(matrix(6,6,
1,   1,   1,   1,   1,   1,
0,   1,   0,  -1,   0,  -1,
1,   2,   0,   3,   2,   0,
1,   0,   3,   2,   0,   2,
1,   2,   0,   4,   1,   5,
1,   0,   0,   2,   3,   4))}}}

{{{abs(matrix(6,6,
1,   1,   1,   1,   1,   1,
0,   1,   0,  -1,   0,  -1,
1,   2,   0,   3,   2,   0,
-2, -3,   0,  -1,  -3,  -1,
1,   2,   0,   4,   1,   5,
1,   0,   0,   2,   3,   4))}}}

Now we expand this 6x6 matrix by
the 3rd column.  Since there is only
1 non-zero element in the 3rd column,
the 1 at the top.  We notice that it
is in row 1 and column 3, so we add
1+3 and get 4, and {{{(-1)^4}}}=+1 we
multiply the 1 in the top row by +1
and so the determinant is +1 times the
5x5 determinant formed by removing
both row 1 and column 3:

{{{abs(matrix(5,5,

0,   1,     -1,   0,  -1,
1,   2,      3,   2,   0,
-2, -3,     -1,  -3,  -1,
1,   2,      4,   1,   5,
1,   0,      2,   3,   4))}}}

The first row has the most 0's, so
we'll expand about it after we make it
have all 0's but one.

Let's make the -1 in the 1st row 3rd 
column a 0 by adding 1 times the 2nd 
column to 1 times the 3rd column:

     1   1
{{{abs(matrix(5,5,
0,   1,     -1,   0,  -1,
1,   2,      3,   2,   0,
-2, -3,     -1,  -3,  -1,
1,   2,      4,   1,   5,
1,   0,      2,   3,   4))}}}

{{{abs(matrix(5,5,
0,   1,      0,   0,  -1,
1,   2,      5,   2,   0,
-2, -3,     -4,  -3,  -1,
1,   2,      6,   1,   5,
1,   0,      2,   3,   4))}}}

Let's make the -1 in the 1st row 5th 
column a 0 by adding 1 times the 2nd 
column to 1 times the 5th column:

     1          1 
{{{abs(matrix(5,5,
0,   1,      0,   0,  -1,
1,   2,      5,   2,   0,
-2, -3,     -4,  -3,  -1,
1,   2,      6,   1,   5,
1,   0,      2,   3,   4))}}}

{{{abs(matrix(5,5,
0,   1,      0,   0,   0,
1,   2,      5,   2,   2,
-2, -3,     -4,  -3,  -4,
1,   2,      6,   1,   7,
1,   0,      2,   3,   4))}}}

Now we expand this 5x5 matrix by
the 1st row.  Since there is only
1 non-zero element in the 1st row,
the 1 at the 1st row 2nd column.
Since it is in row 1 and column 2, so we add
1+2 and get 3, and {{{(-1)^3}}}=-1 we
multiply the 1 in the top row by -1
and so the determinant is -1 times the
4x4 determinant formed by removing
both row 1 and column 2:

{{{-1abs(matrix(4,4,
1,    5,   2,   2,
-2,  -4,  -3,  -4,
1,    6,   1,   7,
1,    2,   3,   4))}}}

We can simplify by multiplying the -1
by the 2nd row, changing their signs
to positive:

{{{abs(matrix(4,4,
1,    5,   2,   2,
2,    4,   3,   4,
1,    6,   1,   7,
1,    2,   3,   4))}}}

Unfortunately there are no 0's at
all.  So we'll make some.  The
simplest row or column is the 1st
column.  To make the 2 in the
1st column a 0, add -2 times the
1st row to 1 times the 2nd row:
{{{matrix(4,1,-2,1,"","")}}}{{{abs(matrix(4,4,
1,    5,   2,   2,
2,    4,   3,   4,
1,    6,   1,   7,
1,    2,   3,   4))}}}

{{{abs(matrix(4,4,
1,    5,   2,   2,
0,   -6,  -1,   0,
1,    6,   1,   7,
1,    2,   3,   4))}}}

Now since that 0 popped up
over there in the second row,
that makes us change plans, for
now we have 2 0's in the 2nd row.

So we change plans and now we 
want to expand about the 2nd
row.  So we can get a 0 where
the -6 is by multiplying the
3rd column by -6 and adding it
to 1 times column 2.

{{{abs(matrix(4,4,
1,   -7,   2,   2,
0,    0,  -1,   0,
1,    0,   1,   7,
1,  -16,   3,   4))}}}

Now we expand this 4x4 matrix by
the 2nd row.  Since there is only
1 non-zero element in the 1st row,
the -1 at the 2nd row 3rd column.
Since it is in row 2 and column 3, we add
2+3 and get 5, and {{{(-1)^5}}}=-1, we
multiply the -1 in the 2nd row 3rd column
by -1 and so the determinant is 1 times the
3x3 determinant formed by removing
both row 2 and column 3:

{{{abs(matrix(3,3,
1,   -7,   2,
1,    0,   7,
1,  -16,   4))}}}

The
simplest row or column is the 2nd
row.  To make the 7 a 0, we add
-7 times the 1st column to 1 times 
the 3rd column:

-7   1
{{{abs(matrix(3,3,
1,   -7,   2,
1,    0,   7,
1,  -16,   4))}}}

{{{abs(matrix(3,3,
1,   -7,  -5,
1,    0,   0,
1,  -16,  -3))}}}

Now we expand this 3x3 matrix by
the 2nd row.  Since there is only
1 non-zero element in the 2nd row,
the 1 at the 2nd row 1st column.
Since it is in row 2 and column 1, we add
2+1 and get 3, and {{{(-1)^3}}}=-1, we
multiply the 1 in the 2nd row 1st column
by -1 and so the determinant is -1 times the
2x2 determinant formed by removing
both row 2 and column 1:

{{{-1abs(matrix(2,2,
   -7,  -5,
  -16,  -3))}}}

We can simplify by multiplying the -1
by the 1st row, which changing the signs:

{{{abs(matrix(2,2,
   7,  5,
  -16,  -3))}}}


To expand, it's just the difference of
the product of the upper left to lower
right diagonal and the product of the
upper right to lower left diagonal.

(7)(-3)-(5)(-16)=-21+80=59

So the value of the determinant is 59.

Edwin</pre>