SOLUTION: | 3 1 -1 2 | | 0 0 -3 3 | | 9 5 0 1 | |-7 3 -2 0 | Evaluate.

Algebra ->  Systems-of-equations -> SOLUTION: | 3 1 -1 2 | | 0 0 -3 3 | | 9 5 0 1 | |-7 3 -2 0 | Evaluate.       Log On


   

Question 601376: | 3 1 -1 2 |
| 0 0 -3 3 |
| 9 5 0 1 |
|-7 3 -2 0 |

Evaluate.
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
| 3  1 -1  2 |
| 0  0 -3  3 |
| 9  5  0  1 |
|-7  3 -2  0 |

There are already 2 zeros in the 2nd row.
So we need to get one more 0 in that row and
then we can expand about it.

To get a zero where then -3 is add the 4th column to it:

-1 + 2 =  1
-3 + 3 =  0
 0 + 1 =  1
-2 + 0 = -2

Now the only element in the 2nd row that isn't 0 is the -3,
which I've colored red:

| 3  1 -1  1 |
| 0  0 -3  0 |
| 9  5  0  1 |
|-7  3 -2 -2 |

So we expand by the 2nd row.  First we have to decide if the
-3 is in a + or a - position in the
"checkerboard" 4×4 sign-scheme determinant, which is

| +  -  +  - |
| -  +  -  + |
| +  -  +  - |
| -  +  -  + |

So we see the -3 is in a - position, so we will have to multiply
the -3 by a -1, and use a +3 multiplier by the 3×3 minor determinant.  
To get the minor 3×3 determinant, we erase all the elements in the same row
and column with the -3, and we have this:

| 3  1     1 |
|      -3    |
| 9  5     1 |
|-7  3    -2 |
 
And we put a +3 out front and we have this 3×3 determinant
 
  | 3  1  1 |
3×| 9  5  1 |
  |-7  3 -2 |

To get a 0 where the -2 is, multiply the first row by 2 and add
that to the 3rd row:

    3  1  1 
         ×2
    6  2  2
   -7  3 -2
   -1  5  0

  | 3  1  1 |
3×| 9  5  1 |
  |-1  5  0 |  

 To get a 1 where the 1 is in the second row 3rd column, 
multiply the first row by -1 and add that to the 2nd row:

    3  1  1 
        ×-1
   -3 -1 -1
    9  5  1
    6  4  0

Now the only element in the 3rd column that isn't 0 is the 1,
which I've colored red:

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

So we expand by the 3rd column since it has all 0's but one.
First we have to decide if the -3 is in a
+ or a - position in the "checkerboard" 3×3 sign-scheme determinant, 
which is

| +  -  + |
| -  +  - |
| +  -  + |

So we see the 1 is in a + position, so we just multiply the red 1
by +1, which will not change it and use a +1 multiplier by the minor 2×2
determinant.  To get the minor 2×2 determinant, we erase all the elements
in the same row and column with the red 1, and we have this:

  |       1 |
3×| 6  4    |
  |-1  5    |
  
And we put a +1 out front with the 3 multuiplier that's already out there
and we have this 2×2 determinant

3×1×| 6  4 |
    |-1  5 |

To get a 0 where the 5 is multiply the 1st column by 5 and add that
the the first column:

 6(5) + 4  = 30 + 4 = 34
-1(5) + 5  = -5 + 5 =  0

3×1×| 6 34 |
    |-1  0 |

So we expand by the 2nd column since it has all 0's but one.
First we have to decide if the 34 is in a
+ or a - position in the "checkerboard" 2×2 sign-scheme determinant, 
which is

| +  - |
| -  + |

So we see the red 34 is in a - position, so we multiply the red 34
by -1, and use a -34 multiplier by the minor 1×1
determinant.  To get the minor 1×1 determinant, we erase all the 
elements in the same row and column with the red 34, and we have this:

3×1×|   34 |
    |-1    |
  
And we put a -34 out front with the 3×1 multiplier that's already out there
and we have this 1×1 determinant

3×1×(-34)×|-1|

And a 1×1 determinant is equal to its 1 element, so we have

3×1×(-34)×(-1) = 102.

[Notice, we don't have to go all the way to a 1×1 determinant.
Some people know a way to expand a 3×3 determinant and stop there
and expand by the method they know. Most people stop with a 2×2 
determinant and expand it because they know just to subtract the 
product of the diagonals.  I just wanted to show you that it could 
be taken all the way down to a 1×1 determinant.  The answer will 
be 102 regardless of whether you stop along the way and expand or 
go all the way to a 1×1 determinant]

Edwin