# Homework Help: Matrix fun

1. Apr 13, 2009

### Derill03

Any help solving this determinant:

0 1 1 1
1 0 1 1
1 1 0 1
1 1 1 0

My calc says the answer is -3 but there is supposed to be a quicker way than doing all the individual calculations, I did all the calculations and got -3 but there is supposed to be a quicker way. Anyone?

2. Apr 13, 2009

### dx

What way did you use?

3. Apr 13, 2009

### Derill03

a11a22a33a44
- a11a22a34a43
+ a11a23a34a42
- a11a23a32a44
+ a11a24a32a43
- a11a24a33a42
- a12a23a34a41
+ a12a23a31a44
- a12a24a31a43
+ a12a24a33a41
- a12a21a33a44
+ a12a21a34a43
+ a13a24a31a42
- a13a24a32a41
+ a13a21a32a44
- a13a21a34a42
+ a13a22a34a41
- a13a22a31a44
- a14a21a32a43
+ a14a21a33a42
- a14a22a33a41
+ a14a22a31a43
- a14a23a31a42
+ a14a23a32a41

4. Apr 13, 2009

### dx

You can use row and column operations to simplify the determinant. For example,

R2 --> R2 - R3
R3 --> R3 - R4

This makes the first column 0 0 0 1.

5. Apr 13, 2009

### Derill03

I get:

0 1 1 1
0 -1 1 0
0 0 -1 1
1 1 1 0

which then a co-factor expansion would give:

0+0+0+0+1*determinant of

1 1 1
-1 1 0
0 -1 1

wheres the -1 come from cause i get an answer of 3? is it supposed to be 0+0+0+0-1?

6. Apr 13, 2009

### dx

Yes.

7. Apr 13, 2009

### Dunkle

Don't forget that the equation for the co-factor expansion includes the term $$(-1)^{i+j}$$ where i is the row and j is the column.

In this case, $$i = 4$$ and $$j = 1$$, so this term is $$(-1)^{4+1} = (-1)^5 = -1$$.