# How do I simplify 4x4 det before solving? I am doing somethingwrong

1. Nov 23, 2012

### Lenjaku

1. The problem statement, all variables and given/known data

solve the det
3 1 1 0
-4 5 0 -1
0 7 2 8
2 -1 0 3

2. Relevant equations

3. The attempt at a solution
I am trying to solve it and get a different answer each time so I am doing something I shouldn't obviously.

Ok I am adding rows to other rows
(ex 2*row1 +row2)
Same I do for columns.

When I expend the matrix I pay attention to the + and -
So what is wrong :S

This is what I did (I didn't choose the shortest way because I wanted to make sure I am solving it right but hence something is off :S)

I did 2*R4+r2,R3-2*R1
I opened the det by a3,1
And got to this by this point:
1*
-4 5 -1​
-6 5 8​
0 3 5 ​
then I did
R1-R2,3*R1+R2
then opened it by a2,1
I got the det:
6*
5 -19​
3 5​
Which is 492 (should b 82)
Now in all the cases I tried I reached 82 before multiplying the number outside the 2x2 det (6 in this case)

Well it is kinda hard showing the steps of the det though it is a simple one but mainly is there something wrong with the steps I take?

2. Nov 23, 2012

### HallsofIvy

Okay, that's a good start. It gets a 0 in the first column, second row.

But I don't understand why you did that? I thought you were trying to get 0s in the first column. This gives you
$$\begin{bmatrix}3 & 1 & 1 & 0 \\ 0 & 3 & 0 & 5 \\ -6 & 5 & 0 & 8 \\ 2 & -1 & 0 & 3\end{bmatrix}$$

Perhaps I misunderstood what you meant before. Because you already have two 0s in the third column, if you subtract 2 times the first row from the third row, Replacing R3 with R3- 2R1, you get
$$\left|\begin{array}{cccc}3 & 1 & 1 & 0 \\-4 & 5 & 0 & -1\\ -6 & 5 & 0 & 8 \\ 2 & -1 & 0 & 3 \end{array}\right|$$
Expanding on the third column, you get
$$1\left|\begin{array}{ccc}4 & 5 & -1 \\ 6 & 5 & 8 \\ 2 & -1 & 3\end{array}\right|$$
Although I still don't see how you got that third row.

3. Nov 23, 2012

### HallsofIvy

Okay, that's a good start. It gets a 0 in the first column, second row.

But I don't understand why you did that? I thought you were trying to get 0s in the first column. This gives you
$$\begin{bmatrix}3 & 1 & 1 & 0 \\ 0 & 3 & 0 & 5 \\ -6 & 5 & 0 & 8 \\ 2 & -1 & 0 & 3\end{bmatrix}$$

Perhaps I misunderstood what you meant before. Because you already have two 0s in the third column, if you subtract 2 times the first row from the third row, Replacing R3 with R3- 2R1, you get
$$\left|\begin{array}{cccc}3 & 1 & 1 & 0 \\-4 & 5 & 0 & -1\\ -6 & 5 & 0 & 8 \\ 2 & -1 & 0 & 3 \end{array}\right|$$
Expanding on the third column, you get
$$1\left|\begin{array}{ccc}4 & 5 & -1 \\ 6 & 5 & 8 \\ 2 & -1 & 3\end{array}\right|$$
Although I still don't see how you got that third row.

4. Nov 23, 2012

### Lenjaku

Not everything I did had to have a meaning I was playing around with it.
I figured out what I was doing wrong though...I treated it as matrix .
When I write 2R2+R3
I actually mean R2=2R2+R3
I didn't know I can't do that with dets or I change its value:S

5. Nov 24, 2012

### Millennial

Row operations of course do change the value of the determinant.
Why don't you try expansion by minors to solve it? Might be a bit tedious, but would work after all.