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

  • Thread starter Lenjaku
  • Start date
  • #1
16
0

Homework Statement



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

Homework Equations





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?
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
961

Homework Statement



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

Homework Equations





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

,R3-2*R1
But I don't understand why you did that? I thought you were trying to get 0s in the first column. This gives you
[tex]\begin{bmatrix}3 & 1 & 1 & 0 \\ 0 & 3 & 0 & 5 \\ -6 & 5 & 0 & 8 \\ 2 & -1 & 0 & 3\end{bmatrix}[/tex]

I opened the det by a3,1
And got to this by this point:
1*
-4 5 -1​
-6 5 8​
0 3 5​
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
[tex]\left|\begin{array}{cccc}3 & 1 & 1 & 0 \\-4 & 5 & 0 & -1\\ -6 & 5 & 0 & 8 \\ 2 & -1 & 0 & 3 \end{array}\right|[/tex]
Expanding on the third column, you get
[tex]1\left|\begin{array}{ccc}4 & 5 & -1 \\ 6 & 5 & 8 \\ 2 & -1 & 3\end{array}\right|[/tex]
Although I still don't see how you got that third row.

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?
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,833
961

Homework Statement



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

Homework Equations





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

,R3-2*R1
But I don't understand why you did that? I thought you were trying to get 0s in the first column. This gives you
[tex]\begin{bmatrix}3 & 1 & 1 & 0 \\ 0 & 3 & 0 & 5 \\ -6 & 5 & 0 & 8 \\ 2 & -1 & 0 & 3\end{bmatrix}[/tex]

I opened the det by a3,1
And got to this by this point:
1*
-4 5 -1​
-6 5 8​
0 3 5​
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
[tex]\left|\begin{array}{cccc}3 & 1 & 1 & 0 \\-4 & 5 & 0 & -1\\ -6 & 5 & 0 & 8 \\ 2 & -1 & 0 & 3 \end{array}\right|[/tex]
Expanding on the third column, you get
[tex]1\left|\begin{array}{ccc}4 & 5 & -1 \\ 6 & 5 & 8 \\ 2 & -1 & 3\end{array}\right|[/tex]
Although I still don't see how you got that third row.

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?
 
  • #4
16
0
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
296
0
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.
 

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