# 3x3 matrix a1+2a2-a3=0

1. Feb 7, 2010

### Dustinsfl

1. The problem statement, all variables and given/known data
If A is a 3 x 3 matrix a1+2a2-a3=0, then A must be singular.
I have the answering being true but how do I prove it?

2. Feb 7, 2010

### l'Hôpital

Assuming a_i are columns or rows, then sure. Let's assume you row/column reduce something as far as you can.

We have two cases:
Case 1:
You end up with the identity. Taking the determinant, you'll get a nonzero number which will be affected by the row/column operations you did (like if you swapped rows and whatnot).

Case 2:
You have a row/col of zeroes somewhere.
Taking the determinant of this will give you zero (hence singular).

Since a_1 + 2a_2 - a_3 = 0, this implies that a_1 + 2a_2 = a_3.

What case is the matrix in? Why?

3. Feb 7, 2010

### Dustinsfl

We can't prove it using determinants. The equation is in the form Ax=b. Where x is the column vector 1,2,-1.

4. Feb 7, 2010

### l'Hôpital

What does that have to do with anything? According your original post, this deals only with the matrix A. What does that particular equation have anything to do with whether A is singular or not?

5. Feb 7, 2010

### Dustinsfl

Because that column vector is used in proving the singularity but I don't know how to do it.

6. Feb 7, 2010

### l'Hôpital

I don't understand the question. The vector in question has nothing to do with that.

What are the ai? If they are columns/rows, you don't need that column vector. What is b, anyways?

7. Feb 7, 2010

### Dustinsfl

Column vector b is the 0 vector. It has to do with homogeneous equations have trivial and solutions. That is how A is suppose to be proving for the question.

8. Feb 7, 2010

### l'Hôpital

Oooooh! I got you. This seems simple enough.

Consider any general system Ax = b.

If A was invertible, what does this tell you about x?

9. Feb 7, 2010

### Dustinsfl

x is the inverse if and only if b is the I

10. Feb 7, 2010

### l'Hôpital

No no. x can't be the inverse of anything because x is a column vector. x is a solution to this equation. What kind of solution, though? Hint: Is it the only solution?

11. Feb 7, 2010

### Dustinsfl

Nontrivial solution

12. Feb 7, 2010

### l'Hôpital

There are no "trivial" solutions if b is not the zero vector. So, if A is invertible and b is not the zero vector, does there exist a y such that Ay = b and x =/= y ? That is to say, is x unique?

13. Feb 7, 2010

### Dustinsfl

It is giving that b is the 0 vector. I am not sure if it is unique or how to show if it isn't.

14. Feb 7, 2010

### l'Hôpital

The fact b is the zero vector is the key part in the proof, because it guarantees you a solution. It guarantees you the trivial solution of x = 0. According to the problem at hand, you also have another solution of the form x = <1,-2,1>. Therefore, x is not unique.

Is there any way of connection uniqueness of solutions with whether A is invertible?

15. Feb 7, 2010

### Dustinsfl

The system Ax=b of n linear equations in n unknowns has a unique solution if and only if A is nonsingular. Since x can be the 0 vector and vector <1,2,-1>, the solution isn't unique; therefore, A must be singular.

16. Feb 7, 2010

### l'Hôpital

Correct!

17. Feb 7, 2010

### Dustinsfl

Thanks

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