Proving Singularity of 3x3 Matrix with a1+2a2-a3=0 | A is Singular

[Dustinsfl]

Homework Statement

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?

 

[l'Hôpital]

Homework Statement

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?

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?

 

[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.

 

[l'Hôpital]

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

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?

 

[Dustinsfl]

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

 

[l'Hôpital]

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

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

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

 

[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.

 

[l'Hôpital]

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.

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?

 

[Dustinsfl]

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

 

[l'Hôpital]

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

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?

 

[Dustinsfl]

Nontrivial solution

 

[l'Hôpital]

Nontrivial solution

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?

 

[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.

 

[l'Hôpital]

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.

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?

 

[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.

 

[l'Hôpital]

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.

Correct!

 

[Dustinsfl]

Thanks