Multiplication/division of matrices and vectors

[Jhenrique]

1) Let A a square matrix, x a colum vector and b another colum vector. So, I want solve for x the following equation: Ax=b
So: x=b÷A = b×A^-1 And this is the answer! Or would be this the correct answer x = A^-1×b ?

2) Is possible to solve the equation above for A ? How?

 

[micromass]

1) Let A a square matrix, x a colum vector and b another colum vector. So, I want solve for x the following equation: Ax=b
So: x=b÷A = b×A^-1 And this is the answer! Or would be this the correct answer x = A^-1×b ?

First of all, you shouldn't use the division symbol ÷ for matrices. Why not? Matrices are noncommutative. This is, it can happen that $AB \neq BA$. The division ÷ is ambiguous in the noncommutative case, because it is unclear whether $A$÷$B$ means $AB^{-1}$ or $B^{-1}A$. So you should always use the $B^{-1}$ notation.

Anyway, you want to solve $A\mathbf{x} = \mathbf{b}$. First of all, $A$ might not be an invertible matrix, in which case, you can't always solve this system (and if you can, the solution might not be unique!). If your matrix is invertible, then you can multiply the equation on the left with $A^{-1}$ and you get

\mathbf{x} = A^{-1}A\mathbf{x} = A^{-1}\mathbf{b}

Multiplying the equation on the right doesn't work since you'll get

A\mathbf{x}A^{-1} = \mathbf{b}A^{-1}

We can't simplify the left-hand side due to noncommutativity.

2) Is possible to solve the equation above for A ? How?

If you mean that $\mathbf{x}$ and $\mathbf{b}$ are known, then no. You don't have enough data for a unique solution. That is, there will be many solutions to this problem.

 

[Mark44]

2) Is possible to solve the equation above for A ? How?

If you mean that $\mathbf{x}$ and $\mathbf{y}$ are known, then no. You don't have enough data for a unique solution. That is, there will be many solutions to this problem.

The equation that was referred to was Ax = b, so if x and b are known, there is not a unique matrix A.

 

[D H]

1) Let A a square matrix, x a colum vector and b another colum vector. So, I want solve for x the following equation: Ax=b
So: x=b÷A = b×A^-1 And this is the answer!

That is not the answer. The only way what you wrote would make sense is if A is a 1×1 matrix; i.e., a scalar.

Or would be this the correct answer x = A^-1×b ?

That's correct -- if A is invertible.

2) Is possible to solve the equation above for A ? How?

Only if A is a 1×1 matrix. For an N×N matrix, where N>1, the answer is no (not uniquely). Ax=b represents N equations, but you have N² unknowns.

 

[Mark44]

How can I reach the same conclusion?

micromass answered this question in post #2.

If Ax = b, and A is invertible, then multiply on the left by A^-1.

 

[micromass]

The equation that was referred to was Ax = b, so if x and b are known, there is not a unique matrix A.

Aah, thanks you very much! I really need to proof read my posts better. I'll change my post.

 

[Jhenrique]

micromass answered this question in post #2.

If Ax = b, and A is invertible, then multiply on the left by A^-1.

Yeah! sorry... I not seen, I'm reading now!

Edit: thanks for everyone from topic!

Edit2: So, Independent of the uniqueness, you are saying that is not possible to isolate A in Ax=b ?