1. ## Matrix manipulation

How does the fact that A isn't a square n x n matrix change how to do this question? i know that if A is an invertible n x n matrix, the solution would be $x = A^{-1}b$...

Suppose A is a 3 x 4 matrix and there exists a 4 x 3 matrix C such that $AC = I_{3}$ (the 3x3 identity matrix). Let b be an arbitrary vector in $\mathbb{R}^3$. Produce a solution of $Ax=b$

2. Hello,
Originally Posted by Robb
How does the fact that A isn't a square n x n matrix change how to do this question? i know that if A is an invertible n x n matrix, the solution would be $x = A^{-1}b$...

Suppose A is a 3 x 4 matrix and there exists a 4 x 3 matrix C such that $AC = I_{3}$ (the 3x3 identity matrix). Let b be an arbitrary vector in $\mathbb{R}^3$. Produce a solution of $Ax=b$
A matrix that is non square is not invertible.
That's the main difference.

You know that $AC=I_3$
Hence $Ax=b=I_3b=ACb$

By the associativity of matrix multiplication, this is : $Ax=A(Cb)$

Hence, a (*) solution is $x=Cb$

(*) The reason why I say "a" and not "the", and your question states "a", is that it's not necessarily the only one. This is only one possible solution.

Does it look clear to you ?