Prove: $A$ is $3\times4$ matrix of rank $3$, then there exists a $4\times3$ matrix $B$ such that $AB=I$

Question

$A$ is $3\times4$ matrix of rank $3$, then how do we prove that there exists a $3\times4$ matrix $B$ such that $AB=I$

Clearly since the rank$(A)=3$, the reduced echelon form of $A (A')$ should have $3$ non-zero rows.

We obtain $A'$ by performing certain row operations say $r_{i}$

But we know that general form of $A'$ should be: $$\begin{pmatrix}1 & 0 & 0 & a\\\ 0 & 1 & 0 & b \\\ 0 & 0 & 1 & c\end{pmatrix}$$ where $a,b,c$ are real numbers

I am facing trouble with how to proceed further. How do we: a) Use the fact that $A'$ is basically an identity matrix augmented with a real column matrix b) Come up with a general method to obtain $B$

Answer 1

For the matrix A, we can do the row/column operations to make it into the standard form:

$$L_k...L_1AR_1...R_m=\begin{pmatrix}1 & 0 & 0 & a\\\ 0 & 1 & 0 & b \\\ 0 & 0 & 1 & c\end{pmatrix}$$

Where $L_i, i=1,2...,k$ are elementary matrices for row operations, and $R_j, j=1,2...,m$ are elementary matrices of column operations. Let $T$ be the following matrix:

$$T=\begin{pmatrix}1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ 0 & 0 & 0 \end{pmatrix}$$

So we have:

$$\begin{align} L_k...L_1AR_1...R_mT&=I\\ \\ AR_1...R_mT&=L_1^{-1}...L_k^{-1}I\\ \\ AR_1...R_mTL_k...L_1&=I\\ \\ \Rightarrow B&=R_1...R_mTL_k...L_1 \end{align}$$

Answer 2

Since the $3 \times 4 $ matrix $A$ is of rank $3$, then the $3 \times 3$ matrix $A A^T$ is invertible, therefore, we can take $B$ to be

$ B = A^T (A A^T)^{-1} $