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I want to find the inverse of the product of $2$ non-square matrices, but is this even possible?

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The rank of an $m \times n$ matrix is at most $\min(m,n)$, and $\text{rank}(AB) \le \min(\text{rank}(A), \text{rank}(B))$. So if $m > n$ and $A$ is $m \times n$ while $B$ is $n \times m$, the $m \times m$ matrix

$AB$ can’t be invertible, but the $n \times n$ matrix $BA$ can.

Fact 1: The product of a $4 \times 3$ matrix and a $3 \times 4$ matrix is a $4 \times 4$ matrix.

Fact 2: The rank of an $m \times n$ matrix is no bigger than the smallest of its dimensions, i.e. it is $\leq \min{m, n}$. In this case it means each of the two given matrices have a maximum rank of 3.

Fact 3: The rank of the product of matrices is no bigger than the smallest of the rank of the two. Here, it means the rank of the product is $\leq 3$.

Fact 4: If the rank of a square matrix is less than its dimension, it is singular. Since our product matrix’s rank is $\leq 3 < 4$, it is not invertible.

Let $A=\begin{pmatrix}u_1&v_1&w_1\\u_2&v_2&w_2\\u_3&v_3&w_3\\u_4&v_4&w_4\end{pmatrix}$ and $B=\begin{pmatrix}s_1&s_2&s_3&s_4\\t_1&t_2&t_3&t_4\\x_1&x_2&x_3&x_4\end{pmatrix}$, since $A$ is a $4\times 3$ matrix and $AB$ is a $4\times 4$ matrix it follows $\text{rank}(AB)\le\text{rank}(A)\le 3$, then $AB$ is a singular matrix.

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