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Given a $3\times 4$ matrix $A$ such as $$ \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ \end{pmatrix} ,$$ find a matrix $B_{4\times 3}$ such that
$$AB = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} $$ Apart from simply multiplying $A$ with $B$ and generating a $12$ variable system of equations, is there any simpler way of finding $B$ ?

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  • $\begingroup$ Use products with 3x3 elementary matrices as row operations to reduce the original matrix to row echelon form. Then maybe something can be worked out. $\endgroup$ – jdods Oct 19 '15 at 11:43
  • $\begingroup$ I'm not looking for the inverse, meaning I don't want BA = I to be true. Just a matrix B that satisfies AB=I. $\endgroup$ – EuxhenH Oct 19 '15 at 11:49
  • $\begingroup$ B could be a 4x3 matrix which leads to a 3x3 I with 1s on the main diagonal. $\endgroup$ – EuxhenH Oct 19 '15 at 11:51
  • $\begingroup$ It would be clear for everyone to specific the specific matrix you want and I would suggest that you call it something other than $I$ to eliminate confusion. $\endgroup$ – NoChance Oct 19 '15 at 11:53
  • $\begingroup$ Changed it. Thanks $\endgroup$ – EuxhenH Oct 19 '15 at 11:56
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Your matrix $$A=\begin{bmatrix}1&1&1&1\\0&1&1&0\\0&0&1&1\end{bmatrix}$$ has rank=$3$ and you can see that the square matrix $AA^T$ is invertible. Now note that $AA^T(AA^T)^{-1}=I$ so the matrix $B=A^T(AA^T)^{-1}$ is a right inverse of $A$ (but it is not the unique).

in this case we have: $$ AA^T=\begin{bmatrix}4&2&2\\2&2&1\\2&1&2\end{bmatrix} $$ $$ (AA^T)^{-1}=\dfrac{1}{4}\begin{bmatrix}3&-2&-2\\-2&4&0\\-2&0&4\end{bmatrix} $$ $$ B=\dfrac{1}{4}\begin{bmatrix}3&-2&-2\\1&2&-2\\-1&2&2\\1&-2&2\end{bmatrix} $$

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Consider the following case:

$$\begin{bmatrix}1&1&1&1\\0&1&1&0\\0&0&1&1\end{bmatrix}\times \begin{bmatrix}0&0&0\\?&?&?\\?&?&?\\?&?&?\end{bmatrix}=I$$

Removing the first row, the remaining matrix is square:

$$\begin{bmatrix}?&?&?\\?&?&?\\?&?&?\end{bmatrix}=\begin{bmatrix}1&1&1\\1&1&0\\0&1&1\end{bmatrix}^{-1}=\begin{bmatrix}1&0&-1\\-1&1&1\\1&-1&0\end{bmatrix}$$

$$B=\begin{bmatrix}0&0&0\\1&0&-1\\-1&1&1\\1&-1&0\end{bmatrix}$$

Of course there are infinite number of solutions

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