Let's say I have the matrix called Delta,

$$ \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{matrix} $$

What would I have to multiply the delta by to obtain the matrices below?

In brief, is there a way to backsolve matrices through $matrix inversion$ so I can easily find out what I should multiply matrix N by to get a specific result?

1. $$ \begin{matrix} -a & b & c \\ d & -e & f \\ g & h & -i \\ \end{matrix} $$ 2. $$ \begin{matrix} a & b & -c \\ d & -e & f \\ -g & h & i \\ \end{matrix} $$

Just to satisfy my curiosity, what would I have to multiply delta by to get the matrices below?
1. $$ \begin{matrix} a & -b & c \\ -d & -e & -f \\ g & -h & i \\ \end{matrix} $$

  1. $$ \begin{matrix} -a & b & c \\ -d & e & f \\ -g & -h & -i \\ \end{matrix} $$


Let $v=\begin{bmatrix}1\\1\\0\end{bmatrix}$

Let $\Delta = \begin{bmatrix}1&1&1\\1&1&1\\0&0&0\end{bmatrix}$

$A\cdot \Delta = A\cdot \begin{bmatrix}1&1&1\\1&1&1\\0&0&0\end{bmatrix} = \begin{bmatrix}Av&Av&Av\end{bmatrix}$

In particular, all columns of $A\cdot \Delta$ will be identical for any $A$. As such, there is absolutely no way for there to exist some $A$ such that $A\cdot \Delta = \begin{bmatrix}-1&1&1\\1&-1&1\\0&0&0\end{bmatrix}$ since the columns of this matrix are not all identical.

This same counterexample should work against the other requests as well.

This counterexample shows that the existence of such a matrix $A$ is not guaranteed to exist and will largely depend on what your matrix $\Delta$ is.

What will be true is that if the matrix $\Delta$ is invertible (i.e. $\Delta^{-1}$ exists), then you can find a desired matrix that when left-multiplied to $\Delta$ will result in a target matrix (assuming dimensions match).

I.e. if $B$ is your target matrix and $\Delta$ is invertible, then to find an $A$ such that $A\Delta = B$, you will find $A=B\Delta^{-1}$ will suffice.

Alternatively, if you wish to leave the realm of usual matrix multiplication (which corresponds to composition of linear transformations and is incredibly useful), there is such a thing known as the Hadamard Product which will easily accomplish what you want.

The Hadamard product is the entry-wise multiplication.

You have for example $\begin{bmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{bmatrix}\circ \begin{bmatrix} a&b&c\\d&e&f\\g&h&i\end{bmatrix} = \begin{bmatrix}-a&b&c\\d&-e&f\\g&h&-i\end{bmatrix}$


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