Solving matrix equation $ A X - 3 \left( A^T A^{-1} \right)^{-1} = A $ I am doing some exercises on matrix equations for my upcoming exam and I have some questions the answers to which I cannot find online. I do study in Portuguese and maybe I don't know the right vocabulary for the search, so I'm trying this community.


Prove that the solution to the following matrix equation $$ A X - 3 \left( A^T A^{-1} \right)^{-1} = A $$ is $$ X = I_3 + 3 \left(A^{T}\right)^{-1}$$


I have tried 3 different ways and I always get stuck at $A^{-1} 3(A^T)^{-1}$. Can someone help me, please?
 A: Good afternoon, José! I believe this may help you:
First of all, notice that in $ AX − 3(A^TA^{-1})^{-1} = A $ it is already implicit that $ A $ is invertible. From that, you can deduce that $ A $ is a nonsingular matrix (its determinant is non-zero), for a matrix is invertible if, and only if, its determinant is non-zero.
Furthermore, notice that a determinant does not change with transposition, so the matrix $ A^T $ must also be nonsingular an thus invertible.
From now on, we can go towards your problem with the $ A^{−1}3(A^T)^{−1}$ matrix, which I will rewrite as $3 A^{−1}(A^T)^{−1}$, knowing that while multiplying a matrix by a scalar, the order doesn't matter.
Let's begin with the manipulations:
Starting with $AX − 3(A^TA^{-1})^{-1} = A$, we can multiply both sides by the inverse of $ A $, which gives
$$ X − 3A^{-1}(A^TA^{-1})^{-1} = I_3. $$
Following that, we can rearrange the expression to get:
$$ X = I_3 + 3A^{-1}(A^TA^{-1})^{-1}. $$
Now, we can use the following property, which will help us to solve the problem you pointed out: 'For any invertible n-by-n matrices $A$ and $B$, $(AB)^{−1} = B^{−1}A^{−1}$'.
(More on that can be found in the fifth bullet point here)
Note that both $ A $ and $A^T$ are invertible, as I have already pointed out. Consequently, the property holds and $$ (A^TA^{−1})^{−1} = (A^{−1})^{−1}(A^T)^{−1} = A(A^T)^{−1}. $$
Now, we can return to $ X = I_3 + 3A^{-1}(A^TA^{-1})^{-1} $ and solve the problem!
$$ X = I_3 + 3A^{-1}(A^TA^{-1})^{-1} $$
$$ X = I_3 + 3A^{-1}(A(A^T)^{−1}) $$
$$ X = I_3 + 3(A^{-1}A)(A^T)^{−1} $$
$$ X = I_3 + 3(I_3)(A^T)^{−1} $$
$$ X = I_3 + 3(A^T)^{−1} $$
Leaving us with only a particular form of notation for $ (A^T)^{−1} $ : $ A^{-T} $
(Also, the fourth bullet point of the last link may help visualizing this)
$$ X = I_3 + 3A^{-T.} $$
A: You already are given a form for $X$ that can be proven simply by plugging the value in and verifying that the left hand side of the equation equals the right hand side.

One key identity (for invertible matrices) will be $(MN)^{-1}=N^{-1}M^{-1}$.
Try to apply that to the $3(A^{T}A^{-1})^{-1}$ part of your equation and see how far you can get from there.
