Solving matrix equation $ A X - 3 \left( A^T A^{-1} \right)^{-1} = A $

Question

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.

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

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I have tried 3 different ways and I always get stuck at $A^{-1} 3(A^T)^{-1}$. Can someone help me, please?

Answer 1

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)

$$ X = I_3 + 3A^{-T.} $$

Answer 2

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.

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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.