Assume that $A^\mathrm{T}A$ is invertible and that $A$ is $3\times 3$, $B$ is $3\times 1$, $\alpha$ is $3\times 1$. How does the expression above reduce?
just to $A^\mathrm{T}B$ ?
Assume that $A^\mathrm{T}A$ is invertible and that $A$ is $3\times 3$, $B$ is $3\times 1$, $\alpha$ is $3\times 1$. How does the expression above reduce?
just to $A^\mathrm{T}B$ ?
If $A$ is square and $A^tA$ is invertible then of course $A$ is invertible, and your expression simplifies to $A^{-1}B$.
The expression itself does not reduce (though there are more complicated methods that are efficient at solving it). I believe your equation may be based on the desire to solve $A\vec{\alpha}=B$ given $A$ and $B$, with $A$ underdetermined, meaning it has fewer columns than rows, which is consistent with your problem statement of $5$ rows and $3$ columns for $A$. This equation has no exact solution, thus the next best thing is sought.
As there is no exact solution, consider that a solution that is the "closest" of all possible. With the notion of "close" as the Euclidean distance, it becomes what is called a Least Squares equation. To give you the very abbreviated version of Least Squares, left multiply both sides of the equation to get $A^TA\vec{\alpha}=A^TB$ If you are not yet familiar with Least Squares, for now know that the reason for this is to have something invertible to work with ($A^TA$ will have an inverse) and to keep things in terms of the range of $A$. Thus, to solve it, again left multiply both sides of the equation to get $(A^TA)^{-1}A^TA\vec{\alpha}=(A^TA)^{-1}A^TB$ which reduces to $\vec{\alpha}=(A^TA)^{-1}A^TB$
This equation then has a solution, and it can be shown to be the closest solution of $\vec{\alpha}$ to the original equation $A\vec{\alpha}=B$
Back again to your original question if it is reduceable; the answer is no-- though there are ways of going about solving this without calculating every middle step as I have shown here, and that would take quite a bit more explaining than I have done thus far.