Special products

Show that

Variant 1
Matrix \(\mathbf A\mathbf A^T\) is always symmetric.
Resolution
Elements \((\mathbf A\mathbf A^T)_{ij}\) and \((\mathbf A\mathbf A^T)_{ji}\) are results of the same computation on the \(i\)-th and the \(j\)-th row of \(\mathbf A\), hence are the same.
Variant 2
Every matrix of type \(m \times n\) satisfies \(\mathbf I_m\mathbf A=\mathbf A=\mathbf A\mathbf I_n\).
Resolution
The product \((\mathbf I_m\mathbf A)_{ij}\) chooses from the \(j\)-th column of \(\mathbf A\) the \(i\)-th element, i.e. \(a_{ij}\).

The other argument is analogous.
Variant 3
The product of regular matrices is regular.
Resolution
If \(\mathbf A\) and \(\mathbf B\) are regular, then exist \(\mathbf A^{-1}\) and \(\mathbf B^{-1}\). Also holds

\((\mathbf A\mathbf B)(\mathbf B^{-1}\mathbf A^{-1})= \mathbf A(\mathbf B\mathbf B^{-1})\mathbf A^{-1}= \mathbf A(\mathbf I_n)\mathbf A^{-1}= \mathbf A\mathbf A^{-1}= \mathbf I_n\).

I.e. \(\mathbf B^{-1}\mathbf A^{-1}\) is the inverse for \(\mathbf A\mathbf B\), hence \(\mathbf A\mathbf B\) is a regular matrix.