I have two questions (True/False and why):
1. If: $$Ax=b$$ has Infinitely many solutions, does necessarily $$A^2x=b$$ has Infinitely many solutions too?
2. If det($A_{n \times n}$)=2, det($B_{n \times n}$)=-2, where $n=4$, does it mean that: $$(A+B)x=0$$ has Infinitely many solutions?
Thank you.
No: if $A = \begin{pmatrix}0 & 1\\\ 0 & 0\end{pmatrix}$ and $b = \begin{pmatrix}1\\\ 0\end{pmatrix}$, then for any scalar $\lambda$, the vector $x = \begin{pmatrix}\lambda\\\ 1\end{pmatrix}$ is solution of $Ax=b$, but $A^2=0$, hence the equation $A^2 x = b$ has no solution.
No: take $$A =\begin{pmatrix}2 & 0 & 0 & 0\\\ 0&1&0&0\\\ 0&0&1&0 \\\ 0&0&0 & 1\end{pmatrix} \qquad B = \begin{pmatrix}1 & 0 & 0 & 0\\\ 0&-2&0&0\\\ 0&0&1&0 \\\ 0&0&0 & 1\end{pmatrix} $$