Exercise 2.5.38

$\mathit{Ax}=b$ has many solutions when $A=$ ones $(4,4)=$ singular and $b=$ ones $(4,1)$. $A\setminus b$ in MATLAB will pick the shortest solution $x=(1,1,1,1)∕4$. This is the only solution that is a combination of the rows of $A$ (later it comes from the "pseudoinverse" $A^{\text{+}}=\mathrm{pinv}<!--\; FUNCTION\; APPLICATION\; -->(A)$ which replaces $A^{-1}$ when $A$ is singular). Any vector that solves $\mathit{Ax}=0$ could be added to this particular solution $x$.