## College Algebra (6th Edition)

$B$ is not the multiplicative inverse of $A$.
(See p.629) Let A be an n\times n square matrix. If there is a square matrix $A^{-1}$ such that $AA^{-1}=I_{n}$ and $A^{-1}A=I_{n},$ then $A^{-1}$ is the multiplicative inverse of $A$. -------------- The products AB and BA should both be $I_{2}=\left[\begin{array}{ll} 1 & 0\\ 0 & 1 \end{array}\right].$ $AB=\left[\begin{array}{ll} -2(1)+(-1)(1) & -2(1)+(-1)(2)\\ -1(1)+1)(1) & -1(1)+1(2) \end{array}\right]=\left[\begin{array}{ll} -3 & ...\\ ... & ... \end{array}\right]\neq I_{2}$ Since at least one of the products is NOT $I_{2}$, $B$ is not the multiplicative inverse of $A$.