## College Algebra (6th Edition)

B is the multiplicative inverse of A. $(B=A^{-1})$
(See p.629) Let A be an $\mathrm{n}\times \mathrm{n}$ 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 $I_{2}=\left[\begin{array}{ll} 1 & 0\\ 0 & 1 \end{array}\right].$ $AB=\left[\begin{array}{ll} -2(1)+1(3) & -2(2)+1(4)\\ \frac{3}{2}(1)+(-\frac{1}{2})(3) & \frac{3}{2}(2)+(-\frac{1}{2})(4) \end{array}\right]=\left[\begin{array}{ll} 1 & 0\\ 0 & 1 \end{array}\right]=I_{2}$ $BA=\left[\begin{array}{ll} 1(-2)+2(\frac{3}{2}) & 1(1)+2(-\frac{1}{2})\\ 3(-2)+4(\frac{3}{2}) & 3(1)+4(-\frac{1}{2}) \end{array}\right]=\left[\begin{array}{ll} 1 & 0\\ 0 & 1 \end{array}\right]=I_{2}$ $AB=BA=I_{2},$ so B is the multiplicative inverse of A. $(B=A^{-1})$