## College Algebra (6th Edition)

Sample: B(AC)=(BA)C LHS: $AC=\left[\begin{array}{ll} 1 & 0\\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} -1 & 0\\ 0 & 1 \end{array}\right]$ $=\left[\begin{array}{ll} -1+0 & 0+0\\ 0+0 & 0+1 \end{array}\right]=\left[\begin{array}{ll} -1 & 0\\ 0 & 1 \end{array}\right]$ $B(AC)=\left[\begin{array}{ll} 1 & 0\\ 0 & -1 \end{array}\right]\left[\begin{array}{ll} -1 & 0\\ 0 & 1 \end{array}\right]$ $=\left[\begin{array}{ll} -1+0 & 0+0\\ 0+0 & 0-1 \end{array}\right]=\left[\begin{array}{ll} -1 & 0\\ 0 & -1 \end{array}\right]$ RHS: $BA=\left[\begin{array}{ll} 1 & 0\\ 0 & -1 \end{array}\right]\left[\begin{array}{ll} 1 & 0\\ 0 & 1 \end{array}\right]$ $=\left[\begin{array}{ll} 1+0 & 0+0\\ 0+0 & 0-1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0\\ 0 & -1 \end{array}\right]$ $(BA)C=\left[\begin{array}{ll} 1 & 0\\ 0 & -1 \end{array}\right]\left[\begin{array}{ll} -1 & 0\\ 0 & 1 \end{array}\right]$ $=\left[\begin{array}{ll} -1+0 & 0+0\\ 0+0 & 0-1 \end{array}\right]=\left[\begin{array}{ll} -1 & 0\\ 0 & -1 \end{array}\right]$ Thus, B(AC)=(BA)C