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 Matrices Multiplication properties Property 1: Associative Property of Multiplication Given Amxn , Bnxp and Cpxq , then A·(B·C)=(A·B)·C . $\fs2Given\;A=\left(\begin{matrix}2&-1\\0&-3\end{matrix}\right),\;B=\left(\begin{matrix}3&-1\\1&0\end{matrix}\right)\;and\;C=\left(\begin{matrix}1&2\\-1&0\end{matrix}\right)$ $\fs2A\cdot\;(B\cdot\;C)=\left(\begin{matrix}2&-1\\0&-3\end{matrix}\right)\cdot\left[\left(\begin{matrix}3&-1\\1&0\end{matrix}\right)\cdot\left(\begin{matrix}1&2\\-1&0\end{matrix}\right)\\right]=\left(\begin{matrix}2&-1\\0&-3\end{matrix}\right)\cdot\left(\begin{matrix}4&6\\1&2\end{matrix}\right)=\left(\begin{matrix}7&10\\-3&-6\end{matrix}\right)$ $\fs2(A\cdot\;B)\cdot\;C=\left[\left(\begin{matrix}2&-1\\0&-3\end{matrix}\right)\cdot\left(\begin{matrix}3&-1\\1&0\end{matrix}\right)\\right]\cdot\left(\begin{matrix}1&2\\-1&0\end{matrix}\right)=\left(\begin{matrix}5&-2\\-3&0\end{matrix}\right)\cdot\left(\begin{matrix}1&2\\-1&0\end{matrix}\right)=\left(\begin{matrix}7&10\\-3&-6\end{matrix}\right)$ Property 2: Multiplication of Matrices is not Commutative Given Amxn and Bnxm, then A·B ≠ B·A. (AB does not have to = BA) $\fs2Given\;A=\left(\begin{matrix}2&-1\\0&-3\end{matrix}\right)\;and\;B=\left(\begin{matrix}3&-1\\1&0\end{matrix}\right)$ $\fs2A\cdot\;B=\left(\begin{matrix}2&-1\\0&-3\end{matrix}\right)\cdot\left(\begin{matrix}3&-1\\1&0\end{matrix}\right)=\left(\begin{matrix}5&-2\\-3&0\end{matrix}\right)$ $\fs2B\cdot\;A=\left(\begin{matrix}3&-1\\1&0\end{matrix}\right)\cdot\left(\begin{matrix}2&-1\\0&-3\end{matrix}\right)=\left(\begin{matrix}6&0\\2&-1\end{matrix}\right)$ Property 3: Distributive Property of Multiplication Given Amxn , Bnxp and Cnxp , then A·(B+C)=A·B +A·C Given Apxq , Bnxp and Cnxp, then (B+C)·A=B·A +C·A Property 4: Multiplicative identity Given Amxn, There are unique matrices In and Im with Amxn·In=Amxn and Im·Amxn=Amxn The matrix In, called the identity matrix or unit matrix of size n is the n-by-n square matrix with ones on the main diagonal and zeros elsewhere. $\fs2A\cdot\;I=\left(\begin{matrix}2&-1\\0&-3\end{matrix}\right)\cdot\left(\begin{matrix}1&0\\0&1\end{matrix}\right)=\left(\begin{matrix}2&-1\\0&-3\end{matrix}\right)$ Property 5: Null Matrices may have Non-null Divisors. The matrix product AB can be zero although A ≠ 0 and B ≠0. Similar, it is possible that A ≠0, A2 ≠ 0, . . . , but Ap = 0. $\fs2Given\;A=\left(\begin{matrix}1&1\\-2&-2\end{matrix}\right)\;and\;\;B=\left(\begin{matrix}3&-1\\-3&1\end{matrix}\right)$ $\fs2then\;A\cdot\;B=0\;;\;\;\left(\begin{matrix}1&1\\-2&-2\end{matrix}\right)\cdot\left(\begin{matrix}3&-1\\-3&1\end{matrix}\right)=\left(\begin{matrix}0&0\\0&0\end{matrix}\right)$ No se verifica la ley de cancelación