## Precalculus (6th Edition) Blitzer

Prove that, $A\left( B+C \right)=AB+AC$ and $\left( A+B \right)C=AC+BC$ holds for any matrix $A,B$ and $C$. So, take the three matrices, $A=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right],B=\left[ \begin{matrix} 1 & 0 \\ 0 & -1 \\ \end{matrix} \right]\text{ and }C=\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$. First we will consider $A\left( B+C \right)$ , That is, \begin{align} & A\left( B+C \right)=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\left( \left[ \begin{matrix} 1 & 0 \\ 0 & -1 \\ \end{matrix} \right]+\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \right) \\ & =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\left( \left[ \begin{matrix} 1-1 & 0+0 \\ 0+0 & -1+1 \\ \end{matrix} \right] \right) \\ & =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right] \end{align} Next we will find, $AB+AC$. For this consider, \begin{align} & AB=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 \\ 0 & -1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1\left( 1 \right)+0\left( 0 \right) & 1\left( 0 \right)+0\left( -1 \right) \\ 0\left( 1 \right)+1\left( 0 \right) & 0\left( 0 \right)+1\left( -1 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 \\ 0 & -1 \\ \end{matrix} \right] \end{align} And, \begin{align} & AC=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1\left( -1 \right)+0\left( 0 \right) & 1\left( 0 \right)+0\left( 1 \right) \\ 0\left( -1 \right)+1\left( 0 \right) & 0\left( 0 \right)+1\left( 1 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \end{align} Now we will add $AB\text{ and }AC$ as below: \begin{align} & AB+AC=\left[ \begin{matrix} 1 & 0 \\ 0 & -1 \\ \end{matrix} \right]+\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right] \end{align} This verifies that $A\left( B+C \right)=AB+AC=\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right]$. Next verify that $\left( A+B \right)C=AC+BC$. For this we will consider, \begin{align} & \left( A+B \right)C=\left( \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]+\left[ \begin{matrix} 1 & 0 \\ 0 & -1 \\ \end{matrix} \right] \right)\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & \left( \left[ \begin{matrix} 1+1 & 0+0 \\ 0+0 & 1-1 \\ \end{matrix} \right] \right)\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 2 & 0 \\ 0 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \end{align} After further simplification, we get, \begin{align} & \left[ \begin{matrix} 2 & 0 \\ 0 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 2\left( -1 \right)+0\left( 0 \right) & 2\left( 0 \right)+0\left( 1 \right) \\ 0\left( -1 \right)+\left( 0 \right)0 & 0\left( 0 \right)+0\left( 1 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -2 & 0 \\ 0 & 0 \\ \end{matrix} \right] \end{align} Therefore, $\left( A+B \right)C=\left[ \begin{matrix} -2 & 0 \\ 0 & 0 \\ \end{matrix} \right]$ Now find $AC$ as follows: \begin{align} & AC=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1\left( -1 \right)+0\left( 0 \right) & 1\left( 0 \right)+0\left( 1 \right) \\ 0\left( -1 \right)+1\left( 0 \right) & 0\left( 0 \right)+1\left( 1 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \end{align} And $BC$ as follows: \begin{align} & BC=\left[ \begin{matrix} 1 & 0 \\ 0 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1\left( -1 \right)+0\left( 0 \right) & 1\left( 0 \right)+0\left( 1 \right) \\ 0\left( -1 \right)+\left( -1 \right)0 & 0\left( 0 \right)+\left( -1 \right)1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -1 & 0 \\ 0 & -1 \\ \end{matrix} \right] \end{align} Add $AC+BC$ as follows: \begin{align} & AC+BC=\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right]+\left[ \begin{matrix} -1 & 0 \\ 0 & -1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -1-1 & 0+0 \\ 0+0 & 1-1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -2 & 0 \\ 0 & 0 \\ \end{matrix} \right] \end{align} Therefore, $AC+BC=\left[ \begin{matrix} -2 & 0 \\ 0 & 0 \\ \end{matrix} \right]$ This verifies that, $\left( A+B \right)C=AC+BC=\left[ \begin{matrix} -2 & 0 \\ 0 & 0 \\ \end{matrix} \right]$ Hence, the distributive property is verified.