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# Verify the distributive $\mid$ law $A$ ($B$+$C$)= $AB$+$AC$ when$\mathbf{A}=\left(\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right), \quad \mathbf{B}=\left(\begin{array}{rrrr} 2 & -1 & 1 & 0 \\ 3 & -1 & 2 & 1 \end{array}\right), \quad \mathbf{C}=\left(\begin{array}{rrrr} -1 & 1 & 1 & 2 \\ -2 & 2 & 0 & -1 \end{array}\right)$

Solution

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In order to verify the distributive law, we will compute $\boldsymbol{A}(\boldsymbol{B}+\boldsymbol{C})$ and $\boldsymbol{A}\boldsymbol{B}+\boldsymbol{A}\boldsymbol{C}$ separately, then we will compare them. Let's begin by $\boldsymbol{A}(\boldsymbol{B}+\boldsymbol{C})$ :

\begin{aligned} \boldsymbol{B}+\boldsymbol{C}&= \begin{pmatrix} 2&-1&1&0\\3&-1&2&1 \end{pmatrix}+ \begin{pmatrix} -1&1&1&2\\-2&2&0&-1 \end{pmatrix}\\ &=\begin{pmatrix} 2-1&-1+1&1+1&0+2\\3-2&-1+2&2+0&1-1 \end{pmatrix}\\ &=\begin{pmatrix} 1&0&2&2\\1&1&2&0 \end{pmatrix} \end{aligned}

then,

\begin{aligned} \boldsymbol{A}(\boldsymbol{B}+\boldsymbol{C})&= \begin{pmatrix} 1&2\\3&4 \end{pmatrix} \begin{pmatrix} 1&0&2&2\\1&1&2&0 \end{pmatrix}\\ &= \begin{pmatrix} 1\cdot 1+2\cdot 1&1\cdot 0+2\cdot 1&1\cdot 2+2\cdot 2&1\cdot 2+2\cdot 0\\ 3\cdot 1+4\cdot 1&3\cdot 0+4\cdot 1&3\cdot 2+4\cdot 2&3\cdot 2+4\cdot 0\\ \end{pmatrix}\\ &= \begin{pmatrix} 3&2&6&2\\7&4&14&6 \end{pmatrix} \end{aligned}

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