User:Eml4500.f08.bottle.brockmiller/hw5

$$ \begin{bmatrix}1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 1 \cdot \bigg[ \sum_{j=1}^{6}{K_{1j}d_{j}} - F_{1} \bigg]+ 0\cdot \bigg[\sum_{j=1}^{6}{K_{2-6j}d_{j}} - F_{2-6} \bigg]=0 $$ Therefore,

$$ \begin{bmatrix}0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1j}d_{j}} - F_{1} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{2j}d_{j}} - F_{2} \bigg]+ 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{3-6j}d_{j}} - F_{3-6} \bigg]=0 $$ Therefore,

$$ \begin{bmatrix}0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1-2j}d_{j}} - F_{1-2} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{3j}d_{j}} - F_{3} \bigg]+ 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{4-6j}d_{j}} - F_{4-6} \bigg]=0 $$ Therefore,

$$ \begin{bmatrix}0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1-3j}d_{j}} - F_{1-3} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{4j}d_{j}} - F_{4} \bigg]+ 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{5-6j}d_{j}} - F_{5-6} \bigg]=0 $$ Therefore,

$$ \begin{bmatrix}0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1-4j}d_{j}} - F_{1-4} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{5j}d_{j}} - F_{5} \bigg]+ 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{6j}d_{j}} - F_{6} \bigg]=0 $$ Therefore,

$$ \begin{bmatrix}0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1-5j}d_{j}} - F_{1-5} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{6j}d_{j}} - F_{6} \bigg]=0 $$ Therefore,