User:Eml4500.f08.lulz.abcd/HW5

Prove Kd=F is equivalent to W(kd-F)=0
$$\mathbf{W}\cdot(\mathbf{K}\mathbf{d}-\mathbf{F})=\begin{Bmatrix} w_3\\ w_4\\ \end{Bmatrix}\cdot(\mathbf{\bar{K}}\mathbf{\bar{d}}-\mathbf{\bar{F}})=0 $$ for all $$ \begin{Bmatrix} w_3\\ w_4\\ \end{Bmatrix}$$

Remember and apply our previous derivations for K, d, and F:

k_{33} & k_{34}\\ k_{43} & k_{44}\\ \end{bmatrix}$$ d_3\\ d_4\\ \end{bmatrix}$$ F_3\\ F_4\\ \end{bmatrix}$$
 * $$\mathbf{K}=\begin{bmatrix}
 * $$\mathbf{d}=\begin{bmatrix}
 * $$\mathbf{F}=\begin{bmatrix}

$$\begin{Bmatrix} w_3\\ w_4\\ \end{Bmatrix}\cdot(\begin{bmatrix} k_{33} & k_{34}\\ k_{43} & k_{44}\\ \end{bmatrix}\begin{bmatrix} d_{3}\\ d_{4} \\ \end{bmatrix}-\begin{bmatrix} F_{3}\\ F_{4} \\ \end{bmatrix})=0$$

This matrix can be converted by the dot product and simple matrix math to a single algebraic expression:

$$w_3(k_{33}d_3+k_{34}d_4-F_3)+w_4(k_{43}d_3+k_{44}d_4-F_4)=0$$

This equation will have multiple solutions depending on $$w_3$$ and $$w_4$$. Since by definition the relationships apply for any $$w_3$$ and $$w_4$$ then any arbitrary value can be applied to solve the expression created above. Picking different sets of values will create a system of equations that, when written in matrix form, are identical to $$\mathbf{W}\cdot(\mathbf{K}\mathbf{d}-\mathbf{F})=0$$ and other values will also prove identity.

If we pick 0 for our values of w:

$$w_3=0, w_4=0$$

$$(0)(k_{33}d_3+k_{34}d_4-F_3+(0)(k_{43}d_3+k_{44}d_4-F_4)=0$$

$$0=0$$

If we pick 0 for one value of w and 1 for the other:

$$w_3=0, w_4=1$$

$$(0)(k_{33}d_3+k_{34}d_4-F_3+(1)(k_{43}d_3+k_{44}d_4-F_4)=0$$

$$k_{43}d_3+k_{44}d_4-F_4=0$$

$$F_4=k_{43}d_3 + k_{44}d_4$$

If we pick 1 for our values of w:

$$w_3=1, w_4=1$$

$$(1)(k_{33}d_3+k_{34}d_4-F_3+(1)(k_{43}d_3+k_{44}d_4-F_4)=0$$

$$k_{33}d_3+k_{34}d_4+k_{43}d_3+k_{44}d_4=F_3+F_4$$

$$\begin{bmatrix} k_{33}d_3 + k_{34}d_4\\ k_{43}d_3 + k_{44}d_4\\ \end{bmatrix}=\begin{bmatrix} F_3\\ F_4\\ \end{bmatrix}$$

These three examples each bring us back to our proof that:

\bold{Kd}=\bold{F}