could you help me to figure out why the row 1 & 2 are eliminated in here:
$ \begin{Bmatrix} F_1 \\ F_2 \\ F_3 \\ F_4 \\ \end{Bmatrix}= \begin{bmatrix} k & -k & k &-k\\ -k & k & -k &k\\ k & -k & k &-k\\ -k & k & -k &k\\ \end{bmatrix} * \begin{Bmatrix} 0 \\ 0 \\ u_3 \\ u_4 \\ \end{Bmatrix}$ (Assuming k is some scalar)
Since the $u_1=0$ and $u_2=0$ the rows 1 & 2 and columns 1 &2 are eliminated. I understand why the columns 1 and 2 are gone. Since the variable $u_1$ and $u_2$ are zeros the algebraic form is:
$0k-0k+u_3k-u_4k=F_1$
$-0k+0k-u_3k+u_4k=F_2$
$0k-0k+u_3k-u_4k=F_3$
$-0k+0k-u_3k+u_4k=F_4$
But I don't understand why the prof crosses out the rows 1 and 2 as well.
P.s. any edits are welcome. Thank you in advance!
I'm still not totally clear on your question but here's an attempt.
First, deleting the first two columns of the matrix with $k$s gives $$\begin{Bmatrix} F_1 \\ F_2 \\ F_3 \\ F_4 \\ \end{Bmatrix} =\begin{bmatrix} k &-k\\ -k &k\\ k &-k\\ -k &k\\ \end{bmatrix} * \begin{Bmatrix} 0 \\ 0 \\ u_3 \\ u_4 \\ \end{Bmatrix}\ .$$ This doesn't work as the sizes of the matrices on the RHS don't match ($4\times2$ times $4\times1$). You can fix this by deleting the first two rows of the $\bf u$ vector, $$\begin{Bmatrix} F_1 \\ F_2 \\ F_3 \\ F_4 \\ \end{Bmatrix} =\begin{bmatrix} k &-k\\ -k &k\\ k &-k\\ -k &k\\ \end{bmatrix} * \begin{Bmatrix}u_3 \\ u_4 \\ \end{Bmatrix}\ ,$$ and this doesn't change anything important as you have merely stopped adding some zeros. Now if I understood your comment correctly, you have also deleted the first two rows in other matrices, $$\begin{Bmatrix} F_3 \\ F_4 \\ \end{Bmatrix} =\begin{bmatrix} k &-k\\ -k &k\\ \end{bmatrix} * \begin{Bmatrix}u_3 \\ u_4 \\ \end{Bmatrix}\ .$$ This gives you exactly the same information regarding $F_3$ and $F_4$, but you have lost the information about $F_1$ and $F_2$.
As to why... the only guess I can make is that $F_1$ and $F_2$ are irrelevant to the problem you are discussing, perhaps because they are in fact equal to $F_3$ and $F_4$ respectively.