# Why row and column eliminated if variable is zero in matrix operations

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!

• Which rows 1 and 2 are crossed out? The LHS? First factor on the RHS? Second factor on the RHS? More than one of these? Commented Sep 11, 2018 at 2:14
• uhm, all. row 1 and 2. Commented Sep 11, 2018 at 11:26
• I'm just wondering if there is some matrix operation here that I'm missing out on. Commented Sep 11, 2018 at 11:28

## 1 Answer

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.

• Thank you for your answer! F1 and F2 are in fact what we are looking for. And F3 and F4 are usually known. So when we get this new 2X2 matricies then we can find the u3 and u4. And then! Here is a kicker. Then we revive the 4X4 matrix, and solve for F1 and F2. Because, now F1 & F2 are the only 2 unknowns. So basically we dropped them, for convenience (as Prof said) and then reinstated them and solved them. Is this any clearer? What do you think? Commented Oct 2, 2018 at 19:10
• Actually, I'm much more confused now. If you look at the first and third rows of your matrix product you get$$F_1=ku_3-ku_4\ ,\quad F_3=ku_3-ku_4\ ,$$so $F_1=F_3$. I don't understand how $F_3$ can be known and $F_1$ unknown when they are in fact the same! Commented Oct 3, 2018 at 0:13
• You forgot to flip the sign. They may be equal (given k are the same) but opposite in sign. Commented Oct 3, 2018 at 12:46
• No, look more carefully at the first matrix equation in your question: $F_1=F_3=-F_2=-F_4$. So if you know one of the $F$s then you know them all. Commented Oct 4, 2018 at 0:01
• This may be a stupid question (sorry abut that) but why are you equating the forces? P.s. corrected the signs in the equation form. Commented Oct 5, 2018 at 21:43