# Gauss elimination applied to a $4\times 2$ system of equations

I have used Gauss elimination just for a while, and I am still not comfortable, and most of the times I do mistakes and I do not arrive the the right upper triangular matrix.

I have problem which says:

What conditions on all $b$’s must hold (clearly mark your findings), that each of these systems has a solution? Hint. Apply Gauss’ method and see what happens to the right-hand side.

$$x - 2y = b_1 \\ 2x - 9y = b_2 \\ 7x - 3y = b_3 \\ 2x - y = b_4$$

So far, I have just apply the Gaussian elimination method to $3\times 3$ matrices, so, in this case, I would not know how to do it, if it is even possible...

$$\begin{cases} x - 2y = b_1 \\ 2x - 9y = b_2 \\ 7x - 3y = b_3 \\ 2x - y = b_4\end{cases} \iff \left[\begin{array}{cc|c} 1 & -2 & b_1 \\ 2 & -9 & b_2 \\ 7 & -3 & b_3 \\ 2 & -1 & b_4\end{array}\right] \sim \left[\begin{array}{cc|c} 1 & -2 & b_1 \\ 0 & -5 & b_2-2b_1 \\ 0 & 11 & b_3-7b_1 \\ 0 & 3 & b_4-2b_1\end{array}\right] \sim \left[\begin{array}{cc|c} 1 & 0 & b_1+2(\frac{2b_1-b_2}{5}) \\ 0 & 1 & \frac{2b_1-b_2}{5} \\ 0 & 0 & b_3-7b_1 - 11(\frac{2b_1-b_2}{5}) \\ 0 & 0 & b_4-2b_1 - 3(\frac{2b_1-b_2}{5})\end{array}\right]$$

So what does this tell you? What it tells you is that the conditions on the $b_i$'s are $0=b_3-7b_1 - 11(\frac{2b_1-b_2}{5}) \implies -57b_1+11b_2+5b_3=0$ and $0=b_4-2b_1 - 3(\frac{2b_1-b_2}{5}) \implies -16b_1+3b_2+5b_4=0$. The only way your original equations are consistent is if these two equations in only the $b_i$'s hold.

It might be easier if set up a new matrix for these equations:

$$\left[\begin{array}{cccc|c} -57 & 11 & 5 & 0 & 0 \\ -16 & 3 & 0 & 5 & 0\end{array}\right] \sim \left[\begin{array}{cccc|c} 1 & 0 & 3 & -11 & 0 \\ 0 & 1 & 16 & -57 & 0\end{array}\right]$$

Let $b_4=t$ and $b_3=s$ to get $b_1 = -3s+11t$ and $b_2 = -16s+57t$.

So as long as $\pmatrix{b_1 \\ b_2 \\ b_3 \\ b_4} = s\pmatrix{-3 \\ -16 \\ 1 \\ 0} + t\pmatrix{11 \\ 57 \\ 0 \\1}$ for any $s,t \in \Bbb R$, your system will be consistent.

P.S. I've checked my answer and it is correct. Just to let you know, though: your question asks you to solve this via Gaussian reduction, so that's what I did, BUT, there's a much simpler solution that requires $0$ calculation. Can you figure out what that method is?

• Ok, I understood what you have done. But I am not seeing how can I further apply Gaussian elimination to this example. Usually to find the solution, we can go further until we have one 1 per line (and the rest $0$s) except for the constants. For example: $$\left[\begin{array}{ccc|c} 1 & 0 & 0 & b_1\\ 0 & 1 & 0 & b_2\\ 0 & 0 & 1 & b3\\ \end{array}\right]$$ In our case, we have just 2 variables, and I would not know how to proceed… – nbro Mar 8 '15 at 16:41
• You could just convert the four equations back to scalar equations as is, but it'd be easier to do one more step in the Gauss-Jordan elimination. There are $2$ columns, so if the columns are linearly independent (they are), you can find $2$ pivots (a row with a leading $1$, where all of the positions above and below the $1$ are $0$s). I've given you $1$, now do the same process again to get another. Then converting that augmented matrix to scalar equations will give you your answer much quicker. – user137731 Mar 8 '15 at 16:54
• I could try to make $0$ the $-2$ of the first equation, but then what? I have $x$ equals to something in the first equation... – nbro Mar 8 '15 at 17:00
• Ok, I just did it for you. Look at my edit. – user137731 Mar 8 '15 at 17:03
• Oh my god, I don't know if am really retarded, or it's you mathematicians and physicist that are genius :( I am getting crazy with these stupid matrices, I lost a lot of time on internet to find a complete guide, and I have read a lot of things, but I am still retarded – nbro Mar 8 '15 at 17:35

here is another way to look at this complementing Bye-World's solution. you have an over determined system $Ax = b.$ that is you are looking for vectors $b$ in the plane spanned by the first two columns of $A.$ what that requires is $b$ ito be orthogonal to vectors spanned by the first two columns. you can find the orthogonal vectors by row reducing $$\pmatrix{1&2&7&2\\-2&-9&-3&-1} \to \pmatrix{1&2&7&2\\0&-5&11&0}$$ if you set $z = 0, w = 1$ you get $y = 0, x = -2$ and another solution you get by setting $z = 5, w = 0$ you get $y = 11, x = -57.$

so the consistency conditions are $$-2b_1 +b_4 = 0, -57b_1 + 11b_2 + 5b_3 = 0$$