0
$\begingroup$

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...

$\endgroup$
1
$\begingroup$

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

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?

$\endgroup$
  • $\begingroup$ 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… $\endgroup$ – nbro Mar 8 '15 at 16:41
  • $\begingroup$ 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. $\endgroup$ – user137731 Mar 8 '15 at 16:54
  • $\begingroup$ 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... $\endgroup$ – nbro Mar 8 '15 at 17:00
  • $\begingroup$ Ok, I just did it for you. Look at my edit. $\endgroup$ – user137731 Mar 8 '15 at 17:03
  • $\begingroup$ 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 $\endgroup$ – nbro Mar 8 '15 at 17:35
0
$\begingroup$

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 $$-2b_1 +b_4 = 0, -57b_1 + 11b_2 + 5b_3 = 0 $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.