# What am I doing wrong solving this system of equations?

$$\begin{cases} 2x_1+5x_2-8x_3=8\\ 4x_1+3x_2-9x_3=9\\ 2x_1+3x_2-5x_3=7\\ x_1+8x_2-7x_3=12 \end{cases}$$

From my elementary row operations, I get that it has no solution. (Row operations are to be read from top to bottom.)

$$\left[\begin{array}{ccc|c} 2 & 5 & -8 & 8 \\ 4 & 3 & -9 & 9 \\ 2 & 3 & -5 & 7 \\ 1 & 8 & -7 & 12 \end{array}\right] \overset{\overset{\large{R_1\to R_1-R_3}}{{R_2\to R_2-2R_3}}}{\overset{R_3\to R_3-2R_4}{\large\longrightarrow}} \left[\begin{array}{ccc|c} 0 & 2 & -3 & 1 \\ 0 & -3 & 1 & -5 \\ 0 & -13 & 9 & -17 \\ 1 & 8 & -7 & 12 \end{array}\right] \overset{\overset{\large{R_3\,\leftrightarrow\, R_4}}{R_2\,\leftrightarrow\, R_3}}{\overset{R_1\,\leftrightarrow\,R_2}{\large\longrightarrow}} \left[\begin{array}{ccc|c} 1 & 8 & -7 & 12 \\ 0 & 2 & -3 & 1 \\ 0 & -3 & 1 & -5 \\ 0 & -13 & 9 & -17 \end{array}\right]$$

$$\overset{R_4\to R_4-R_3}{\large\longrightarrow} \left[\begin{array}{ccc|c} 1 & 8 & -7 & 12 \\ 0 & 2 & -3 & 1 \\ 0 & -3 & 1 & -5 \\ 0 & 10 & 8 & -12 \end{array}\right] \overset{\overset{\large{R_3\to R_3+R_2}}{R_4\to R_4-5R_2}}{\large\longrightarrow} \left[\begin{array}{ccc|c} 1 & 8 & -7 & 12 \\ 0 & 2 & -3 & 1 \\ 0 & -1 & -2 & -4 \\ 0 & 0 & 23 & -17 \end{array}\right] \overset{\overset{\large{R_2\to R_2+2R_3}}{R_3\to-R_3}}{\large\longrightarrow}$$

$$\left[\begin{array}{ccc|c} 1 & 8 & -7 & 12 \\ 0 & 0 & -7 & -7 \\ 0 & 1 & 2 & 4 \\ 0 & 0 & 23 & -17 \\ \end{array}\right] \overset{R_2\,\leftrightarrow\,R_3}{\large\longrightarrow} \left[\begin{array}{ccc|c} 1 & 8 & -7 & 12 \\ 0 & 1 & 2 & 4 \\ 0 & 0 & -7 & -7 \\ 0 & 0 & 23 & -17 \\ \end{array}\right]$$

However, the answer in the book $$(3, 2, 1)$$ fits the system.

Was there an arithmetical mistake, or do I misunderstand something fundamentally?

• It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures. – José Carlos Santos Nov 7 '18 at 9:59
• I would also recommend simply checking answers via a computer, you can quickly see your arithmetic mistake matrix.reshish.com/gauss-jordanElimination.php . Even the most experienced mathematicians still make minus sign errors or simple multiplication errors. We're mathematicians after all, not primary school teachers :P – Wesley Strik Nov 7 '18 at 10:23

Hint: Try inputing the solution $$(3,2,1)$$ into every step. That will allow you to identify the step where you went wrong.

• Note: this is probably the simplest answer I ever gave that recieved a score of 10 :) – 5xum Nov 7 '18 at 11:56
• it's also a great and remarkably salient answer that fully solves the problem and many others like it. In other words: it's the best kind of answer. Well-done. – The Count Nov 7 '18 at 22:45
• In addition, it is a nice little case where it makes sense to do a binary search... – Sebastian Schoennenbeck Nov 8 '18 at 15:26
• Hint: There is only one equation in the last set that does not work with that solution, so it is not necessary to input the solution in EVERY equation, just working backwards from the wrong one reveals the problem relatively quickly. – kutschkem Nov 8 '18 at 16:08

You do (in the third matrix): $$L3-L4=(0, -3, 1 \mid -5)-(0, -13, 9 \mid -19)=(0, 10, -8 \mid 12)$$ but you have $$(0, 10, 8 \mid -12)$$ instead.

Note 1: Referring to 5xum's answer, in case you don't know the final answer, any solution will do, e.g.: $$(x_1,x_2,x_3)=(0,0,-1);(0,0,-1);(1,0,-1);(12,0,0) \ \ \text{(respectively)}$$ Note 2: In step $$3$$, you can reduce column $$3$$ instead of column $$2$$: $$\left[ \begin{array}{ccc|c} 1&8&-7&12\\ 0&2&-3&1\\ 0&-3&1&-5\\ 0&-13&9&-17\\ \end{array} \right] \Rightarrow \left[ \begin{array}{ccc|c} 1&8&-7&12\\ 0&-7&0&-14\\ 0&-3&1&-5\\ 0&14&0&28\\ \end{array} \right] \stackrel{\frac{-R_2}{7};\\ \frac{-3R_2}{7}+R_3}{=}\Rightarrow \left[ \begin{array}{ccc|c} 1&8&-7&12\\ 0&1&0&2\\ 0&0&1&1\\ 0&14&0&28\\ \end{array} \right]$$ The second and fourth equations are dependent, therefore they produce the same solution $$x_2=2$$. You can finish the problem now.

Note 3. When you can not find your mistake (sometimes the brain gets blocked/accustommed and can not see obvious mistakes), leave it for some time (1 hour, 1 day) and return with fresh mind. You will be surprised to easily spot the error.

• +1 for the last paragraph: Keep calm & try again later is a good approach – Barranka Nov 7 '18 at 18:20
• can you explain Note 1? How can I use this four tuples you show us to find the error? – miracle173 Nov 8 '18 at 6:48
• @miracle173, unfortunately all four cannot be used at once, but pairwise solutions (which is relatively simple to find by putting $0$ into one of three unknowns) must be checked. For example, equations $1$ and $2$ have common solution $(0,0,-1)$, so any row operation of the two equations must be satisfied by it. Or $(8/3,0,-1/3)$ is a common solution of the equations $1$ and $3$, so the OP's first result $[0 \ 2 \ -3 \ | \ 1]$ must be satisfied by it. – farruhota Nov 8 '18 at 18:52
• I understand now, but this seems not to be useful to me. – miracle173 Nov 8 '18 at 20:38

Usually you don't know the solution of your equation. So you have to use other methods for check your calculation. One way is to add checksums to your calculation.

Don't start your calculations with the original matrix

$$\left[\begin{array}{ccc|c} 2 & 5 & -8 & 8 \\ 4 & 3 & -9 & 9 \\ 2 & 3 & -5 & 7 \\ 1 & 8 & -7 & 12 \end{array}\right]$$

but start with the matrix augmented by a column at the right. An element of this column is equal to the sum of the other elements of this row. So the augmented matrix is

$$\left[\begin{array}{ccc|c|c} 2 & 5 & -8 & 8 & 7 \\ 4 & 3 & -9 & 9 & 7 \\ 2 & 3 & -5 & 7 & 7 \\ 1 & 8 & -7 & 12 & 14 \end{array}\right]$$

You do the same row operations as with the original matrices.

$$\left[\begin{array}{ccc|c} 2 & 5 & -8 & 8 \\ 4 & 3 & -9 & 9 \\ 2 & 3 & -5 & 7 \\ 1 & 8 & -7 & 12 \end{array}\right] \overset{\overset{\large{R_1\to R_1-R_3}}{{R_2\to R_2-2R_3}}}{\overset{R_3\to R_3-2R_4}{\large\longrightarrow}} \left[\begin{array}{ccc|c} 0 & 2 & -3 & 1 \\ 0 & -3 & 1 & -5 \\ 0 & -13 & 9 & -17 \\ 1 & 8 & -7 & 12 \end{array}\right]$$

you do

$$\left[\begin{array}{ccc|c|c} 2 & 5 & -8 & 8 & 7 \\ 4 & 3 & -9 & 9 & 7 \\ 2 & 3 & -5 & 7 & 7 \\ 1 & 8 & -7 & 12 & 14 \end{array}\right] \overset{\overset{\large{R_1\to R_1-R_3}}{{R_2\to R_2-2R_3}}}{\overset{R_3\to R_3-2R_4}{\large\longrightarrow}} \left[\begin{array}{ccc|c} 0 & 2 & -3 & 1 & 0 \\ 0 & -3 & 1 & -5 & -7 \\ 0 & -13 & 9 & -17 & -21 \\ 1 & 8 & -7 & 12 & 14 \end{array}\right]$$

Then you check if for the matrix you calculated the checksum is still valid. Here the checksum is correct.

Now let's investigate the step where the error occurs. We get

$$\left[\begin{array}{ccc|c|c} 1 & 8 & -7 & 12 & 14 \\ 0 & 2 & -3 & 1 & 0 \\ 0 & -3 & 1 & -5 & -7 \\ 0 & -13 & 9 & -17 & -21 \end{array}\right] \overset{R_4\to R_4-R_3}{\large\longrightarrow} \left[\begin{array}{ccc|c | c} 1 & 8 & -7 & 12 & 14\\ 0 & 2 & -3 & 1 & 0\\ 0 & -3 & 1 & -5 & -7\\ 0 & 10 & 8 & -12 & -14 \end{array}\right]$$

For this result matrix the checksum of row 4 is not valid, so an error must have occurred.

Actually the checksum property means $$(1,1,1,1)^T$$ is a solution of the following system of equations that corresponds to the augmented matrix:

$$\begin{eqnarray} 2x_1&+&5x_2&-&8x_3&+&8x_4&=&7\\ 4x_1&+&3x_2&-&9x_3&+&9x_4&=&7\\ 2x_1&+&3x_2&-&5x_3&+&7x_4&=&7\\ x_1&+&8x_2&-&7x_3&+&12x_4&=&14 \end{eqnarray}$$

So this method is similar to the method in the accepted answer, where the solution given by the textbook is used.

What if this method does not uncover the errors. This can happen if you made more then one error in a step- Then you can use a weighted checksum to try to uncover the error.