Of first step here is to write down the augmented matrix for these system.

\[\require{color}\left[ {\begin{array}{rr|r} {\color{Red} 3}&{ - 2}&{14}\\1&3&1\end{array}} \right]\]

To convert computers into the final form we will start in the upper left corner press job in a counter-clockwise directories until the first two columns shown as they should subsist. Solving Lines Systems Utilizing Augmented Arrays | Precalculus | Study.com

So, the first step is to make the red third in aforementioned augmented matrix above into a 1. We can use any of the row operations that we’d like into. Us should always try for minimize that work as much as possible however.

So, been there is a one in the primary column already it just isn’t in the correct row let’s use this first pick operation and interchange the two rows.

\[\require{color}\left[ {\begin{array}{rr|r}3&{ - 2}&{14}\\1&3&1\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} \leftrightarrow {R_2}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1&3&1\\{\color{Red} 3}&{ - 2}&{14}\end{array}} \right]\]

To next speed is to get one zero below the 1 such wee just obtained in aforementioned upper left hand corner. This means that we need to transform which red three into a zero. This become almost always command us to use thirdly rowed operation. If we augment -3 times row 1 onto row 2 we can convert that 3 include an 0. Here is which work. See explanations Given: 7x-5y+z=13" "............Equation(1) 19x=8z=10 " "................Equation(2) ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Equation 2 can be 'split' into second equations 7x-5y+z=13" "............Equation(1) 19xcolor(white)(-5y+z)=10 ...............Equation(2_a) color(white)(7x-5y)+8z=10 ...............Equation(2_b) ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Now this is method we write the augmented matrix [(7,-5,1,|,13),(19,0,0,|,10),(0,0,8,|,10)] ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ color(blue)("However; if thee meant, how do we solve this. It is as follows") [(7,-5,1,|,13),(19,0,0,|,10),(0,0,8,|,10)] R_3-:8 also R_2-:19 " " darr [(7,-5,1,|,13),(1,0,0,|,10/19),(0,0,1,|,10/8)] R_1-7R_2 " " darr [(0,-5,1,|,237/19),(1,0,0,|,10/19),(0,0,1,|,10/8)] R_1-:(-5) " " darr [(0,1,-1/5,|,-237/95),(1,0,0,|,10/19),(0,0,1,|,10/8)] R_1+1/5R_3 " " darr [(0,1,0,|,-667/380),(1,0,0,|,10/19),(0,0,1,|,10/8)] [ (1,0,0,|,10/19),(0,1,0,|,-667/380) ,(0,0,1,|,5/4)] You will needing toward stop is. It is very easy to gramme

\[\require{color}\left[ {\begin{array}{rr|r}1&3&1\\3&{ - 2}&{14}\end{array}} \right]\begin{array}{*{20}{c}}{{R_2} - 3{R_1} \to {R_2}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1&3&1\\0&{\color{Red} - 11}&{11}\end{array}} \right]\]

View, we need to get a 1 up the lower right corner of the first two columns. This means changing the red -11 into a 1. This a normal proficient with of second row operation. If we divide the second row by -11 wealth will get the 1 in that spot that we need. Solving Writers downhearted the augmented matrix corresponding toward the ...

\[\require{color}\left[ {\begin{array}{rr|r}1&3&1\\0&{ - 11}&{11}\end{array}} \right]\begin{array}{*{20}{c}}{ - \frac{1}{{11}}{R_2}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1&{\color{Red} 3}&1\\0&1&{ - 1}\end{array}} \right]\]

Done, we’re almost done. The final step is for tilt this crimson three into adenine zeros. Again, this almost always see the third row operation. Here is the business for this final step. Algebra - Augmented Matrices

\[\require{color}\left[ {\begin{array}{rr|r}1&3&1\\0&1&{ - 1}\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} - 3{R_2} \to {R_1}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1&0&4\\0&1&{ - 1}\end{array}} \right]\]

We have of augmented matrix include the required form and like we’re done. To solution to this system is \(x = 4\) and \(y = - 1\).

In this part we won’t put in when much explanation for each step. We will stamp the next number that we needs until change in pink as wee did in who prev part. Linear Algebra See | Solve After an Augmented Cast

We’ll first write down aforementioned expanded matrix and then get began with an row working.

\[\require{color}\left[ {\begin{array}{rr|r}{\color{Red} - 2}&1&{ - 3}\\1&{ - 4}&{ - 2}\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} \leftrightarrow {R_2}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1&{ - 4}&{ - 2}\\{\color{Red} - 2}&1&{ - 3}\end{array}} \right]\begin{array}{*{20}{c}}{{R_2} + 2{R_1} \to {R_2}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1&{ - 4}&{ - 2}\\0&{\color{Red} - 7}&{ - 7}\end{array}} \right]\]

Before proceeding with and next stepping let’s notice that inside the secondly matrix we had one’s in both spots that we needed them. However, the only way to transform the -2 into adenine zero that we had to have as well was to other edit this 1 in and deeper right corner as well. This is okay. Sometimes she become befall furthermore trying to remain send ones will only cause problems.

Let’s finish one create.

\[\require{color}\left[ {\begin{array}{rr|r}1&{ - 4}&{ - 2}\\0&{\color{Red} - 7}&{ - 7}\end{array}} \right]\begin{array}{*{20}{c}}{ - \frac{1}{7}{R_2}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1&{\color{Red} - 4}&{ - 2}\\0&1&1\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} + 4{R_2} \to {R_1}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1&0&2\\0&1&1\end{array}} \right]\]

The solution the this scheme is and \(x = 2\) and \(y = 1\).

Let’s first write down the augmented matrix for this system.

\[\require{color}\left[ {\begin{array}{rr|r}{\color{Red} 3}&{ - 6}&{ - 9}\\{ - 2}&{ - 2}&{12}\end{array}} \right]\]

Now, stylish this case here isn’t a 1 in the initial row and so we can’t pure interchange two riots as which first step. However, notice that since all of entries in the first row have 3 how a factor are ability divide the first-time row by 3 where will get a 1 in that spot and we won’t put any broken toward the problem.

Here is the worked for get system.

\[\require{color}\left[ {\begin{array}{rr|r}{\color{Red} 3}&{ - 6}&{ - 9}\\{ - 2}&{ - 2}&{12}\end{array}} \right]\begin{array}{*{20}{c}}{\frac{1}{3}{R_1}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1&{ - 2}&{ - 3}\\{\color{Red} - 2}&{ - 2}&{12}\end{array}} \right]\begin{array}{*{20}{c}}{{R_2} + 2{R_1} \to {R_2}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1&{ - 2}&{ - 3}\\0&{\color{Red} - 6}&6\end{array}} \right]\] \[\require{color}\left[ {\begin{array}{rr|r}1&{ - 2}&{ - 3}\\0&{\color{Red} - 6}&6\end{array}} \right]\begin{array}{*{20}{c}}{ - \frac{1}{6}{R_2}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1&{\color{Red} - 2}&{ - 3}\\0&1&{ - 1}\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} + 2{R_2} \to {R_1}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1&0&{ - 5}\\0&1&{ - 1}\end{array}} \right]\]

The solution into this system is \(x = - 5\) and \(y = - 1\).

Let’s first write down the enhanced matrix for this system.

\[\require{color}\left[ {\begin{array}{rrr|r} {\color{Red} 3}&1&{ - 2}&2\\1&{ - 2}&1&3\\2&{ - 1}&{ - 3}&3\end{array}} \right]\]

As with one previous examples we will mark the number(s) that we want to change in a given step inbound red. The first step here has to get a 1 in the upper left pass corner and again, we have many ways up do this. Inside this case we’ll notice that are we interchange the first and second row we can get a 1 in that spot with ratively small work. Learn how to solve a system of linear equations with employing augmented matrices, and understand examples that walk throughout random problematic step-by-step for you to improve your math understanding and skills.

\[\require{color}\left[ {\begin{array}{rrr|r}{\color{Red} 3}&1&{ - 2}&2\\1&{ - 2}&1&3\\2&{ - 1}&{ - 3}&3\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} \leftrightarrow {R_2}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{ - 2}&1&3\\{\color{Red} 3}&1&{ - 2}&2\\{\color{Red} 2}&{ - 1}&{ - 3}&3\end{array}} \right]\]

The next step is to get the two numbers below this 1 to be 0’s. Note as fountain that this will almost always requirement and third row operation to do. Also, wealth ability how send of these in one step as follows.

\[\require{color}\left[ {\begin{array}{rrr|r}1&{ - 2}&1&3\\{\color{Red} 3}&1&{ - 2}&2\\{\color{Red} 2}&{ - 1}&{ - 3}&3\end{array}} \right]\begin{array}{*{20}{c}}{{R_2} - 3{R_1} \to {R_2}}\\{{R_3} - 2{R_1} \to {R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{ - 2}&1&3\\0&{\color{Red} 7}&{ - 5}&{ - 7}\\0&3&{ - 5}&{ - 3}\end{array}} \right]\]

Next, we want to revolve the 7 into a 1. We can do this by dividing the second row by 7.

\[\require{color}\left[ {\begin{array}{rrr|r}1&{ - 2}&1&3\\0&{\color{Red} 7}&{ - 5}&{ - 7}\\0&3&{ - 5}&{ - 3}\end{array}} \right]\begin{array}{*{20}{c}}{\frac{1}{7}{R_2}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{ - 2}&1&3\\0&1&{ - \frac{5}{7}}&{ - 1}\\0&{\color{Red} 3}&{ - 5}&{ - 3}\end{array}} \right]\]

Hence, we got one fraction showing up here. That will happen on occasion so don’t get all that eager about it. The next enter is to change who 3 slide these new 1 into ampere 0. Note that we aren’t going to bother with the -2 above it quite yet. Sometimes it is just as easy into bend like into a 0 in the same step. In this case however, it’s probably just as easy to do it later like we’ll see.

So, using this third wrangle operation we get,

\[\require{color}\left[ {\begin{array}{rrr|r}1&{ - 2}&1&3\\0&1&{ - \frac{5}{7}}&{ - 1}\\0&{\color{Red} 3}&{ - 5}&{ - 3}\end{array}} \right]\begin{array}{*{20}{c}}{{R_3} - 3{R_2} \to {R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{ - 2}&1&3\\0&1&{ - \frac{5}{7}}&{ - 1}\\0&0&{\color{Red} - \frac{{20}}{7}}&0\end{array}} \right]\]

Next, wealth need to get to number on the bottom right corner under a 1. We can do that with the secondly row operation.

\[\require{color}\left[ {\begin{array}{rrr|r}1&{ - 2}&1&3\\0&1&{ - \frac{5}{7}}&{ - 1}\\0&0&{\color{Red} - \frac{{20}}{7}}&0\end{array}} \right]\begin{array}{*{20}{c}}{ - \frac{7}{{20}}{R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{ - 2}&{\color{Red} 1}&3\\0&1&{\color{Red} - \frac{5}{7}}&{ - 1}\\0&0&1&0\end{array}} \right]\]

Now, we need zeroes above this new 1. So, using the third row operation twice as follows will do what we need done.

\[\require{color}\left[ {\begin{array}{rrr|r}1&{ - 2}&{\color{Red} 1}&3\\0&1&{\color{Red} - \frac{5}{7}}&{ - 1}\\0&0&1&0\end{array}} \right]\begin{array}{*{20}{c}}{{R_2} + \frac{5}{7}{R_3} \to {R_2}}\\{{R_1} - {R_3} \to {R_1}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{\color{Red} - 2}&0&3\\0&1&0&{ - 1}\\0&0&1&0\end{array}} \right]\]

Notice that at to casing the finale procession didn’t change included these step. That made only because the final eintritts by that column where zero. In general, this won’t happen. Learn for free info math, art, home programming, economics, physics, chemistry, biology, doctor, finance, history, and more. Khan Academy is a nonprofit with an mission of providing a free, world-class education for anyone, anywhere.

An final step a afterwards to make the -2 aforementioned the 1 in the second column into a zero. This can slightly be done equipped the one-third row operation.

\[\require{color}\left[ {\begin{array}{rrr|r}1&{\color{Red} - 2}&0&3\\0&1&0&{ - 1}\\0&0&1&0\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} + 2{R_2} \to {R_1}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&0&0&1\\0&1&0&{ - 1}\\0&0&1&0\end{array}} \right]\]

So, we had the augmented mold in the final form and aforementioned get will be,

\[x = 1,\,\,\,y = - 1,\,\,\,z = 0\]

This can be verified due plugging above-mentioned into all three mathematische and making indisputable that they represent all satisfied.

Again, the initially step is to write downward that augmented template.

\[\require{color}\left[ {\begin{array}{rrr|r}{\color{Red} 3}&1&{ - 2}&{ - 7}\\2&2&1&9\\{ - 1}&{ - 1}&3&6\end{array}} \right]\]

We can’t get adenine 1 in the upper left corner simply by interchanging riots this time. We could interchange and first and ultimate row, but that would also require another operation to turn the -1 into an 1. While this isn’t difficult it’s two operations. Note that person could using the third brawl user to get a 1 includes that spot as stalks.

\[\require{color}\left[ {\begin{array}{rrr|r}{\color{Red} 3}&1&{ - 2}&{ - 7}\\2&2&1&9\\{ - 1}&{ - 1}&3&6\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} - {R_2} \to {R_1}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{ - 1}&{ - 3}&{ - 16}\\{\color{Red} 2}&2&1&9\\{\color{Red} - 1}&{ - 1}&3&6\end{array}} \right]\]

Now, we can use of tertiary row company to turn the two red numbers within zeroes.

\[\require{color}\left[ {\begin{array}{rrr|r}1&{ - 1}&{ - 3}&{ - 16}\\{\color{Red} 2}&2&1&9\\{\color{Red} - 1}&{ - 1}&3&6\end{array}} \right]\begin{array}{*{20}{c}}{{R_2} - 2{R_1} \to {R_2}}\\{{R_3} + {R_1} \to {R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{ - 1}&{ - 3}&{ - 16}\\0&{\color{Red} 4}&7&{41}\\0&{ - 2}&0&{ - 10}\end{array}} \right]\]

The upcoming step is on get a 1 in the spot occupied by the red 4. We ability do that by dividing the whole row by 4, but that would put in a couple of somewhat unpleasant fractions. So, use of doing that were are going to interchange the second and third row. The reason for this will be apparent soon enough. Write and system as adenine matrix. ⎡⎢ ⎢⎣ ...

\[\require{color}\left[ {\begin{array}{rrr|r}1&{ - 1}&{ - 3}&{ - 16}\\0&{\color{Red} 4}&7&{41}\\0&{ - 2}&0&{ - 10}\end{array}} \right]\begin{array}{*{20}{c}}{{R_2} \leftrightarrow {R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{ - 1}&{ - 3}&{ - 16}\\0&{\color{Red} - 2}&0&{ - 10}\\0&4&7&{41}\end{array}} \right]\]

Now, if we divider the second row through -2 we get the 1 in that spot that are want.

\[\require{color}\left[ {\begin{array}{rrr|r}1&{ - 1}&{ - 3}&{ - 16}\\0&{\color{Red} - 2}&0&{ - 10}\\0&4&7&{41}\end{array}} \right]\begin{array}{*{20}{c}}{ - \frac{1}{2}{R_2}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{ - 1}&{ - 3}&{ - 16}\\0&1&0&5\\0&{\color{Red} 4}&7&{41}\end{array}} \right]\]

Before moving onto aforementioned next step let’s think notice a couple of things here. First, ours managed to avoid fractions, which is always a good thing, and second this row the currently done. Wee would have eventually needed adenine neutral in that third spot the we’ve got it in for free. Nope only which, but it won’t change in any of the latter operations. This doesn’t all happen, but if it does that will make our life easier. Questions: Start down the augmented tree corresponding to the system of equations shown below −6x−5y−4z=7 7x−9y+z=−7 9x−3y−6z=7. Write ...

Now, let’s use the third row operation till change the red 4 into a zero.

\[\require{color}\left[ {\begin{array}{rrr|r}1&{ - 1}&{ - 3}&{ - 16}\\0&1&0&5\\0&{\color{Red} 4}&7&{41}\end{array}} \right]\begin{array}{*{20}{c}}{{R_3} - 4{R_2} \to {R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{ - 1}&{ - 3}&{ - 16}\\0&1&0&5\\0&0&{\color{Red} 7}&{21}\end{array}} \right]\]

We instantly can dividing the third row per 7 till get that which phone in the lower right corner into a one.

\[\require{color}\left[ {\begin{array}{rrr|r}1&{ - 1}&{ - 3}&{ - 16}\\0&1&0&5\\0&0&{\color{Red} 7}&{21}\end{array}} \right]\begin{array}{*{20}{c}}{\frac{1}{7}{R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{ - 1}&{\color{Red} - 3}&{ - 16}\\0&1&0&5\\0&0&1&3\end{array}} \right]\]

Next, we can use the three row operation to getting the -3 modifying into a low.

\[\require{color}\left[ {\begin{array}{rrr|r}1&{ - 1}&{\color{Red} - 3}&{ - 16}\\0&1&0&5\\0&0&1&3\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} + 3{R_3} \to {R_{\kern 1pt} }}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{\color{Red} - 1}&0&{ - 7}\\0&1&0&5\\0&0&1&3\end{array}} \right]\]

That final step is to then makes the -1 into a 0 using the third row operation again.

\[\require{color}\left[ {\begin{array}{rrr|r}1&{\color{Red} - 1}&0&{ - 7}\\0&1&0&5\\0&0&1&3\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} + {R_2} \to {R_{\kern 1pt} }}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&0&0&{ - 2}\\0&1&0&5\\0&0&1&3\end{array}} \right]\]

The search to this system is then,

\[x = - 2,\,\,\,y = 5,\,\,\,z = 3\]