\begin{array}{rrrrrrr}
x & + & y & + & z & = & 4 \\
x & - & y & - & 2z & = & 1 \\
2x & + & y & - & z & = & 2
\end{array}
\begin{array}{rrrrrrr}
x & + & y & + & z & = & 4 \\
x & - & y & - & 2z & = & 1 \\
2x & + & y & - & z & = & 2
\end{array}
Solving a system of linear equation
by elimination method
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
1 & -1 & -2 & 1 \\
2 & 1 & -1 & 2
\end{array}
\right)
\begin{array}{rrrrrrr}
x & + & y & + & z & = & 4 \\
x & - & y & - & 2z & = & 1 \\
2x & + & y & - & z & = & 2
\end{array}
1
1
1
1
1
-1
-1
-2
2
1
2
4
1
2
4
1
1
1
1
1
1
1
2
2
-
-
-
Augmented matrix
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
1 & -1 & -2 & 1 \\
2 & 1 & -1 & 2
\end{array}
\right)
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
1 & -1 & -2 & 1 \\
2 & 1 & -1 & 2
\end{array}
\right)
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
1 & -1 & -2 & 1 \\
2 & 1 & -1 & 2
\end{array}
\right)
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
0 & 2 & 3 & 3 \\
2 & 1 & -1 & 2
\end{array}
\right)
R1 - R2
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
0 & 2 & 3 & 3 \\
2 & 1 & -1 & 2
\end{array}
\right)
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
1 & -1 & -2 & 1 \\
2 & 1 & -1 & 2
\end{array}
\right)
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
0 & 2 & 3 & 3 \\
2 & 1 & -1 & 2
\end{array}
\right)
R1 - R2
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
0 & 2 & 3 & 3 \\
2 & 1 & -1 & 2
\end{array}
\right)
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
1 & -1 & -2 & 1 \\
2 & 1 & -1 & 2
\end{array}
\right)
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
0 & 2 & 3 & 3 \\
2 & 1 & -1 & 2
\end{array}
\right)
R1 - R2
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
0 & 2 & 3 & 3 \\
0 & 1 & 3 & 6
\end{array}
\right)
2R1-R3
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
0 & 2 & 3 & 3 \\
0 & 1 & 3 & 6
\end{array}
\right)
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
1 & -1 & -2 & 1 \\
2 & 1 & -1 & 2
\end{array}
\right)
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
0 & 2 & 3 & 3 \\
2 & 1 & -1 & 2
\end{array}
\right)
R1 - R2
\left(
\begin{array}{ccc|c}
1 & 1 & 1 & 4 \\
0 & 2 & 3 & 3 \\
0 & 1 & 3 & 6
\end{array}
\right)
2R1-R3
\left(
\begin{array}{ccc|c}
1 & 1 & 1 & 4 \\
0 & 2 & 3 & 3 \\
0 & 1 & 3 & 6
\end{array}
\right)
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
1 & -1 & -2 & 1 \\
2 & 1 & -1 & 2
\end{array}
\right)
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
0 & 2 & 3 & 3 \\
2 & 1 & -1 & 2
\end{array}
\right)
R1 - R2
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
0 & 2 & 3 & 3 \\
0 & 1 & 3 & 6
\end{array}
\right)
2R1-R3
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
0 & 2 & 3 & 3 \\
0 & 0 & -3 & -9
\end{array}
\right)
R2-2R3
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
0 & 2 & 3 & 3 \\
0 & 0 & -3 & -9
\end{array}
\right)
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
1 & -1 & -2 & 1 \\
2 & 1 & -1 & 2
\end{array}
\right)
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
0 & 2 & 3 & 3 \\
2 & 1 & -1 & 2
\end{array}
\right)
R1 - R2
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
0 & 2 & 3 & 3 \\
0 & 1 & 3 & 6
\end{array}
\right)
2R1-R3
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
0 & 2 & 3 & 3 \\
0 & 0 & -3 & -9
\end{array}
\right)
R2-2R3
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
0 & 2 & 3 & 3 \\
0 & 0 & -3 & -9
\end{array}
\right)
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
0 & 2 & 3 & 3 \\
0 & 0 & -3 & -9
\end{array}
\right)
Upper triangular matrix
\left(
\begin{array}{rrr|r}
1 & 1 & 1 & 4 \\
0 & 2 & 3 & 3 \\
0 & 0 & -3 & -9
\end{array}
\right)
\begin{array}{ccccccc}
x & + & y & + & z & = & 4 \\
& & 2y & + & 3z & = & 3 \\
& & & - & 3z & = & -9
\end{array}
-
Start from the last equation to find zzz.
-
Substitute zzz into the second equation to find yyy.
-
Substitute yyy and zzz into the first equation to find xxx.
Now we just use back-substitution:
\begin{array}{ccccccc}
x & + & y & + & z & = & 4 \\
& & 2y & + & 3z & = & 3 \\
& & & - & 3z & = & -9
\end{array}
x = 4,\; y = -3,\; z = 3
Then
\begin{array}{rrrrrrr}
x & + & y & + & z & = & 4 \\
x & - & y & - & 2z & = & 1 \\
2x & + & y & - & z & = & 2
\end{array}
which is also solution for
x = 4,\; y = -3,\; z = 3
\begin{array}{rrrrrrr}
x & + & y & + & z & = & 4 \\
x & - & y & - & 2z & = & 1 \\
2x & + & y & - & z & = & 2
\end{array}
The solution of the system
is
\begin{array}{rrrrrrr}
(4) & + & (-3) & + & (3) & = & 4 \\
(4) & - & (-3) & - & 2(3) & = & 1 \\
2(4) & + & (-3) & - & (3) & = & 2
\end{array}
Checking the solution
Made by
Juan Carlos Ponce Campuzano
School of Environment and Science