To solve the set of linear equations below, let’s try using this method:
\[\Large
x_{1} + 2x_{2} – 3x_{3} = 2
\]
\[\Large
2x_{1} – x_{2} – x_{3} = 11
\]
\[\Large
3x_{1} + 2x_{2} + x_{3} = -5
\]
Step 1: Write in Matrix Form
\[\Large
\begin{bmatrix}
1 & 2 & -3 \\
2 & -1 & -1 \\
3 & 2 & 1 \\
\end{bmatrix} ·
\begin{bmatrix}
x_{1} \\
x_{2} \\
x_{3}
\end{bmatrix} =
\begin{bmatrix}
3 \\
11 \\
-5
\end{bmatrix}
\]
Step 2: Write in Augmented Matrix Form
\[\Large
\begin{bmatrix}
1 & 2 & -3 & \big| & 3 \\
2 & -1 & -1 & \big| & 11 \\
3 & 2 & 1 & \big| & -5
\end{bmatrix}
\]
Step 3: Row Operations
Now subtract 2 /1 times the first row from the second row and 3 / 1 times the first row from the third row. This gives:
\[\Large
\begin{bmatrix}
1 & 2 & -3 & \big| & 3 \\
0 & -5 & 5 & \big| & 5 \\
0 & -4 & 10 & \big| & -14
\end{bmatrix}
\]
Now subtract -4 / -5, i.e. 4 / 5 times the second row from the third row. The matrix then becomes:
\[\Large
\begin{bmatrix}
1 & 2 & -3 & \big| & 3 \\
0 & -5 & 5 & \big| & 5 \\
0 & 0 & 6 & \big| & -18
\end{bmatrix}
\]
Note that as a result of these steps, the matrix of coefficients of x has been reduced to a triangular matrix.
Finally, we detach the right-hand column back to its original position:
\[\Large
\begin{bmatrix}
1 & 2 & -3 \\
0 & -5 & 5 \\
0 & 0 & 6 \\
\end{bmatrix} ·
\begin{bmatrix}
x_{1} \\
x_{2} \\
x_{3}
\end{bmatrix} =
\begin{bmatrix}
3 \\
5 \\
-18
\end{bmatrix}
\]
Step 4: Back Substitution
\[\Large
6x_{3} = -18 ~~~~ \therefore \ x_{3} = -3
\]
\[\Large -5x_{2} + 5x_{3} = 5 ~~~~\therefore \ -5x_{2} = 5 + 15 = 20 ~~~~ \therefore \ x_{2} = -4 \]
\[\Large x_{1} + 2x_{2} -3x_{3} = 3 ~~~~ \therefore \ x_{1} – 8 + 9 = 3 ~~ \therefore \ ~~ x_{1} = -4 \]
\[\Large \therefore \ x_{1} = 2 ;~~ x_{2} = -4 ;~~ x_{3} = -3 \]
\[\Large -5x_{2} + 5x_{3} = 5 ~~~~\therefore \ -5x_{2} = 5 + 15 = 20 ~~~~ \therefore \ x_{2} = -4 \]
\[\Large x_{1} + 2x_{2} -3x_{3} = 3 ~~~~ \therefore \ x_{1} – 8 + 9 = 3 ~~ \therefore \ ~~ x_{1} = -4 \]
\[\Large \therefore \ x_{1} = 2 ;~~ x_{2} = -4 ;~~ x_{3} = -3 \]
Note that when dealing with augmented matrix, we may, if we wish:
- interchange two rows
- multiply any row by a non-zero factor
- add (or subtract) a constant multiple of any one row to (or from) another.
These operations are permissible since we are really dealing with the coefficients of both sides of the equations