Simultaneous Linear Equations/Examples/Arbitrary System 1
Jump to navigation
Jump to search
Example of Simultaneous Linear Equations
The system of simultaneous linear equations:
| \(\text {(1)}: \quad\) | \(\ds x_1 - 2 x_2 + x_3\) | \(=\) | \(\ds 1\) | |||||||||||
| \(\text {(2)}: \quad\) | \(\ds 2 x_1 - x_2 + x_3\) | \(=\) | \(\ds 2\) | |||||||||||
| \(\text {(3)}: \quad\) | \(\ds 4 x_1 + x_2 - x_3\) | \(=\) | \(\ds 1\) |
has as its solution set:
| \(\ds x_1\) | \(=\) | \(\ds -\dfrac 1 2\) | ||||||||||||
| \(\ds x_2\) | \(=\) | \(\ds \dfrac 1 2\) | ||||||||||||
| \(\ds x_3\) | \(=\) | \(\ds \dfrac 3 2\) |
Proof
Subtract $2 \times$ equation $(1)$ from equation $(2)$.
Subtract $4 \times$ equation $(1)$ from equation $(3)$.
This gives us:
| \(\text {(1)}: \quad\) | \(\ds x_1 - 2 x_2 + x_3\) | \(=\) | \(\ds 1\) | |||||||||||
| \(\text {(2')}: \quad\) | \(\ds 3 x_2 - x_3\) | \(=\) | \(\ds 0\) | |||||||||||
| \(\text {(3')}: \quad\) | \(\ds 9 x_2 - 5 x_3\) | \(=\) | \(\ds -3\) |
Divide equation $(2')$ by $3$ to get $(2)$.
Add $2 \times$ equation $(2)$ to equation $(1)$.
Subtract $9 \times$ equation $(2)$ from equation $(3')$.
This gives us:
| \(\text {(1')}: \quad\) | \(\ds x_1 + \dfrac {x_3} 3\) | \(=\) | \(\ds 1\) | |||||||||||
| \(\text {(2)}: \quad\) | \(\ds x_2 - \dfrac {x_3} 3\) | \(=\) | \(\ds 0\) | |||||||||||
| \(\text {(3)}: \quad\) | \(\ds -2 x_3\) | \(=\) | \(\ds -3\) |
From equation $(3)$ we have directly that $x_3 = \dfrac 3 2$.
Substituting for $x_3$ in equation $(1')$ and equation $(2)$ gives the single solution:
| \(\ds x_1\) | \(=\) | \(\ds -\dfrac 1 2\) | ||||||||||||
| \(\ds x_2\) | \(=\) | \(\ds \dfrac 1 2\) | ||||||||||||
| \(\ds x_3\) | \(=\) | \(\ds \dfrac 3 2\) |
$\blacksquare$
Sources
- 1982: A.O. Morris: Linear Algebra: An Introduction (2nd ed.) ... (next): Chapter $1$: Linear Equations and Matrices: $1.1$ Introduction: Example $\text {(i)}$