linear algebra

Linear Algebra - Exercises

9 Linear Systems. Method of Elimination
1. Determine whether
a) x = 1, y = −1
b) x = 2, y = −3
are solutions to the linear equation

2x + 3y = −1.

Solution. a)
2 · 1 + 3 · (−1) = −1,
−1 = −1,
so x = 1, y = −1 is a solution.
b)
2 · 2 + 3 · (−3) = −1,
−5 6= −1,
so x = 2, y = −3 is not a solution.
2.Determine if
a) x = 0, y = 2
b) x = −1, y = −3
are solutions to the system

−x + 3y = −8,
2x + y = −5.

Solution. a) Substituting x = 0, y = 2 into Equation 1, we obtain

−1 · 0 + 3 · 2 = −8,
6 6= −8,

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,LINEAR ALGEBRA - EXERCISES 48

so x = 0, y = 2 is not a solution.
b) Substituting x = −1, y = −3 into Equation 1, we obtain

−1 · (−1) + 3 · (−3) = −8,
−8 = −8,

and into Equation 2, we obtain

2 · (−1) + (−3) = −5,
−5 = −5.

Therefore, x = −1, y = −3 is a solution.
3. Solve the system
2x + y = 2,
4x − y = 1.
by the method of elimination.
Solution.
1
2
· Equation 1
1
x + 2
y= 1
4x − y = 1
Equation 2 − 4 · Equation 1

x + 12 y = 1
− 3y = −3

>From Equation 2
y=1
Substituting y = 1 into Equation 1
1
x+ =1
2
1
x=
2
The system has the unique solution
1
x=
2
y=1

, LINEAR ALGEBRA - EXERCISES 49

4. Solve the system
2x + y = 1,
x + 2y = −1,
x − y = 2.
by the method of elimination.
Solution.
Interchange Equation 1 and Equation 2
x + 2y = −1
2x + y = 1
x − y = 2
Equation 2 − 2 · Equation 1
Equation 3 − Equation 1
x + 2y = −1
− 3y = 3
− 3y = 3
− 13 · Equation 2
− 13 · Equation 3
x + 2y = −1
y = −1
y = −1
Equation 3 − Equation 2
x + 2y = −1
y = −1
0 = 0
We substitute y = −1 into Equation 1
x + 2 · (−1) = −1
x=1
The system has the unique solution
x=1
y = −1

5. Solve the system
x + y + z = 1,
x + 2y + 2z = 1,
x − y + z = 3.