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Math 105
Semester Review - Solutions
5. Solve the following systems of linear equations, if possible.
a) 2x
− y + 2z
= 1
b)
c)
x +
5x
−
y −
y +
3z
=
z =
4
6
x + 4y
− 11z
= 11
⎡2 −1 2 1⎤ ⎡2 −1 2 1 ⎤
⎢ R2 →2R2 −R1
1 1 −3 4
⎥ ⎢
0 3 −8 7
⎥
⎢ ⎥R3 →2R3−5R
⎢
⎥
1
⎢5 −1 1 6⎥ ⎢0 3 −8 7 ⎥
⎢ ⎥R4 →2R4 −R1
⎢
1 4 −11 11
⎥⎦
⎣ ⎦ ⎣0 9 −24 21
−1 1
⎡2 −1 2 1⎤
⎡1 2
1
2⎤
R3 R3 R
⎢
2
0 3 8 7
⎥ 1
→ − − R
8 7
1
→
⎢
⎥
2
R
⎢
−
3 0 1
⎥
⎢
3 3⎥
R4 → R4 −3R
⎢
2
0 0 0 0⎥
1
R2 3
R ⎢
2
0 0 0 0⎥
⎢
⎥
→
⎢ ⎥
⎣0 0 0 0⎦
⎣0 0 0 0⎦
5 1 7 8
+ t, + t,
t
4x
−
2x
− 5y
+
2x
+ 4y
+
2x
−
Solution: ( )
y + 2z
=
z
=
3
9
z = − 6
3 3 3 3
⎡4 −1 2 3 ⎤ ⎡4 −1 2 3 ⎤
⎢ R2 →2R2 −R1
2 −5 1 9
⎥ ⎢
0 −9 0 15
⎥
⎢ ⎥R3 →2R3−R
⎢
⎥
1
⎢2 4 1 −6⎥
⎣ ⎦ ⎢⎣0 9 0 −15⎥⎦
−1 1 3
⎡4 −1 2 3⎤
⎡1
4 2 4
⎤
1
R3 → R3+ R
⎢
2
0 −9 0 15
⎥ R1 →
4
R1
⎢ −5
⎢
⎥
0 1 0
⎥
−1
3
R2 9
R ⎢
⎥
2
⎢⎣
0 0 0 0⎥
→
⎦ ⎢⎣
0 0 0 0⎥⎦
1 1 5
− t, − , t
Solution: ( )
y +
6x
+ 3y
− 2z
+ 3w
+
z
3 2 3
−
5t
= 12
t =
2x
+ 5y
− 4z
+ 3w
+ 11t
=
1
8
⎡2 −1 1 0 −5 12⎤ ⎡2 −1 1 0 −5 12 ⎤
⎢ 2 2
3
1
6 3 2 3 1 1
⎥R →R − R ⎢
⎢
−
⎥
0 6 5 3 16 35
R3 R3 R ⎢
− −
⎥
→ −
⎥⎥
1
⎢2 5 −4 3 11 8⎥
⎣ ⎦ ⎢⎣0 6 −5 3 16 − 4 ⎦
c
c
c
3
c
c
c
3
= t
= +
7 8
2 3 3
= +
5 1
1 3 3
= t
=
−5
2 3
= −
1 1
1 3 2
t
t
t
⎡2 −1 1 0 −5 12 ⎤
R3 R3 R
⎢
2
0 6 5 3 16 35
⎥
→ −
⎢
− −
⎥
⎢⎣0 0 0 0 0 31 ⎥⎦
No solutions
1
R
1 1
5
1
→
2
R
−
−
1 ⎡1 2 2
0
2
6 ⎤
1 −5 1 8 −35
R2 →
6
R
⎢
2
0 1
⎥
⎢
6 2 3 6 ⎥
1
R 0 0 0 0 0 1
3 31
R ⎢
⎥
→
3⎣ ⎦
Winter 2006 Martin Huard 4