Name:
Student number:
1.1 Exercises: 4, 8, 10, 12, 16, 20, 22, 24, 26
4
question. find the point of intersection of the lines x1 + 5x2 = 1 and 3x1 + 7x2 = 5
answer. solve quickly. multiply the first eqn by 3 and subtract (3x1 + 15x2 ) − (3x1 + 7x2 ) =
3 − 5 ⇒ 8x2 = −2 ⇒ x2 = − 14 . then x1 = 1 − 5x2 = 1 − 5(− 41 ) = 49 . so the point is ( 94 , − 41 )
8
question. think of each matrix as an augmented matrix. say the next two row ops to keep
solving
1 −4 5 0 7
0 1 −3 0 6
(a)
0 0
1 0 2
0 0
0 1 −5
answer. clear above the pivot in col 3. do R2 ← R2 + 3R3 . then clear the 5 in row 1 with
R1 ← R1 − 5R3
1 −6 4 0 −1
0 2 −7 0 4
(b)
0 0
1 2 −3
0 0
3 1 6
answer. kill the 3 under the pivot in col 3. do R4 ← R4 − 3R3 . then scale that new row so
the pivot is 1, R4 ← (− 51 )R4
10
1
0
question. row reduce to solve
0
0
2
1
0
0
0
0
1
0
0
0
0
0
3
4
0
1
1
answer. clear above pivots: R1 ← R1 − 2R2 , then R2 ← R2 − 4R4 , then R1 ← R1 + 5R4 .
left side becomes the identity and the rhs is all zeros, so x1 = x2 = x3 = x4 = 0
12
x1 + 3x2 + 4x3 = 4
question. solve the system 3x1 + 7x2 + 7x3 = 8
4x1 + 6x2 + 2x3 = 4
1 3 4
4
1 3
4 | 4
1 3 4 | 4
8 → 0 −2 −5 | −4 → 0 1 52 | 2 →
answer. quick rref path 3 7 7
4 6 2 4
0 −6 −14 | −12
0 0 1 | 0
1 3 0 | 4
1 0 0 | −2
0 1 0 | 2 → 0 1 0 | 2 so (x1 , x2 , x3 ) = (−2, 2, 0)
0 0 1 | 0
0 0 1 | 0
16
question. check the solution from 12 by plugging it back
−2 + 3(2) + 4(0) = 4 ok
answer. put (−2, 2, 0) into each equation 3(−2) + 7(2) + 7(0) = 8 ok all good
4(−2) + 6(2) + 2(0) = 4 ok
20
x1 + 2x4 = 3
2x2 + 2x3 = 0
question. solve the system
x3 + 3x4 = 1
2x1 + 3x2 + 2x3 + x4 = 5
answer. reduce and one row becomes zero, so one free variable. let x4 = t. then x3 = 1 − 3t,
x2 = −1 + 3t, x1 = 3 − 2t. solution set is {(3 − 2t, −1 + 3t, 1 − 3t, t) | t ∈ R}
22
question. do the planes x1 + 2x2 + x3 = 4, x2 + x3 = 1, x1 + 3x2 = 0 meet in at least one
common point
answer. yes they meet in one point. from x1 + 3x2 = 0 get x1 = −3x2 . plug into the first
to get −x2 + x3 = 4 ⇒ x3 = 4 + x2 . then use the middle one x2 + (4 + x2 ) = 1 ⇒ 2x2 =
2
−3 ⇒ x2 = − 23 . so x3 = 52 and x1 = 92 . intersection is ( 92 , − 23 , 52 )
24
1 h 3
question. for what h is
a consistent augmented matrix
2 4 6
answer. do R2 ← R2 − 2R1 ⇒ [ 0 4 − 2h | 0 ]. never get a contradiction since rhs is 0. so
consistent for all real h. if h ̸= 2 you get a unique solution, if h = 2 you get infinitely many
26
3 −4
question. for what h is
−6 8
h
consistent
9
answer. do R2 ← R2 + 2R1 ⇒ [ 0 0 | 9 + 2h ]. need 9 + 2h = 0 ⇒ h = − 92 . otherwise
inconsistent
3
0
