MATRICES: SYSTEMS OF EQUATIONS
5 minute review.
Remind students about the different types of systems of
equations: homogeneous versus non-homogeneous, and singular versus non-singular.
Recall how many solutions there are in each case: non-singular = exactly one (which
is 0 if homogeneous), singular + non-homogeneous = zero (inconsistent) or infinitely
many (use parameters to describe), singular + homogeneous = infinitely many (since
0 is a solution).
Class warm-up. Discuss what kinds of systems each of
the nature of their solution sets. Find the solutions.
x1 − x2 + x3
2x − 5y = 1
(b) 3x1 + x2 − 4x3
(a)
x + 3y = 6
5x1 − x2 − 2x3
these are, and therefore
= 0
= 0
= 0
Problems. Choose from the below.
1. Standard systems of equations. Solve the following systems of equations.
(a)
2x1 + 7x3 = 0
x1 − x2 + 3x3 = 0
4x1 + 5x2 + 2x3 = 0
2x1 + 3x3 = 1
(b) x1 − x2 + 5x3 = 1
2x2 − 4x3 = −2
(c)
2x1 + 7x3 = 29
x1 − x2 + 3x3 = 0
4x1 + 5x2 + 2x3 = −29
2x1 + 3x3 = 0
(d) x1 − x2 + 5x3 = 0
2x2 − 4x3 = 0
x1 + 2x2 + 6x3 + 5x4
x2 − 2x3 + 4x4
(e)
x3 + 3x4
3x1 + x4
=
=
=
=
1
0
0
2
8x1 − x2 + 3x3
5x1 + 2x2 + x3
(f)
x1 + x2 + 7x3
2x1 − 4x2 − 5x3
=
=
=
=
7
0
0
7
2. Singular systems. Find the value(s) of α for which the following systems
of equations have infinitely many solutions and then find those solutions.
−x1 + 4x2 + αx3 = −2
2x1 − x2 = −1
(a)
5x1 + x2 − x3 = −5
(c)
6x1 + 3x2 − 7x3 = 1
x1 + 4x2 + x3 = −1
(b)
9x1 + 15x2 − 4x3 = α
(α + 1)x1 + 6x2 + 5x3 = 6
(α − 2)x2 + x3 = −3
(α − 3)x3 = 0
3. Linear combinations revisited.
(a) Let A and B be any 2 × 2 matrices. Investigate whether you can find
scalars α, β, γ, δ such that
αA + βAT + γB + δB T = I2 .
1
2
MATRICES: SYSTEMS OF EQUATIONS
For the warm-up, (a) is non-singular, solution [3
[ 43 λ 74 λ λ]T .
1]T ; (b) is singular, solutions are
Selected answers and hints.
1. (a) [0
(b) [1
0]T
0
−1 T
3 ]
−5
3
(c) [−24
9
(d) [0
0]T
0
67
(e) [ 100
1
10
(f) [ 13
21
−34
21
11]T
3
100
−1 T
100 ]
1 T
7]
2. (a) α = −1, solutions are x1 = λ, x2 = 2λ + 1, x3 = 7λ + 6.
(b) α = −2, solutions are x1 = λ, x2 =
−13λ−6
, x3
31
=
21λ−7
31
(c) Possible values of α are −1, 2, 3. There are no solutions when α = 2.
• α = −1 solutions are [λ
1
0]T .
• α = 3 solutions are [λ 21 − 4λ − 24 + 4λ]T .
a b
e f
3. Let A =
and B =
. Writing out the four equations that
c d
g h
result gives


  
a a e e
α
1
 b c f g  β   0 


  
 c b g f  γ  =  0 ,
d d h h
δ
1
which is a singular system. This means there is either no solution or infinitely
many solutions depending on the matrices A and B.
For more details, start a thread on the discussion board.