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Ch 06 matrices v29 Page 1 of 54 www.utstat.utoronto.ca/sharp
Data Projector Frameworks: MATA33 W11 Sharp
(partly stolen from Haeussler et al 13e Ch 6)
Matrix ‘size’
• A matrix consisting of m horizontal rows and n vertical columns is called an m×n matrix or a matrix of size m×n.
• For the entry aij, we call i the row subscript
and j the column subscript.
mnmm
n
n
aaa
aaa
aaa
...
......
......
......
...
...
21
21221
11211
A
b
c
d
h
Ch 06 matrices v
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Ch 06 matrices v29 Page 3 of 54 www.utstat.utoronto.ca/sharp
Equality
Matrices A = [aij
] and B = [bij] are equal if
they have the same size and aij
= bij for each i
and j.
Ch 06 matrices v29 Page 4 of 54 www.utstat.utoronto.ca/sharp
Transpose of a Matrix
A transpose matrix is denoted by AT
Roughly, rotate it around its diagonal if it had one. Note that (AT)T = A
654
321A
63
52
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MSuco
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Ch 06 matrices v
Matrixum A orrespo
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Ch 06 matrices v29 Page 6 of 54 www.utstat.utoronto.ca/sharp
Properties of Scalar Multiplication:
k(A+B) =kA + kB
(k+l)A=kA+lA k(lA)=(kl)A
0A=0 k0=0 (A+B)T = AT +BT
(kA)T= k AT
Ch 06 matrices v29 Page 7 of 54 www.utstat.utoronto.ca/sharp
Matrix Subtraction:
(a)
(b)
13
08
84
3203
1144
2662
30
14
26
23
14
62
52
80
42
66
10
262BAT
Ch 06 matrices v29 Page 8 of 54 www.utstat.utoronto.ca/sharp
Matrix of 'size' 2 by 4 3 2 5 6
4 3 7 1
Matrix of 'size' 2 by 3 3 2 6
4 3 1
Maybe this 2 by 3 matrix indicates games of soccer and of chess played by
Rob, Sal and Ted last week:
Rob Sal Ted
Soccer 3 2 6
Chess 4 3 1
Call the matrix A. If they play the same number of games 4 weeks in a row,
then total for the short month is:
12 8 24
16 12 4
We have multiplied the matrix A by the scalar 4 in the obvous way. Note 4A=A4 (commutative
Let's do a matrix addition of a fifth week's data, B, also in the obvious way:
2 2 3 12 8 24
3 1 3 16 12 4
14 10 27
19 13 7
Working element by element, since x+y=y+x for scalars, then matrix addition is commutative:
Also, since( x +y)+z=x+ (y+z) for scalars, then matrix addition also is associative
)
( )
( )
( )
+
=A
(
)
( )
( )
B+A= A+B
(A+B)+C= A+(B+C)
4A=
B+4A=
= (
M
•
Siz
A
B
AB C D CD
ijc
Ch 06 matrices v
Matrix
• AB
give
zes an
= 3 ×
= 5 ×
B = 3
= 3 ×= 7 ×D = u
n
kj
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Multi
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3 ma
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plicat
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atrix
atrix
matrix
atrix
atrix
ned bu
kj ab
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ion
p mat
:
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BA = 5
C = 7 ×
i ba 2
C who
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ose en
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ij is
Ch 06 matrices v29 Page 10 of 54 www.utstat.utoronto.ca/sharp
Matrix by matrix multiplication examples
2 3 3 2 5 6 18 13 31 15
0 1 4 3 7 1 4 3 7 1
2*6+3*1=15
2 2 ‐2 3 3 2 5 2 13 0 36 11
0 1 ‐3 1 4 3 7 4 ‐1 2 18 ‐1
2 ‐1 ‐3 2
1 ‐4 2 1
5 0 0 0 ‐1 0.2 0.0 0.0 0.0
0 2 0 0 = 0.0 0.5 0.0 0.0
0 0 ‐1 0 0.0 0.0 ‐1.0 0.0
0 0 0 1 0.0 0.0 0.0 1.0)
( )( )
( ) (
( )* =
( )) )( (* =
Ch 06 matrices v29 Page 11 of 54 www.utstat.utoronto.ca/sharp
More matrix products
a)
b)
c)
d)
32
6
5
4
321
183
122
61
61
3
2
1
1047
0110
11316
212
315
201
401
122
031
2222122121221121
2212121121121111
2221
1211
2221
1211
babababa
babababa
bb
bb
aa
aa
Ch 06 matrices v29 Page 12 of 54 www.utstat.utoronto.ca/sharp
Example: Cost Vector
Given the price and the quantities, calculate the total cost.
Solution:
The cost vector is
432P
C of units
B of units
A of units
11
5
7
Q
73
11
5
7
432
PQ
Ch 06 matrices v29 Page 13 of 54 www.utstat.utoronto.ca/sharp
Example: Matrix Mult is Associative
compute ABC in two ways
Associative: A(BC)=(AB)C
11
20
01
211
103
43
21CBA
1318
94
43
12
43
21
11
20
01
211
103
43
21BCA
1318
94
11
20
01
5413
521
11
20
01
211
103
43
21CAB
Ch 06 matrices v29 Page 14 of 54 www.utstat.utoronto.ca/sharp
Example: Raw materials and cost:
Quantities of each of 3 house types:
Use by each type of 5 raw materials:
Price of products:
Materials cost for 1 copy:
Materials cost for the quantities produced:
1275Q
1500
150
800
1200
2500
C
71650
81550
75850
1500
150
800
1200
2500
1358256
21912187
17716205
RC
900,809,1
71650
81550
75850
1275
RCQQRC
1358256
21912187
17716205
R
Ch 06 matrices v29 Page 15 of 54 www.utstat.utoronto.ca/sharp
Examples using I and 0 matrices:
00
00
10
01
41
23
103
101
51
52
OIBA
31
22
41
23
10
01 a.
AI
63
63
20
02
41
233
10
012
41
23323 b. IA
OAO
00
00
41
23 c.
IAB
10
01
41
23 d.
103
101
51
52
Ch 06 matrices v29 Page 16 of 54 www.utstat.utoronto.ca/sharp
7
4
38
52
2
1
x
x
BAX
Matrix form of linear equations:
Can be written:
738
452
21
21
xx
xx
1. 2. 3.
a
Ch 06 matrices v2
InteMulAdd
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Ch 06 matrices v29 Page 18 of 54 www.utstat.utoronto.ca/sharp
Properties of a reduced matrix
• All zero-rows at the bottom. • For each nonzero-row, leading entry is 1 and
the rest in that column are zeros. • Leading entry in each row is to the right of the
leading entry in any row above it.
Ch 06 matrices v29 Page 19 of 54 www.utstat.utoronto.ca/sharp
Reduced matrices: Not red (3 sb 1) Reduced Not reduced
Reduced Not reduced Reduced
0000
2100
3010
f.
010
000
001
e. 000
000d.
01
10c.
010
001b.
30
01 a.
Ch 06 matrices v29 Page 20 of 54 www.utstat.utoronto.ca/sharp
Evil Nerd 1: he likes to make things complex
He knows that x=3 y=5 but he likes to baffle classmates so instead he says: x =3 2x +y =11 or even better 3x + y = 14 4x+2y=22
Classmates say ‘go away to Evil Nerd’ and disentangle it:
3 1 | 144 2 | 22 (start, now R2 /2 to give:)
Typeequationhere.
3 1 | 142 1 | 11 (now R1 – R2 to give:)
Ch 06 matrices v29 Page 21 of 54 www.utstat.utoronto.ca/sharp
1 0 | 32 1 | 11 (now R2 – 2R1 to give)
1 0 | 30 1 | 5 (hence x=3, y=5, ha ha nerd)
So
Exlinpreeqrowdia
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Ch 06 matrices v229
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n by 4))
Ch 06 matrices v29 Page 25 of 54 www.utstat.utoronto.ca/sharp
Evil Nerd 2 x=3 y=2 z=4 Above is what he knows, but he makes it complex: x+2y=7 (adding 2*(y=2)) y=2 z=4 x+2y+z=11 (adding (z=4)) x+3y+0z=9 x+3y+z=13 x+2y+z=1 2x+5y+z=20 x+3y+z=13
Ch 06 matrices v29 Page 26 of 54 www.utstat.utoronto.ca/sharp
Decode Evil Nerd
1 2 1 112 5 1 201 3 1 13
R1 swap R22 5 1 201 2 1 111 3 1 13
R1-R21 3 0 91 2 1 111 3 1 13
R2-R11 3 0 90 -1 1 21 3 1 13
R2*(-1)1 3 0 90 1 -1 -21 3 1 13
R3-R11 3 0 90 1 -1 -20 0 1 4
R2+R31 3 0 90 1 0 20 0 1 4
R1-3R21 0 0 30 1 0 20 0 1 4
x=3, y=1, z=4: Nerd foiled
Ch 06 matrices v29 Page 27 of 54 www.utstat.utoronto.ca/sharp
Section 6.4, Example 3 Matrix Reduction
Example 3
2 3 -12 1 51 1 1
R1-3*R3-1 0 -42 1 51 1 1
(-1)*R11 0 42 1 51 1 1
R2-2*R11 0 40 1 -31 1 1
R3-R1-R21 0 40 1 -30 0 0
x=4y=-3
1
52
132
yx
yx
yx
Ch 06 matrices v29 Page 28 of 54 www.utstat.utoronto.ca/sharp
Section 6.4, Example 5 Parametric Form of Solution
1 3/2 1 3 50 1 2 1 20 -9/2 -6 -3 -6
-(3/2)R2+R11 0 -2 3/2 20 1 2 1 20 -9/2 -6 -3 -6
(9/2)R2+R31 0 -2 3/2 20 1 2 1 20 0 3 3/2 3
R3/31 0 -2 3/2 20 1 2 1 20 0 1 1/2 1
-2R3+R21 0 -2 3/2 20 1 0 0 00 0 1 1/2 1
2R3+R11 0 0 5/2 40 1 0 0 00 0 1 1/2 1
9633
22
06232
431
432
4321
xxx
xxx
xxxx
Ch 06 matrices v29 Page 29 of 54 www.utstat.utoronto.ca/sharp
Section 6.4, Example 5 Parametric Form (Cont)
Where x4 (or r= x4 looks neater) is any real value x1 = 4 – (5/2) r x2 = 0 x3 = 1 – (1/2) r x4 = 0 + r and let r vary.
x1 4 -5/2
x2 = 0 +r 0
x3 1 -1/2
x4 0 1
This is a line in 4-space, going through (4,0,1,0)T
421
3
2
425
1
1
0
4
xx
x
xx
Ch 06 matrices v29 Page 30 of 54 www.utstat.utoronto.ca/sharp
Two Parameter Families of Solutions (Ch 6.5)
These are found when e.g. we use matrix reduction and end up with two free variables, and can express the other variables in terms of them. It makes more sense with an example:
Ch 06 matrices v29 Page 31 of 54 www.utstat.utoronto.ca/sharp
Probs06.05Q06: Solve the set: w + x+ y +2z =4 2w+ x +2y +2z =7 w+ 2x+ y+ 4z = 5 3w -2x + 3y -4z = 7 4w- 3x + 4y -6z = 9
1 1 1 2 4
2 1 2 2 7
1 2 1 4 5
3 ‐2 3 ‐4 7
4 ‐3 4 ‐6 9
1 1 1 2 4
0 ‐1 0 ‐2 ‐1
0 1 0 2 1
0 ‐5 0 ‐10 ‐5
0 ‐7 0 ‐14 ‐7
1 1 1 2 4
0 1 0 2 1
0 1 0 2 1
0 ‐5 0 ‐10 ‐5
0 ‐7 0 ‐14 ‐7
1 0 1 0 3
0 1 0 2 1
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
Last matrix above is the unique reduced form
Ch 06 matrices v29 Page 32 of 54 www.utstat.utoronto.ca/sharp
Probs06.05Q06 (cont)
The rows of zeros mean that y, z can be regarded as taking all real values-it is traditional to say y=r, z=s: w=-r+3 x=-2s +1 y=r z=s
w 3 -1 0
x = 1 +r 0 +s -2y 0 1 0z 0 0 1
Letting r and s vary, this is a plane in 4-space.
Ch 06 matrices v29 Page 33 of 54 www.utstat.utoronto.ca/sharp
Homogeneous Systems
Is called homogeneous if c1 =c2 =….=cm =0 (all the c=0). Reduce it, say that k non-zero rows survive.
If homogeneous then If k<n there’s infinite solutions If k=n there’s a unique solution Unique but trivial: xi =0 for all i
o Example of trivial: x+y=0 3x+y=0 has unique trivial solution x=0, y=0
o Example of infinite: x+y=0 3x+3y=0 has infinite solutions: line y=-x/3
mnmnmm
nn
cxaxaxa
cxaxaxa
...
.
.
.
.
...
2211
11212111
Ch 06 matrices v29 Page 34 of 54 www.utstat.utoronto.ca/sharp
Homogenous: Infinite Solutions or all-zero solution? Probs06.05Q20: Solve the homogeneous set: 2x+ y +z =0 x -y +2z =0 x+ y+ z = 0
Ch 06 matrices v29 Page 35 of 54 www.utstat.utoronto.ca/sharp
2 1 1 0
1 ‐1 2 0
1 1 1 0
Drop last column: all zeros
2 1 1
1 ‐1 2
1 1 1
1 0 0
1 ‐1 21 1 1
1 0 0
0 ‐1 2
0 1 1
1 0 0
0 1 ‐2
0 1 1
1 0 0
0 1 ‐2
0 0 3
1 0 0
0 1 ‐2
0 0 1
1 0 0
0 1 0
0 0 1 Last matrix above is the unique reduced form. k=n=3.It gives us x=0, y=0, z=0: the unique trivial solution.
Ch 06 matrices v29 Page 36 of 54 www.utstat.utoronto.ca/sharp
Homogenous: Infinite Solutions or all-zero solution? Probs06.05Q20 changed a bit: Solve the homogeneous set: 2x+ y +z =0 x -y +2z =0 x + z= 0 (was x+ y+ z = 0 previously) You can maybe see that adding the first and second equations gives the third (time 3). That is, the third equation gives no more info, so we have two equations, three unknowns. But let’s check by reducing it:
Ch 06 matrices v29 Page 37 of 54 www.utstat.utoronto.ca/sharp
2 1 1 0
1 ‐1 2 0
1 0 1 0
Drop last column: all zeros
2 1 1
1 ‐1 2
1 0 1
1 1 0
1 ‐1 21 0 1
1 1 0
0 ‐1 1
1 0 1
1 0 1
0 1 ‐1
1 0 1
1 0 1
0 1 ‐1
0 0 0
1 0 1
0 1 ‐1
0 0 0
Last matrix above is the unique reduced form. k=2, n=3. Setting r=z x=-r y=r z=r
Ch 06 matrices v29 Page 38 of 54 www.utstat.utoronto.ca/sharp
(6.6) Matrix Inverse to solve equation
Say we have a system of equations Ax = B
Here’ some logic:
Matrix inverse A-1 exists =>
A-1 exists such that A-1 A =I =>
x=A-1 B , a unique solution But we already know that in trying to solve for x, there might be 0, 1 or an infinite number of solutions. So the logic above says that a matrix inverse can only exist if there is just 1 solution. It turns out that this can only happen if A is a square matrix.
Ch 06 matrices v29 Page 39 of 54 www.utstat.utoronto.ca/sharp
Matrices A has an inverse?
So only square matrices can have inverses, and not all of them do. But if a square matrix A does have an inverse A-1 then:
A-1 also is square A-1 is unique A-1 is the same size as A A-1 commutes with A: A-1 A = I = AA-1
Ch 06 matrices v29 Page 40 of 54 www.utstat.utoronto.ca/sharp
Example 1: Inverse of Matrix
So C is the inverse of A, A is ‘invertible’.
Which was expected since a matrix commutes with its inverse. AC=CA=I
73
21A
13
27C
ICA
10
01
73
21
13
27
IAC
10
01
13
27
73
21
Ch 06 matrices v29 Page 41 of 54 www.utstat.utoronto.ca/sharp
Finding Inverse by Reduction
[A|I] => => => [R|B]
Where R is the reduced form of A
If R=I, then A is invertible and A-1 =B (we did it‼) If R≠I, then A is not invertible (no-one can do it!)
Example 3: Inversion by Reduction
Can we invert A? Can we reverse the changes that Evil Nerd made? Turns out we can:
So we have
22
01A
10
01
22
01IA
12
01
20
01
2/11
01
10
01
2/11
011A
Ch 06 matrices v29 Page 42 of 54 www.utstat.utoronto.ca/sharp
Example 5 – Use Inverse to Solve a System
1 0 ‐2 1 0 0
4 ‐2 1 0 1 0
1 2 ‐10 0 0 1
1 0 ‐2 1 0 0
0 ‐2 9 ‐4 1 0
0 2 ‐8 ‐1 0 1
1 0 ‐2 1 0 0
0 ‐2 9 ‐4 1 0
0 0 1 ‐5 1 1
1 0 0 ‐9 2 2
0 ‐2 9 ‐4 1 0
0 0 1 ‐5 1 1
1 0 0 ‐9 2 2
0 ‐2 0 41 ‐8 ‐9
0 0 1 ‐5 1 1
1 0 0 ‐9 2 2
0 1 0 ‐20.5 4 4.5
0 0 1 ‐5 1 1
1 3
1 2 3
1 2 3
2 1
4 2 2
2 10 1
x x
x x x
x x x
1021
124
201
A
Ch 06 matrices v29 Page 43 of 54 www.utstat.utoronto.ca/sharp
Example 5 –Use Inverse to Solve a System(cont)
X=A-1 B
115
4
229
29
2411A
4
17
7
1
2
1
115
4
229
29
241
3
2
1
x
x
x
Ch 06 matrices v29 Page 44 of 54 www.utstat.utoronto.ca/sharp
Two easy inverses:
10 1
10 1
c
b
a
D
00
00
00
c
b
a
D
/100
0/10
00/11
Ch 06 matrices v29 Page 45 of 54 www.utstat.utoronto.ca/sharp
Remember about matrices
A is symmetric if AT = A o needs A T ij = Aij i,j o hence needs Akl = Alk k, l.
Homogeneous systems Ax = 0,
n unknowns m equations k=number of nonzero rows in reduced coeff. mat If kn: (always true if m n)
o Infinity solutions, including trivial xi =0 i If k=n: one unique solution, the trivial xi =0 i
Matrix Inverse
A-1 is unique if it exists at all A-1 commutes with A: that is A-1 A= I= A A-1 If A-1 and B-1 exist, then (AB)-1 = B-1 A-1 If A-1 exists then (AT)-1 = (A-1) T
Ch 06 matrices v2
(d) S
Com-9 = 5k2 =
29
MA
Solutio
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ATA3
on
if ,,,,, k = k = ±
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Ch 06 matrices v29 Page 47 of 54 www.utstat.utoronto.ca/sharp
MATA33 Final Summer 2009 LAQ3a Let k be an arbitrary real constant and let x and y be variables. Find all solutions to the system [6 points] kx + 2y = x x - ky = y
Solution (k-1) x + 2y =0 x -(k+1)y=0 Fast: Can e.g. substitute second equation in first and get (k-1)(k+1)y + 2y=0=(k2+1)y, hence y=0, x=0 Slow: Reduce:
1 21 1
1 11 2
1 10 2 1 1
1 10 1
1 00 1
So only the trivial solution x=0, y=0 for all k Hence homogeneous, with trivial solution x=0, y=0
Ch 06 matrices v29 Page 48 of 54 www.utstat.utoronto.ca/sharp
MATA33 Final Summer 2009 LAQ3b
What equation must real constants a, b, and c satisfy in order that the system of linear equations x + 3y + z = a 2x - y - z = b 4x + 5y + z = c has infinitely many solutions ? [6 points] Solution Can only have infinitely many solutions if number of non-zero rows in reduced matrix, k <3=n, number of variables.
1 3 12 1 14 5 1
1 3 10 7 30 7 3
2 4
So if the second and third row are the same, we have two equations in three unknowns, infinite solutions, which needs: b-2a=c-4a or 2a +b-c=0
Ch 06 matrices v2
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Ch 06 matrices v2
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Ch 06 matrices v2
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False:
True: False,
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MATA33 Final Summer 2009 LAQ8
In all of this question let P be an arbitrary n × n matrix where n ≥ 2. (a) Assume P is invertible and prove mathematically that the matrix equation PX = Q has a solution for every n × 1 matrix Q. [3 points] Solution: P invertible => P-1 exists => P-1PX= P-1 Q => X= P-1 Q (b) Assume the matrix equation PX = B has a solution for each n×1 matrix B and prove mathematically that P is invertible. (Note: parts (a) and (b) are independent of each other (they are not related.)) [6 points] Let Mi be the n×1 matrix with all zeros except a 1 in row i. Let Si be the n×1 matrix which solves PX = Mi, so P Si= Mi
Let Q be the n×n matrix with columns the Si
Solution: Q= (S1, S2, …..... Sn) PQ = P(S1, S2, …..... Sn) = (M1, M2, ……..Mn) = I Hence Q is an inverse to P and P is invertible.