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2000 Fall
[Note: if you see symbols like ¹ or Þ in various places, then your browser is not reading the style sheet for this Web page properly (or the Symbol font is not installed on your computer). The translation is:
“Þ” as “Þ” (implies),Find the products AB and BA, if possible, or explain why the product does not exist when
[ 2 6 1 3 ] [ 1 0 -1 2 6 ]
A = [ 0 4 2 5 ] , B = [ 4 2 2 4 4 ]
[ 1 2 2 8 ] [ 3 1 6 8 -2 ]
[ 5 3 1 3 1 ]
A is a (3×4) matrix and B is a (4×5) matrix.
A B is therefore a (3×5) matrix while B A does
not exist (because of the mismatch between the number of columns
of the left-hand matrix B and the number of rows of the
right-hand matrix A).
[ 1 0 -1 2 6 ]
[ 2 6 1 3 ] [ 4 2 2 4 4 ]
AB = [ 0 4 2 5 ] . [ 3 1 6 8 -2 ]
[ 1 2 2 8 ] [ 5 3 1 3 1 ]
As an example of the calculation of this product, its (2,4) element
is
(0×2 + 4×4 + 2×8 + 5×3) = 47
The product is
[ 44 22 19 45 37 ]
AB = [ 47 25 25 47 17 ]
[ 55 30 23 50 18 ]
==================
Determine, if possible, the inverse for each of the following matrices by using both the row reduction technique and the determinant / adjoint method.
Where appropriate also find the product of the inverse and the matrix B.
U = [ 3 -2 ] , B = [ 1 ]
[ -7 5 ] [ 3 ]
The row reduction of [U | I], using the
linsys.exe
program, is:
Program to handle the Arithmetic in the Row Reduction of a Linear System
VisualBASIC version (c)1999, Glyn George.
Linear System:
Row 1: 3/1 -2/1 1/1 0/1
Row 2: -7/1 5/1 0/1 1/1
-----------------------------------------------------------------------------
Selected row operation: Divide row 1 by 3 / 1 .
Linear System:
Row 1: 1/1 -2/3 1/3 0/1
Row 2: -7/1 5/1 0/1 1/1
-----------------------------------------------------------------------------
Selected row operation: Subtract -7 / 1 times row 1 from row 2 .
Linear System:
Row 1: 1/1 -2/3 1/3 0/1
Row 2: 0/1 1/3 7/3 1/1
-----------------------------------------------------------------------------
Selected row operation: Divide row 2 by 1 / 3 .
Linear System:
Row 1: 1/1 -2/3 1/3 0/1
Row 2: 0/1 1/1 7/1 3/1
-----------------------------------------------------------------------------
Selected row operation: Subtract -2 / 3 times row 2 from row 1 .
Linear System:
Row 1: 1/1 0/1 5/1 2/1
Row 2: 0/1 1/1 7/1 3/1
-----------------------------------------------------------------------------
Program execution terminated.
Therefore
-1 [ 5 2 ]
U = [ 7 3 ]
========
The adjoint of any 2×2 matrix is very easy to find:
adj [ a b ] = [ d -b ]
[ c d ] [-c a ]
The determinant is just
(ad - bc) and
the inverse is
(unless
Therefore
-1 1 [ 5 2 ] [ 5 2 ]
U = --------------- [ 7 3 ] = [ 7 3 ]
3*5 - (-7)*(-2) ========
-1 [ 5 2 ] [ 1 ] [ 11 ]
U B = [ 7 3 ] [ 3 ] = [ 16 ]
======
which is the solution to the 2×2 linear system U x = B.
[ -1 2 4 ] [ .7 ]
V = [ 2 -4 1 ] , B = [ .85 ]
[ 5 6 -3 ] [ -.53 ]
The row reduction of [V | I], using the
linsys.exe
program, is:
Program to handle the Arithmetic in the Row Reduction of a Linear System
VisualBASIC version (c)1999, Glyn George.
Linear System:
Row 1: -1/1 2/1 4/1 1/1 0/1 0/1
Row 2: 2/1 -4/1 1/1 0/1 1/1 0/1
Row 3: 5/1 6/1 -3/1 0/1 0/1 1/1
-----------------------------------------------------------------------------
Selected row operation: Divide row 1 by -1 / 1 .
Linear System:
Row 1: 1/1 -2/1 -4/1 -1/1 0/1 0/1
Row 2: 2/1 -4/1 1/1 0/1 1/1 0/1
Row 3: 5/1 6/1 -3/1 0/1 0/1 1/1
-----------------------------------------------------------------------------
Selected row operation: Subtract 2 / 1 times row 1 from row 2 .
Linear System:
Row 1: 1/1 -2/1 -4/1 -1/1 0/1 0/1
Row 2: 0/1 0/1 9/1 2/1 1/1 0/1
Row 3: 5/1 6/1 -3/1 0/1 0/1 1/1
-----------------------------------------------------------------------------
Selected row operation: Subtract 5 / 1 times row 1 from row 3 .
Linear System:
Row 1: 1/1 -2/1 -4/1 -1/1 0/1 0/1
Row 2: 0/1 0/1 9/1 2/1 1/1 0/1
Row 3: 0/1 16/1 17/1 5/1 0/1 1/1
-----------------------------------------------------------------------------
Selected row operation: Swap rows 3 and 2 .
Linear System:
Row 1: 1/1 -2/1 -4/1 -1/1 0/1 0/1
Row 2: 0/1 16/1 17/1 5/1 0/1 1/1
Row 3: 0/1 0/1 9/1 2/1 1/1 0/1
-----------------------------------------------------------------------------
Selected row operation: Divide row 2 by 16 / 1 .
Linear System:
Row 1: 1/1 -2/1 -4/1 -1/1 0/1 0/1
Row 2: 0/1 1/1 17/16 5/16 0/1 1/16
Row 3: 0/1 0/1 9/1 2/1 1/1 0/1
-----------------------------------------------------------------------------
Selected row operation: Subtract -2 / 1 times row 2 from row 1 .
Linear System:
Row 1: 1/1 0/1 -15/8 -3/8 0/1 1/8
Row 2: 0/1 1/1 17/16 5/16 0/1 1/16
Row 3: 0/1 0/1 9/1 2/1 1/1 0/1
-----------------------------------------------------------------------------
Selected row operation: Divide row 3 by 9 / 1 .
Linear System:
Row 1: 1/1 0/1 -15/8 -3/8 0/1 1/8
Row 2: 0/1 1/1 17/16 5/16 0/1 1/16
Row 3: 0/1 0/1 1/1 2/9 1/9 0/1
-----------------------------------------------------------------------------
Selected row operation: Subtract -15 / 8 times row 3 from row 1 .
Linear System:
Row 1: 1/1 0/1 0/1 1/24 5/24 1/8
Row 2: 0/1 1/1 17/16 5/16 0/1 1/16
Row 3: 0/1 0/1 1/1 2/9 1/9 0/1
-----------------------------------------------------------------------------
Selected row operation: Subtract 17 / 16 times row 3 from row 2 .
Linear System:
Row 1: 1/1 0/1 0/1 1/24 5/24 1/8
Row 2: 0/1 1/1 0/1 11/144 -17/144 1/16
Row 3: 0/1 0/1 1/1 2/9 1/9 0/1
-----------------------------------------------------------------------------
Program execution terminated.
Therefore
-1 1 [ 6 30 18 ]
V = --- [ 11 -17 9 ]
144 [ 32 16 0 ]
===================
The adjoint of V is the matrix of cofactors of VT:
T [ -1 2 5 ]
V = [ 2 -4 6 ]
[ 4 1 -3 ]
C = + | -4 6 | = (12 - 6) = 6
11 | 1 -3 |
C = - | 2 6 | = -(-6 - 24) = 30
12 | 4 -3 |
C = + | 2 -4 | = (2 + 16) = 18
13 | 4 1 |
C = - | 2 5 | = -(-6 - 5) = 11
21 | 1 -3 |
C = + | -1 5 | = (3 - 20) = -17
22 | 4 -3 |
C = - | -1 2 | = -(-1 - 8) = 9
23 | 4 1 |
C = + | 2 5 | = (12 + 20) = 32
31 | -4 6 |
C = - | -1 5 | = -(-6 - 10) = 16
32 | 2 6 |
C = + | -1 2 | = (4 - 4) = 0
33 | 2 -4 |
[ 6 30 18 ]
==> adj(V) = [ 11 -17 9 ]
[ 32 16 0 ]
Expanding along the last row,
det V = a31 C13
+ a32 C23
+ a33 C33
= (5×18) + (6×9)
+ (-3×0) = 144
Therefore
-1 1 [ 6 30 18 ]
V = --- [ 11 -17 9 ]
144 [ 32 16 0 ]
===================
-1 1 [ 6 30 18 ] [ 0.7 ] 1 [ 20.16 ]
V B = --- [ 11 -17 9 ] . [ 0.85 ] = --- [ -11.52 ]
144 [ 32 16 0 ] [ -0.53 ] 144 [ 36.00 ]
[ 0.14 ]
= [ -0.08 ]
[ 0.25 ]
=========
which is the solution to the 3×3 linear system V x = B.
[ 2 8 6 ] [ 1 ]
H = [ 3 4 0 ] , B = [ 1 ]
[ 0 16 18 ] [ 1 ]
The row reduction of [H | I], using the
linsys.exe
program, is:
Program to handle the Arithmetic in the Row Reduction of a Linear System VisualBASIC version (c)1999, Glyn George. Linear System: Row 1: 2/1 8/1 6/1 1/1 0/1 0/1 Row 2: 3/1 4/1 0/1 0/1 1/1 0/1 Row 3: 0/1 16/1 18/1 0/1 0/1 1/1 ----------------------------------------------------------------------------- Selected row operation: Divide row 1 by 2 / 1 . Linear System: Row 1: 1/1 4/1 3/1 1/2 0/1 0/1 Row 2: 3/1 4/1 0/1 0/1 1/1 0/1 Row 3: 0/1 16/1 18/1 0/1 0/1 1/1 ----------------------------------------------------------------------------- Selected row operation: Subtract 3 / 1 times row 1 from row 2 . Linear System: Row 1: 1/1 4/1 3/1 1/2 0/1 0/1 Row 2: 0/1 -8/1 -9/1 -3/2 1/1 0/1 Row 3: 0/1 16/1 18/1 0/1 0/1 1/1 ----------------------------------------------------------------------------- Selected row operation: Divide row 2 by -8 / 1 . Linear System: Row 1: 1/1 4/1 3/1 1/2 0/1 0/1 Row 2: 0/1 1/1 9/8 3/16 -1/8 0/1 Row 3: 0/1 16/1 18/1 0/1 0/1 1/1 ----------------------------------------------------------------------------- Selected row operation: Subtract 16 / 1 times row 2 from row 3 . Linear System: Row 1: 1/1 4/1 3/1 1/2 0/1 0/1 Row 2: 0/1 1/1 9/8 3/16 -1/8 0/1 Row 3: 0/1 0/1 0/1 -3/1 2/1 1/1 ----------------------------------------------------------------------------- Program execution terminated.
Note the zero entries across the last row of the echelon form
of matrix H.
Therefore H-1
does not exist.
The adjoint of H is the matrix of cofactors of HT:
T [ 2 3 0 ]
H = [ 8 4 16 ]
[ 6 0 18 ]
C = + | 4 16 | = (72 - 0) = 72
11 | 0 18 |
C = - | 8 16 | = -(144 - 96) = -48
12 | 6 18 |
C = + | 8 4 | = (0 - 24) = -24
13 | 6 0 |
C = - | 3 0 | = -(54 - 0) = -54
21 | 0 18 |
C = + | 2 0 | = (36 - 0) = 36
22 | 6 18 |
C = - | 2 3 | = -(0 - 18) = 18
23 | 6 0 |
C = + | 3 0 | = (48 - 0) = 48
31 | 4 16 |
C = - | 2 0 | = -(32 - 0) = -32
32 | 8 16 |
C = + | 2 3 | = (8 - 24) = -16
33 | 8 4 |
[ 72 -48 -24 ]
==> adj(H) = [ -54 36 18 ]
[ 48 -32 -16 ]
Expanding along the first column of H,
det H = a11 C11
+ a21 C12
+ a31 C13
= (2×72) + (3×-48)
+ (0×-24) = 0
Therefore the inverse does not exist and
the solution to the linear system
[By row reduction, one can show that the solution is the one
parameter family
x = (1/16) × (4 + 24t,
1 - 18t, 16t) ].
Another matrix inverse technique.
By using only the properties of matrix multiplication and the
inverse
Find A-1 when
A2 - 2A = 5I,
and
A = [ 1 2 ]
[ 3 1 ]
Pre-multiply the equation
A2 - 2A = 5I
by A-1, then
A - 2I =
5A-1
Þ
A-1 =
(A - 2I) / 5
Therefore
-1 1 [ (1-2) (2-0) ] 1 [ -1 2 ]
A = --- [ ] = --- [ ]
5 [ (3-0) (1-2) ] 5 [ 3 -1 ]
==============
[By the method of question 2(a) above, it is easy to verify that this matrix is indeed the inverse of A.]
Find A-1 when
A3 - 2A2
- A - 6I = 0,
and
[ 1 3 2 ]
A = [ -1 -2 1 ]
[ 0 -1 3 ]
Again, multiply every term in the equation by
A-1 on the left, to
obtain
A2 - 2A
- I
- 6 A-1
= 0
Þ
A-1 =
(A2 - 2A
- I) / 6
2 [ 1 3 2 ] [ 1 3 2 ] [ -2 -5 11 ]
A = A A = [ -1 -2 1 ] . [ -1 -2 1 ] = [ 1 0 -1 ]
[ 0 -1 3 ] [ 0 -1 3 ] [ 1 -1 8 ]
Therefore
-1 1 ( [ -2 -5 11 ] [ -2 -6 -4 ] [ -1 0 0 ] )
A = ---( [ 1 0 -1 ] + [ 2 4 -2 ] + [ 0 -1 0 ] )
6 ( [ 1 -1 8 ] [ 0 2 -6 ] [ 0 0 -1 ] )
1 [ -5 -11 7 ]
= --- [ 3 3 -3 ]
6 [ 1 1 1 ]
=================
One can check that the product A A-1 = I, so that the matrix above truly is the inverse of A.
The components in a particular electronic circuit can be described as being in one of three states: “low”, “medium” or “high”.
After each cycle of operation, the following changes occur:
20% of all components that were in a “low” state are now in
a “medium” state.
10% of all components that were in a “low” state are now in
a “high” state.
20% of all components that were in a “medium” state are now
in a “low” state.
10% of all components that were in a “medium” state are now
in a “high” state.
10% of all components that were in a “high” state are now in
a “low” state.
10% of all components that were in a “high” state are now in
a “medium” state.
All other components do not change state.
If, before the first cycle,
80% of all components were in a “low” state,
20% of all components were in a “medium” state and
no components were in a “high” state,
then find the proportions of components that are in each of the three
states
Incorporating the notes from the question, the transition matrix is
From:
low medium high
low [ .7 .2 .1 ]
To: medium [ .2 .7 .1 ] = A
high [ .1 .1 .8 ]
[ .7 .2 .1 ] [ .8 ] [ .6 ]
x = A x = [ .2 .7 .1 ] [ .2 ] = [ .3 ]
1 0 [ .1 .1 .8 ] [ 0 ] [ .1 ]
======
Therefore, after the first cycle,
[ .7 .2 .1 ] [ .6 ] [ .49 ]
x = A x = [ .2 .7 .1 ] [ .3 ] = [ .34 ]
2 1 [ .1 .1 .8 ] [ .1 ] [ .17 ]
=======
Therefore, after the second cycle,
The steady state vector x is such that
Þ
(A - I) x = 0.
[ -.3 .2 .1 ]
A - I = [ .2 -.3 .1 ]
[ .1 .1 -.2 ]
We have the additional constraint that the elements of x must add up to 1, represented by the extra row at the bottom of the linear system:
[ -.3 .2 .1 | 0 ]
[ .2 -.3 .1 | 0 ]
[ .1 .1 -.2 | 0 ]
[ 1 1 1 | 1 ]
Upon multiplying rows 1, 2 and 3 by 10, we obtain an equivalent
linear system with integer-only coefficients, suitable for the
linsys.exe program:
Program to handle the Arithmetic in the Row Reduction of a Linear System VisualBASIC version (c)1999, Glyn George. Linear System: Row 1: -3/1 2/1 1/1 0/1 Row 2: 2/1 -3/1 1/1 0/1 Row 3: 1/1 1/1 -2/1 0/1 Row 4: 1/1 1/1 1/1 1/1 ----------------------------------------------------------------------------- Selected row operation: Swap rows 1 and 3 . Linear System: Row 1: 1/1 1/1 -2/1 0/1 Row 2: 2/1 -3/1 1/1 0/1 Row 3: -3/1 2/1 1/1 0/1 Row 4: 1/1 1/1 1/1 1/1 ----------------------------------------------------------------------------- Selected row operation: Subtract 2 / 1 times row 1 from row 2 . Linear System: Row 1: 1/1 1/1 -2/1 0/1 Row 2: 0/1 -5/1 5/1 0/1 Row 3: -3/1 2/1 1/1 0/1 Row 4: 1/1 1/1 1/1 1/1 ----------------------------------------------------------------------------- Selected row operation: Subtract -3 / 1 times row 1 from row 3 . Linear System: Row 1: 1/1 1/1 -2/1 0/1 Row 2: 0/1 -5/1 5/1 0/1 Row 3: 0/1 5/1 -5/1 0/1 Row 4: 1/1 1/1 1/1 1/1 ----------------------------------------------------------------------------- Selected row operation: Subtract 1 / 1 times row 1 from row 4 . Linear System: Row 1: 1/1 1/1 -2/1 0/1 Row 2: 0/1 -5/1 5/1 0/1 Row 3: 0/1 5/1 -5/1 0/1 Row 4: 0/1 0/1 3/1 1/1 ----------------------------------------------------------------------------- Selected row operation: Subtract -1 / 1 times row 2 from row 3 . Linear System: Row 1: 1/1 1/1 -2/1 0/1 Row 2: 0/1 -5/1 5/1 0/1 Row 3: 0/1 0/1 0/1 0/1 Row 4: 0/1 0/1 3/1 1/1 ----------------------------------------------------------------------------- Selected row operation: Divide row 2 by -5 / 1 . Linear System: Row 1: 1/1 1/1 -2/1 0/1 Row 2: 0/1 1/1 -1/1 0/1 Row 3: 0/1 0/1 0/1 0/1 Row 4: 0/1 0/1 3/1 1/1 ----------------------------------------------------------------------------- Selected row operation: Swap rows 3 and 4 . Linear System: Row 1: 1/1 1/1 -2/1 0/1 Row 2: 0/1 1/1 -1/1 0/1 Row 3: 0/1 0/1 3/1 1/1 Row 4: 0/1 0/1 0/1 0/1 ----------------------------------------------------------------------------- Selected row operation: Divide row 3 by 3 / 1 . Linear System: Row 1: 1/1 1/1 -2/1 0/1 Row 2: 0/1 1/1 -1/1 0/1 Row 3: 0/1 0/1 1/1 1/3 Row 4: 0/1 0/1 0/1 0/1 ----------------------------------------------------------------------------- Selected row operation: Subtract 1 / 1 times row 2 from row 1 . Linear System: Row 1: 1/1 0/1 -1/1 0/1 Row 2: 0/1 1/1 -1/1 0/1 Row 3: 0/1 0/1 1/1 1/3 Row 4: 0/1 0/1 0/1 0/1 ----------------------------------------------------------------------------- Selected row operation: Subtract -1 / 1 times row 3 from row 1 . Linear System: Row 1: 1/1 0/1 0/1 1/3 Row 2: 0/1 1/1 -1/1 0/1 Row 3: 0/1 0/1 1/1 1/3 Row 4: 0/1 0/1 0/1 0/1 ----------------------------------------------------------------------------- Selected row operation: Subtract -1 / 1 times row 3 from row 2 . Linear System: Row 1: 1/1 0/1 0/1 1/3 Row 2: 0/1 1/1 0/1 1/3 Row 3: 0/1 0/1 1/1 1/3 Row 4: 0/1 0/1 0/1 0/1 ----------------------------------------------------------------------------- Program execution terminated.
The steady state vector is therefore
x = (1/3) × (1, 1, 1)
and exactly one third of all components are in each of the low, medium and high states after infinitely many cycles.
One can easily verify that A x = x.
See also the associated Excel spreadsheet.
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