Linear equations 65
(2.8) Cofactor expansion for i=3 has the most zero entries, giving
det A=(−1)(−1)(3+1) det
2 1 0 1
−3214
−7234
3 1 0 1
5 2 1 4
−4 3 1 2
3 2 1 1
5−324
−4 3 4 1
which reduces to
2 1 0 1
−3214
3 4 1 2
3 2 0 1
5−314
−4 3 1 2
3 2 1 0
5−321
−4 3 4 1
You could also expand on column j=4 instead. Recall that det A=det A†, so
expansions on columns are the same as expansions on rows.
(2.10) Use co-factor expansion on the last row of U. The only non-zero entry is Un,n. As a
result, the determinant after one co-factor expansion is
det U=Un,n(−1)2ndet Un−1
66 Linear equations
det U=
i=1
Ui,i
(2.11) The matrix determinants are
det A=det
213
121
234
=5
(2.12) The solutions are
x=
det
1 1 -1
4 0 2
0 -2 3
=0
det
2 1 -1
-1 4 2
4 0 3
det
-1 0 2
4 -2 3
det
-1 0 2
4 -2 3
(2.13) We need to compute a bunch of 3 ×3 determinants:
det
3 2 1
1−1 2
1 0 −3
=20
(2.14) 1 To solve the system by Cramers rule we must compute 4 determinants first
323
112
Next
823
912
and
Finally
68 Linear equations
Then we find the values of x1,x2,and x3as follows
|A|
2 For Gauss elimination first we right the augmented matrix:
1129
eliminating the first row yeilds
3119
3
and the second
7
7
forward elimination yeildsx3=3, x2=2, and x1=1. The same solution found
using Cramers rule.
(2.15) The number of operations for each step are:
(a) The determinant of a 2 ×2 matrix is ad −bc. This calculation can be completed
with two multiplications and a subtraction. So this requires 3 operations. For
(2.16) The Gauss elimination steps gives the following set of matrices:
0−4 0 0 0
Although you might want to swap rows at this point, it is not allowed in Gauss elim-
ination unless you are pivoting. So we continue and get
0 0 0 4 −4
Applying the back substitution formulas gives
We could also have done this problem setting the diagonal entries unity. While
you get the same answer, this is not used in the numerical implementation of Gauss
elimination. If asked to do Gauss elimination, you should not do the following even
though it works:
0 0 0 4 −4
Applying the back substitution formulas gives
(2.17)
(2.18) First write the augmented matrix
3 1 −1 0
Then eliminate the first column
0 11 −5−21
Then the second column
0 0 50 100
Now back sub
(2.19) Starting with the augmented matrix
−2 3 −3 5 8
eliminate the first row
then the second
0 0 −14 −28 −14
72 Linear equations
now backwards elimination first for the solution in column 4
and now for column 5
(2.20) The augemented matrix for this problem is
4 2 −6001
eliminating the first row
0−14/5−18/5−4/501
Then the second
0 0 −10/9−10/9 14/9 1
Solving for all three by back substitution give the following
2−3−3/5
1−7/5−9/10
(2.21) The second and fourth rows are switched to give
02318
(2.22) Swap row 6 with row 3
(2.23) There is no pivoting at the first iteration (note that we are adding an extra step to make
the diagonal entries unity; this is not used in the numerical implementation of Gauss
elimination:
0 7/2 5/2 13/4−3/2
Make a partial pivot step and eliminate:
0 0 2 −1/2−8
74 Linear equations
Make another partial pivot step and eliminate:
0 0 0 9/736/7
Applying the back substitution formula gives
(2.24) Need to swap the first and second rows to get
111114
Elimination gives
The biggest entry is on the diagonal, so we can eliminate to get
0 0 1.0571 1.0857 1.2857 3.4571
Linear equations 75
The biggest entry is on the diagonal, so we can eliminate to get
0 0 0 0.7519 1.1930 2.6967
The biggest entry is on the diagonal, so we can eliminate to get
Back substitution gives
z
1
(2.25) For the Gauss elimination with pivoting, first write as an augmented matrix
−2 4 3 1 14
There is no pivoting on the first step, so we just eliminate to get
0 16/3 17/3−7/314
We now pivot by exchanging rows 2 and 4:
0 2 −3 7 8
and do the elimination:
0 0 −16 18 2
76 Linear equations
We need to pivot again:
0 0 87/24 −81/24 1/4
and do the elimination
0 0 0 270/384 270/384
We now use back substitution to get x4=1, x3=1, x2=2 and x1=−1. The
determinant comes from the product of the diagonals and the number of pivots, det A
=(−1)2(−180) =−180.
(2.26) Using the formula for an upper triangular matrix with pivoting gives
i
(2.27) The first step is a pivot to give
464
234
The elimination factors are m21 =1/4 and m31 =1/2. These give the upper triangular
0 0 2
Using the formula for the determinant with 1 pivot,
(2.28) B =n – p
C=k+1
Linear equations 77
(2.29) A matrix is banded with size pand q(and bandwidth p+q+1 if ai j =0 for j>i+p
and i>j+q. Analyze the pand qvalues for each diagonal and take the maximum
values
i p q
The maximum values are p=4 and q=3 so the bandwidth is 8.
(2.30) 12 (p=7, q=4)
(2.33)
111
(2.34)
(2.35) Using Ly =bgives
1L3,21
y3
0
From forward substitution, we get
Using Ux =ygives
003
x3
−5
78 Linear equations
From back substitution, we get
x2=−4x3=20/3
(2.36) The first round of elimination gives
2? 1
The second round of elimination gives
2−3 1
−5
0 0 25
with
211 1
First solve
211 1
y3
0
which gives
y3
50
then solve
0 0 25
x3
50
which gives
x3
2
(2.37) The Umatrix is
1 0.25 −3 1
The solution is
x4y4z4
5 0 −5
(2.38) We need to create Uand Lfrom Gauss elimination:
1 1 1
311
For the first forcing function, we get y=[0,0,−1] and x=[−1,0,1]. For the sec-
ond forcing function, we get y=[9,−3,−4] and x=[2,3,4]. For the third forcing
function, we get y=[0,−3,2] and x=[−1,3−2].
(2.39) The original matrix is
1 0 0
−2−1 1
132
002
which gives
132
−2−4−3
The forcing function is
132
−2−4−3
1
2
80 Linear equations
which gives
So the original problem was
−2−4−3
x3
−4
During LU decomposition, we would have
−2−1 1
y3
−4
The solution by forward substitution gives
y3
4
(2.40) First perform the LU decomposition with the U matrix on the left and the L matrix
on the right:
0 0 0 5
1201
Linear equations 81
For the first forcing function, the equation for yis
1201
y4
15
Using forward substitution, we get
y4
5
The corresponding equation for xis
0 0 0 5
x4
5
Using back substitution, we get
x4
1
For the second forcing function, the equation for yis
1201
y4
42
Using forward substitution, we get
y4
15
The corresponding equation for xis
0 0 0 5
x4
15
82 Linear equations
Using back substitution, we get
x4
3
For the third forcing function, the equation for yis
1201
y4
−2
Using forward substitution, we get
y4
5
The corresponding equation for xis
0 0 0 5
x4
5
Using back substitution, we get
x4
1
(2.41) Starting from the matrix of constants A
4 3 −1
126
The first round of elimination gives
4 3 −1
0 1.25 6.25
with
Linear equations 83
The second round of elimination yields
4 3 −1
008.5
with
1 0 0
0.25 −.5 1
Now we solve Lb=ythe forward elimination step yeilds
Then we back eliminate to find
0.482
0.471
(2.42) The first round of elimination gives
0−5−1
with
2 ? 1
The second round of elimination gives
241
004
with
For the first column of the inverse, we first solve
2−1 1
y3
0
84 Linear equations
which gives
y3
−1
We then solve
004
x3
−1
which gives
x3
−1/4
For the second column of the inverse, we first solve
2−1 1
y3
0
which gives
We then solve
004
x3
1
which gives
x3
1/4
For the third column of the inverse, we first solve
2−1 1
y3
1
which gives
y3
1