the theory of matrices and determinants
TRANSCRIPT
Ch25-H8152.tex 23/6/2006 15: 9 Page 267
Matrices and Determinants
F
25
The theory of matrices and determinants
25.1 Matrix notation
Matrices and determinants are mainly used for thesolution of linear simultaneous equations. The the-ory of matrices and determinants is dealt with in thischapter and this theory is then used in Chapter 26 tosolve simultaneous equations.
The coefficients of the variables for linear simul-taneous equations may be shown in matrix form. Thecoefficients of x and y in the simultaneous equations
x + 2y = 3
4x − 5y = 6
become
(1 24 −5
)
in matrix notation.
Similarly, the coefficients of p, q and r in theequations
1.3p − 2.0q + r = 7
3.7p + 4.8q − 7r = 3
4.1p + 3.8q + 12r = −6
become
(1.3 −2.0 13.7 4.8 −74.1 3.8 12
)
in matrix form.
The numbers within a matrix are called an array andthe coefficients forming the array are called the ele-ments of the matrix. The number of rows in a matrixis usually specified by m and the number of columnsby n and a matrix referred to as an ‘m by n’ matrix.
Thus,
(2 3 64 5 7
)
is a ‘2 by 3’ matrix. Matrices can-
not be expressed as a single numerical value, but theycan often be simplified or combined, and unknownelement values can be determined by comparisonmethods. Just as there are rules for addition, sub-traction, multiplication and division of numbers inarithmetic, rules for these operations can be appliedto matrices and the rules of matrices are such thatthey obey most of those governing the algebra ofnumbers.
25.2 Addition, subtraction andmultiplication of matrices
(i) Addition of matrices
Corresponding elements in two matrices may beadded to form a single matrix.
Problem 1. Add the matrices
(a)
(2 −1
−7 4
)
and
(−3 07 −4
)
and
(b)
(3 1 −44 3 11 4 −3
)
and
(2 7 −5
−2 1 06 3 4
)
(a) Adding the corresponding elements gives:(
2 −1−7 4
)
+(−3 0
7 −4
)
=(
2 + (−3) −1 + 0−7 + 7 4 + (−4)
)
=(−1 −1
0 0
)
(b) Adding the corresponding elements gives:(
3 1 −44 3 11 4 −3
)
+(
2 7 −5−2 1 0
6 3 4
)
=(
3 + 2 1 + 7 −4 + (−5)4 + (−2) 3 + 1 1 + 01 + 6 4 + 3 −3 + 4
)
=(
5 8 −92 4 17 7 1
)
(ii) Subtraction of matrices
If A is a matrix and B is another matrix, then (A−B)is a single matrix formed by subtracting the elementsof B from the corresponding elements of A.
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268 MATRICES AND DETERMINANTS
Problem 2. Subtract
(a)
(−3 07 −4
)
from
(2 −1
−7 4
)
and
(b)
(2 7 −5
−2 1 06 3 4
)
from
(3 1 −44 3 11 4 −3
)
To find matrix A minus matrix B, the elements ofB are taken from the corresponding elements of A.Thus:
(a)
(2 −1
−7 4
)
−(−3 0
7 −4
)
=(
2 − (−3) −1 − 0−7 − 7 4 − (−4)
)
=(
5 −1−14 8
)
(b)
(3 1 −44 3 11 4 −3
)
−(
2 7 −5−2 1 0
6 3 4
)
=(
3 − 2 1 − 7 −4 − (−5)4 − (−2) 3 − 1 1 − 01 − 6 4 − 3 −3 − 4
)
=(
1 −6 16 2 1
−5 1 −7
)
Problem 3. If
A =(−3 0
7 −4
)
, B =(
2 −1−7 4
)
and
C =(
1 0−2 −4
)
find A + B − C.
A + B =(−1 −1
0 0
)
(from Problem 1)
Hence, A + B − C =(−1 −1
0 0
)
−(
1 0−2 −4
)
=(−1 − 1 −1 − 0
0 − (−2) 0 − (−4)
)
=(−2 −1
2 4
)
Alternatively A + B − C
=(−3 0
7 −4
)
+(
2 −1−7 4
)
−(
1 0−2 −4
)
=(−3 + 2 − 1 0 + ( − 1) − 0
7 + (−7) − (−2) −4 + 4 − (−4)
)
=(−2 −1
2 4
)
as obtained previously
(iii) Multiplication
When a matrix is multiplied by a number, calledscalar multiplication, a single matrix results inwhich each element of the original matrix has beenmultiplied by the number.
Problem 4. If A =(−3 0
7 −4
)
,
B =(
2 −1−7 4
)
and C =(
1 0
−2 −4
)
find
2A − 3B + 4C.
For scalar multiplication, each element is multipliedby the scalar quantity, hence
2A = 2
(−3 07 −4
)
=(−6 0
14 −8
)
3B = 3
(2 −1
−7 4
)
=(
6 −3−21 12
)
and 4C = 4
(1 0
−2 −4
)
=(
4 0−8 −16
)
Hence 2A − 3B + 4C
=(−6 0
14 −8
)
−(
6 −3−21 12
)
+(
4 0−8 −16
)
=(−6 − 6 + 4 0 − (−3) + 0
14 − (−21) + (−8) −8 − 12 + (−16)
)
=(−8 3
27 −36
)
When a matrix A is multiplied by another matrix B, asingle matrix results in which elements are obtainedfrom the sum of the products of the correspondingrows of A and the corresponding columns of B.
Two matrices A and B may be multiplied together,provided the number of elements in the rows ofmatrix A are equal to the number of elements in thecolumns of matrix B. In general terms, when multi-plying a matrix of dimensions (m by n) by a matrix ofdimensions (n by r), the resulting matrix has dimen-sions (m by r). Thus a 2 by 3 matrix multiplied by a3 by 1 matrix gives a matrix of dimensions 2 by 1.
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THE THEORY OF MATRICES AND DETERMINANTS 269
F
Problem 5. If A =(
2 31 −4
)
and B =(−5 7
−3 4
)
find A × B.
Let A × B = C where C =(
C11 C12C21 C22
)
C11 is the sum of the products of the first row ele-ments of A and the first column elements of B takenone at a time,
i.e. C11 = (2 × (−5)) + (3 × (−3)) = −19
C12 is the sum of the products of the first row ele-ments of A and the second column elements of B,taken one at a time,
i.e. C12 = (2 × 7) + (3 × 4) = 26
C21 is the sum of the products of the second rowelements of A and the first column elements of B,taken one at a time,
i.e. C21 = (1 × ( − 5)) + (−4 × (−3)) = 7
Finally, C22 is the sum of the products of the secondrow elements of A and the second column elementsof B, taken one at a time,
i.e. C22 = (1 × 7) + ((−4) × 4) = −9
Thus, A × B =(−19 26
7 −9
)
Problem 6. Simplify(
3 4 0−2 6 −3
7 −4 1
)
×(
25
−1
)
The sum of the products of the elements of eachrow of the first matrix and the elements of the secondmatrix, (called a column matrix), are taken one at atime. Thus:(
3 4 0−2 6 −3
7 −4 1
)
×(
25
−1
)
=(
(3 × 2) + (4 × 5) + (0 × (−1))(−2 × 2) + (6 × 5) + (−3 × (−1))(7 × 2) + (−4 × 5) + (1 × (−1))
)
=(
2629−7
)
Problem 7. If A =(
3 4 0−2 6 −3
7 −4 1
)
and
B =(
2 −55 −6
−1 −7
)
, find A × B.
The sum of the products of the elements of each rowof the first matrix and the elements of each column ofthe second matrix are taken one at a time. Thus:(
3 4 0−2 6 −3
7 −4 1
)
×(
2 −55 −6
−1 −7
)
=
⎛
⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝
[(3 × 2) [(3 × (−5))+ (4 × 5) +(4 × (−6))+ (0 × (−1))] +(0 × (−7))]
[(−2 × 2) [(−2 × (−5))+ (6 × 5) +(6 × (−6))+ (−3 × (−1))] +(−3 × (−7))]
[(7 × 2) [(7 × (−5))+ (−4 × 5) +(−4 × (−6))+ (1 × (−1))] +(1 × (−7))]
⎞
⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠
=(
26 −3929 −5−7 −18
)
Problem 8. Determine(
1 0 32 1 21 3 1
)
×(
2 2 01 3 23 2 0
)
The sum of the products of the elements of each rowof the first matrix and the elements of each column ofthe second matrix are taken one at a time. Thus:(
1 0 32 1 21 3 1
)
×(
2 2 01 3 23 2 0
)
=
⎛
⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝
[(1 × 2) [(1 × 2) [(1 × 0)+ (0 × 1) + (0 × 3) + (0 × 2)+ (3 × 3)] + (3 × 2)] + (3 × 0)]
[(2 × 2) [(2 × 2) [(2 × 0)+ (1 × 1) + (1 × 3) + (1 × 2)+ (2 × 3)] + (2 × 2)] + (2 × 0)]
[(1 × 2) [(1 × 2) [(1 × 0)+ (3 × 1) + (3 × 3) + (3 × 2)+ (1 × 3)] + (1 × 2)] + (1 × 0)]
⎞
⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠
=(
11 8 011 11 28 13 6
)
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270 MATRICES AND DETERMINANTS
In algebra, the commutative law of multiplicationstates that a × b = b × a. For matrices, this law isonly true in a few special cases, and in general A×Bis not equal to B × A.
Problem 9. If A =(
2 31 0
)
and
B =(
2 30 1
)
show that A × B �= B × A.
A × B =(
2 31 0
)
×(
2 30 1
)
=(
[(2 × 2) + (3 × 0)] [(2 × 3) + (3 × 1)][(1 × 2) + (0 × 0)] [(1 × 3) + (0 × 1)]
)
=(
4 92 3
)
B × A =(
2 30 1
)
×(
2 31 0
)
=(
[(2 × 2) + (3 × 1)] [(2 × 3) + (3 × 0)][(0 × 2) + (1 × 1)] [(0 × 3) + (1 × 0)]
)
=(
7 61 0
)
Since
(4 92 3
)
�=(
7 61 0
)
, then A × B �= B × A
Now try the following exercise.
Exercise 108 Further problems on addition,subtraction and multiplication of matrices
In Problems 1 to 13, the matrices A to K are:
A =(
3 −1−4 7
)
B =⎛
⎜⎝
1
2
2
3
−1
3−3
5
⎞
⎟⎠
C =(−1.3 7.4
2.5 −3.9
)
D =(
4 −7 6−2 4 0
5 7 −4
)
E =
⎛
⎜⎜⎜⎜⎝
3 61
25 −2
37
−1 03
5
⎞
⎟⎟⎟⎟⎠
F =(
3.1 2.4 6.4−1.6 3.8 −1.9
5.3 3.4 −4.8
)
G =⎛
⎜⎝
3
4
12
5
⎞
⎟⎠
H =(−2
5
)
J =(
4−11
7
)
K =(
1 00 11 0
)
Addition, subtraction and multiplication
In Problems 1 to 12, perform the matrix opera-tion stated.
1. A + B
⎡
⎢⎣
⎛
⎜⎝
31
2−1
3
−41
36
2
5
⎞
⎟⎠
⎤
⎥⎦
2. D + E
⎡
⎢⎢⎢⎢⎣
⎛
⎜⎜⎜⎜⎝
7 −1 61
23 3
1
37
4 7 −32
5
⎞
⎟⎟⎟⎟⎠
⎤
⎥⎥⎥⎥⎦
3. A − B
⎡
⎢⎣
⎛
⎜⎝
21
2−1
2
3
−32
37
3
5
⎞
⎟⎠
⎤
⎥⎦
4. A + B − C
[(4.8 −7.73
−6.83 10.3
)]
5. 5A + 6B
[(18.0 −1.0
−22.0 31.4
)]
6. 2D + 3E − 4F[(4.6 −5.6 −12.1
17.4 −9.2 28.6−14.2 0.4 13.0
)]
7. A × H
[(−1143
)]
8. A × B
⎡
⎢⎣
⎛
⎜⎝
15
62
3
5
−41
3−6
13
15
⎞
⎟⎠
⎤
⎥⎦
9. A × C
[(−6.4 26.122.7 −56.9
)]
10. D × J
[(135−52−85
)]
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THE THEORY OF MATRICES AND DETERMINANTS 271
F
11. E × K
⎡
⎢⎢⎢⎢⎣
⎛
⎜⎜⎜⎜⎝
31
26
12 −2
3−2
50
⎞
⎟⎟⎟⎟⎠
⎤
⎥⎥⎥⎥⎦
12. D × F
[(55.4 3.4 10.1
−12.6 10.4 −20.4−16.9 25.0 37.9
)]
13. Show that A × C �= C × A⎡
⎢⎢⎢⎢⎣
A × C =(−6.4 26.1
22.7 −56.9
)
C × A =(−33.5 −53.1
23.1 −29.8
)
Hence they are not equal
⎤
⎥⎥⎥⎥⎦
25.3 The unit matrix
A unit matrix, I, is one in which all elementsof the leading diagonal (\) have a value of 1 andall other elements have a value of 0. Multiplicationof a matrix by I is the equivalent of multiplying by1 in arithmetic.
25.4 The determinant of a 2 by 2matrix
The determinant of a 2 by 2 matrix,
(a bc d
)
is
defined as (ad − bc).The elements of the determinant of a matrix are
written between vertical lines. Thus, the determinant
of
(3 −41 6
)
is written as
∣∣∣∣3 −41 6
∣∣∣∣ and is equal to
(3 × 6) − (−4 × 1), i.e. 18 − (−4) or 22. Hence thedeterminant of a matrix can be expressed as a single
numerical value, i.e.
∣∣∣∣3 −41 6
∣∣∣∣ = 22.
Problem 10. Determine the value of∣∣∣∣3 −27 4
∣∣∣∣
∣∣∣∣3 −27 4
∣∣∣∣ = (3 × 4) − (−2 × 7)
= 12 − (−14) = 26
Problem 11. Evaluate
∣∣∣∣(1 + j) j2
− j3 (1 − j4)
∣∣∣∣
∣∣∣∣(1 + j) j2
− j3 (1 − j4)
∣∣∣∣ = (1 + j)(1 − j4) − ( j2)(− j3)
= 1 − j4 + j − j24 + j26= 1 − j4 + j − (−4) + (−6)
since from Chapter 23, j2 = −1= 1 − j4 + j + 4 − 6= −1 − j3
Problem 12. Evaluate
∣∣∣∣5∠30◦ 2∠−60◦3∠60◦ 4∠−90◦
∣∣∣∣
∣∣∣∣5∠30◦ 2∠−60◦3∠60◦ 4∠−90◦
∣∣∣∣ = (5∠30◦)(4∠−90◦)
− (2∠−60◦)(3∠60◦)= (20∠−60◦) − (6∠0◦)= (10 − j17.32) − (6 + j0)= (4 − j17.32) or 17.78∠−77◦
Now try the following exercise.
Exercise 109 Further problems on 2 by 2determinants
1. Calculate the determinant of
(3 −1
−4 7
)
[17]2. Calculate the determinant of⎛
⎜⎝
1
2
2
3
−1
3−3
5
⎞
⎟⎠
[
− 7
90
]
3. Calculate the determinant of(−1.3 7.42.5 −3.9
)
[−13.43]
4. Evaluate
∣∣∣∣
j2 −j3(1 + j) j
∣∣∣∣ [−5 + j3]
5. Evaluate
∣∣∣∣
2∠40◦ 5∠−20◦7∠−32◦ 4∠−117◦
∣∣∣∣
[(−19.75 + j19.79)or 27.95∠134.94◦
]
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272 MATRICES AND DETERMINANTS
25.5 The inverse or reciprocal of a2 by 2 matrix
The inverse of matrix A is A−1 such that A × A−1 = I ,the unit matrix.
Let matrix A be
(1 23 4
)
and let the inverse matrix,
A−1 be
(a bc d
)
.
Then, since A × A−1 = I ,(
1 23 4
)
×(
a bc d
)
=(
1 00 1
)
Multiplying the matrices on the left hand side, gives(
a + 2c b + 2d3a + 4c 3b + 4d
)
=(
1 00 1
)
Equating corresponding elements gives:
b + 2d = 0, i.e. b = −2d
and 3a + 4c = 0, i.e. a = −4
3c
Substituting for a and b gives:⎛
⎜⎝
−4
3c + 2c −2d + 2d
3
(
−4
3c
)
+ 4c 3(−2d) + 4d
⎞
⎟⎠ =
(1 00 1
)
i.e.
(2
3c 0
0 −2d
)
=(
1 00 1
)
showing that2
3c = 1, i.e. c = 3
2and −2d = 1, i.e.
d = −1
2
Since b = −2d, b = 1 and since a = −4
3c, a = −2.
Thus the inverse of matrix
(1 23 4
)
is
(a bc d
)
that
is,
(−2 13
2−1
2
)
There is, however, a quicker method of obtainingthe inverse of a 2 by 2 matrix.
For any matrix
(p qr s
)
the inverse may be
obtained by:
(i) interchanging the positions of p and s,(ii) changing the signs of q and r, and
(iii) multiplying this new matrix by the reciprocal of
the determinant of
(p qr s
)
Thus the inverse of matrix
(1 23 4
)
is
1
4 − 6
(4 −2
−3 1
)
=(−2 1
3
2−1
2
)
as obtained previously.
Problem 13. Determine the inverse of(
3 −27 4
)
The inverse of matrix
(p qr s
)
is obtained by inter-
changing the positions of p and s, changing the signsof q and r and multiplying by the reciprocal of the
determinant
∣∣∣∣p qr s
∣∣∣∣. Thus, the inverse of
(3 −27 4
)
= 1
(3 × 4) − ( − 2 × 7)
(4 2
−7 3
)
= 1
26
(4 2
−7 3
)
=
⎛
⎜⎜⎝
213
113
−726
326
⎞
⎟⎟⎠
Now try the following exercise.
Exercise 110 Further problems on theinverse of 2 by 2 matrices
1. Determine the inverse of
(3 −1
−4 7
)
⎡
⎢⎢⎣
⎛
⎜⎜⎝
7
17
1
174
17
3
17
⎞
⎟⎟⎠
⎤
⎥⎥⎦
2. Determine the inverse of
⎛
⎜⎜⎝
1
2
2
3
−1
3−3
5
⎞
⎟⎟⎠
⎡
⎢⎢⎣
⎛
⎜⎜⎝
75
78
4
7
−42
7−6
3
7
⎞
⎟⎟⎠
⎤
⎥⎥⎦
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THE THEORY OF MATRICES AND DETERMINANTS 273
F
3. Determine the inverse of
(−1.3 7.42.5 −3.9
)
⎡
⎣
(0.290 0.5510.186 0.097
)
correct to 3 dec. places
⎤
⎦
25.6 The determinant of a 3 by 3matrix
(i) The minor of an element of a 3 by 3 matrix isthe value of the 2 by 2 determinant obtained bycovering up the row and column containing thatelement.
Thus for the matrix
(1 2 34 5 67 8 9
)
the minor of
element 4 is obtained by covering the row
(4 5 6) and the column
(147
)
, leaving the 2 by
2 determinant
∣∣∣∣2 38 9
∣∣∣∣, i.e. the minor of element
4 is (2 × 9) − (3 × 8) = −6.
(ii) The sign of a minor depends on its posi-tion within the matrix, the sign pattern
being
(+ − +− + −+ − +
)
. Thus the signed-minor
of element 4 in the matrix
(1 2 34 5 67 8 9
)
is
−∣∣∣∣2 38 9
∣∣∣∣ = −(−6) = 6.
The signed-minor of an element is called thecofactor of the element.
(iii) The value of a 3 by 3 determinant is thesum of the products of the elements and theircofactors of any row or any column of thecorresponding 3 by 3 matrix.
There are thus six different ways of evaluating a 3×3determinant—and all should give the same value.
Problem 14. Find the value of∣∣∣∣∣
3 4 −12 0 71 −3 −2
∣∣∣∣∣
The value of this determinant is the sum of the prod-ucts of the elements and their cofactors, of any rowor of any column. If the second row or second col-umn is selected, the element 0 will make the productof the element and its cofactor zero and reduce theamount of arithmetic to be done to a minimum.
Supposing a second row expansion is selected.The minor of 2 is the value of the determinant
remaining when the row and column containing the2 (i.e. the second row and the first column), is cov-
ered up. Thus the cofactor of element 2 is
∣∣∣∣
4 −1−3 −2
∣∣∣∣
i.e. −11. The sign of element 2 is minus, (see (ii)above), hence the cofactor of element 2, (the signed-minor) is +11. Similarly the minor of element 7 is∣∣∣∣3 41 −3
∣∣∣∣ i.e. −13, and its cofactor is +13. Hence the
value of the sum of the products of the elements andtheir cofactors is 2 × 11 + 7 × 13, i.e.,
∣∣∣∣∣
3 4 −12 0 71 −3 −2
∣∣∣∣∣= 2(11) + 0 + 7(13) = 113
The same result will be obtained whichever row orcolumn is selected. For example, the third columnexpansion is
(−1)
∣∣∣∣2 01 −3
∣∣∣∣− 7
∣∣∣∣3 41 −3
∣∣∣∣+ (−2)
∣∣∣∣3 42 0
∣∣∣∣
= 6 + 91 + 16 = 113, as obtained previously.
Problem 15. Evaluate
∣∣∣∣∣
1 4 −3−5 2 6−1 −4 2
∣∣∣∣∣
Using the first row:
∣∣∣∣∣
1 4 −3−5 2 6−1 −4 2
∣∣∣∣∣
= 1
∣∣∣∣
2 6−4 2
∣∣∣∣− 4
∣∣∣∣−5 6−1 2
∣∣∣∣+ (−3)
∣∣∣∣−5 2−1 −4
∣∣∣∣
= (4 + 24) − 4(−10 + 6) − 3(20 + 2)
= 28 + 16 − 66 = −22
Using the second column:
∣∣∣∣∣
1 4 −3−5 2 6−1 −4 2
∣∣∣∣∣
= −4
∣∣∣∣−5 6−1 2
∣∣∣∣+ 2
∣∣∣∣
1 −3−1 2
∣∣∣∣−(−4)
∣∣∣∣
1 −3−5 6
∣∣∣∣
= −4(−10 + 6) + 2(2 − 3) + 4(6 − 15)
= 16 − 2 − 36 = −22
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274 MATRICES AND DETERMINANTS
Problem 16. Determine the value of∣∣∣∣∣
j2 (1 + j) 3(1 − j) 1 j
0 j4 5
∣∣∣∣∣
Using the first column, the value of the determinant is:
( j2)
∣∣∣∣1 jj4 5
∣∣∣∣− (1 − j)
∣∣∣∣(1 + j) 3
j4 5
∣∣∣∣
+ (0)
∣∣∣∣(1 + j) 3
1 j
∣∣∣∣
= j2(5 − j24) − (1 − j)(5 + j5 − j12) + 0
= j2(9) − (1 − j)(5 − j7)
= j18 − [5 − j7 − j5 + j27]
= j18 − [−2 − j12]
= j18 + 2 + j12 = 2 + j30 or 30.07∠86.19◦
Now try the following exercise.
Exercise 111 Further problems on 3 by 3determinants
1. Find the matrix of minors of(
4 −7 6−2 4 0
5 7 −4
)
[(−16 8 −34−14 −46 63−24 12 2
)]
2. Find the matrix of cofactors of(
4 −7 6−2 4 0
5 7 −4
)
[(−16 −8 −3414 −46 −63
−24 −12 2
)]
3. Calculate the determinant of(
4 −7 6−2 4 0
5 7 −4
)
[−212]
4. Evaluate
∣∣∣∣∣
8 −2 −102 −3 −26 3 8
∣∣∣∣∣
[−328]
5. Calculate the determinant of(
3.1 2.4 6.4−1.6 3.8 −1.9
5.3 3.4 −4.8
)
[−242.83]
6. Evaluate
∣∣∣∣∣
j2 2 j(1 + j) 1 −3
5 −j4 0
∣∣∣∣∣
[−2 − j]
7. Evaluate
∣∣∣∣∣
3∠60◦ j2 10 (1 + j) 2∠30◦0 2 j5
∣∣∣∣∣
[26.94∠−139.52◦ or(−20.49 − j17.49)
]
8. Find the eigenvalues λ that satisfy the follow-ing equations:
(a)
∣∣∣∣(2 − λ) 2
−1 (5 − λ)
∣∣∣∣= 0
(b)
∣∣∣∣∣
(5 − λ) 7 −50 (4 − λ) −12 8 (−3 − λ)
∣∣∣∣∣= 0
(You may need to refer to chapter 1, pages8–11, for the solution of cubic equations).
[(a) λ = 3 or 4 (b) λ = 1 or 2 or 3]
25.7 The inverse or reciprocal of a3 by 3 matrix
The adjoint of a matrix A is obtained by:
(i) forming a matrix B of the cofactors of A, and
(ii) transposing matrix B to give BT , where BT isthe matrix obtained by writing the rows of Bas the columns of BT . Then adj A = BT .
The inverse of matrix A, A−1 is given by
A−1 = adj A|A|
where adj A is the adjoint of matrix A and |A| is thedeterminant of matrix A.
Ch25-H8152.tex 23/6/2006 15: 9 Page 275
THE THEORY OF MATRICES AND DETERMINANTS 275
F
Problem 17. Determine the inverse of the
matrix
⎛
⎝3 4 −12 0 71 −3 −2
⎞
⎠
The inverse of matrix A, A−1 = adj A
|A|The adjoint of A is found by:
(i) obtaining the matrix of the cofactors of theelements, and
(ii) transposing this matrix.
The cofactor of element 3 is +∣∣∣∣
0 7−3 −2
∣∣∣∣ = 21.
The cofactor of element 4 is −∣∣∣∣2 71 −2
∣∣∣∣ = 11, and
so on.
The matrix of cofactors is
(21 11 −611 −5 1328 −23 −8
)
The transpose of the matrix of cofactors, i.e. theadjoint of the matrix, is obtained by writing the rows
as columns, and is
(21 11 2811 −5 −23−6 13 −8
)
From Problem 14, the determinant of∣∣∣∣∣
3 4 −12 0 71 −3 −2
∣∣∣∣∣
is 113.
Hence the inverse of
(3 4 −12 0 71 −3 −2
)
is
(21 11 2811 −5 −23−6 13 −8
)
113or
1113
(21 11 2811 −5 −23−6 13 −8
)
Problem 18. Find the inverse of(
1 5 −23 −1 4
−3 6 −7
)
Inverse = adjoint
determinant
The matrix of cofactors is
(−17 9 1523 −13 −2118 −10 −16
)
The transpose of the matrix of cofactors (i.e. the
adjoint) is
(−17 23 189 −13 −10
15 −21 −16
)
The determinant of
(1 5 −23 −1 4
−3 6 −7
)
= 1(7 − 24) − 5(−21 + 12) − 2(18 − 3)
= −17 + 45 − 30 = −2
Hence the inverse of
(1 5 −23 −1 4
−3 6 −7
)
=
(−17 23 189 −13 −10
15 −21 −16
)
−2
=(
8.5 −11.5 −9−4.5 6.5 5−7.5 10.5 8
)
Now try the following exercise.
Exercise 112 Further problems on theinverse of a 3 by 3 matrix
1. Write down the transpose of(
4 −7 6−2 4 0
5 7 −4
)
[(4 −2 5
−7 4 76 0 −4
)]
2. Write down the transpose of⎛
⎝3 6 1
25 − 2
3 7−1 0 3
5
⎞
⎠
⎡
⎣
⎛
⎝3 5 −16 − 2
3 012 7 3
5
⎞
⎠
⎤
⎦
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276 MATRICES AND DETERMINANTS
3. Determine the adjoint of(
4 −7 6−2 4 0
5 7 −4
)
[(−16 14 −24−8 −46 −12
−34 −63 2
)]
4. Determine the adjoint of⎛
⎝3 6 1
25 − 2
3 7−1 0 3
5
⎞
⎠
⎡
⎢⎣
⎛
⎜⎝
− 25 −3 3
5 42 13
−10 2 310 −18 1
2
− 23 −6 −32
⎞
⎟⎠
⎤
⎥⎦
5. Find the inverse of(
4 −7 6−2 4 0
5 7 −4
)
[
− 1
212
(−16 14 −24−8 −46 −12
−34 −63 2
)]
6. Find the inverse of
⎛
⎝3 6 1
25 − 2
3 7−1 0 3
5
⎞
⎠
⎡
⎢⎣− 15
923
⎛
⎜⎝
− 25 −3 3
5 42 13
−10 2 310 −18 1
2
− 23 −6 −32
⎞
⎟⎠
⎤
⎥⎦
Ch26-H8152.tex 23/6/2006 15: 9 Page 277
F
Matrices and Determinants
26
The solution of simultaneousequations by matrices anddeterminants
26.1 Solution of simultaneousequations by matrices
(a) The procedure for solving linear simultaneousequations in two unknowns using matrices is:
(i) write the equations in the form
a1x + b1y = c1
a2x + b2y = c2
(ii) write the matrix equation corresponding tothese equations,
i.e.
(a1 b1a2 b2
)
×(
xy
)
=(
c1c2
)
(iii) determine the inverse matrix of
(a1 b1a2 b2
)
i.e.1
a1b2 − b1a2
(b2 −b1−a2 a1
)
(from Chapter 25)
(iv) multiply each side of (ii) by the inversematrix, and
(v) solve for x and y by equating correspondingelements.
Problem 1. Use matrices to solve the simulta-neous equations:
3x + 5y − 7 = 0 (1)4x − 3y − 19 = 0 (2)
(i) Writing the equations in the a1x+b1y = c formgives:
3x + 5y = 74x − 3y = 19
(ii) The matrix equation is(
3 54 −3
)
×(
xy
)
=(
719
)
(iii) The inverse of matrix
(3 54 −3
)
is
1
3 × (−3) − 5 × 4
(−3 −5−4 3
)
i.e.
⎛
⎜⎝
3
29
5
294
29
−3
29
⎞
⎟⎠
(iv) Multiplying each side of (ii) by (iii) and remem-bering that A × A−1 = I , the unit matrix, gives:
(1 00 1
)(xy
)
=⎛
⎜⎝
3
29
5
294
29
−3
29
⎞
⎟⎠×
(7
19
)
Thus
(xy
)
=⎛
⎜⎝
21
29+ 95
2928
29− 57
29
⎞
⎟⎠
i.e.
(xy
)
=(
4−1
)
(v) By comparing corresponding elements:
x = 4 and y = −1
Checking:
equation (1),
3 × 4 + 5 × (−1) − 7 = 0 = RHS
equation (2),
4 × 4 − 3 × (−1) − 19 = 0 = RHS
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278 MATRICES AND DETERMINANTS
(b) The procedure for solving linear simulta-neous equations in three unknowns usingmatrices is:
(i) write the equations in the form
a1x + b1y + c1z = d1
a2x + b2y + c2z = d2
a3x + b3y + c3z = d3
(ii) write the matrix equation correspondingto these equations, i.e.(
a1 b1 c1a2 b2 c2a3 b3 c3
)
×(
xyz
)
=(
d1d2d3
)
(iii) determine the inverse matrix of(
a1 b1 c1a2 b2 c2a3 b3 c3
)
(see Chapter 25)
(iv) multiply each side of (ii) by the inversematrix, and
(v) solve for x, y and z by equating thecorresponding elements.
Problem 2. Use matrices to solve the simulta-neous equations:
x + y + z − 4 = 0 (1)2x − 3y + 4z − 33 = 0 (2)3x − 2y − 2z − 2 = 0 (3)
(i) Writing the equations in the a1x + b1y + c1z =d1 form gives:
x + y + z = 42x − 3y + 4z = 333x − 2y − 2z = 2
(ii) The matrix equation is(
1 1 12 −3 43 −2 −2
)
×(
xyz
)
=(
433
2
)
(iii) The inverse matrix of
A =(
1 1 12 −3 43 −2 −2
)
is given by
A−1 = adj A
|A|
The adjoint of A is the transpose of the matrix ofthe cofactors of the elements (see Chapter 25).The matrix of cofactors is
(14 16 5
0 −5 57 −2 −5
)
and the transpose of this matrix gives
adj A =(
14 0 716 −5 −2
5 5 −5
)
The determinant of A, i.e. the sum of the prod-ucts of elements and their cofactors, using a firstrow expansion is
1
∣∣∣∣−3 4−2 −2
∣∣∣∣− 1
∣∣∣∣2 43 −2
∣∣∣∣+ 1
∣∣∣∣2 −33 −2
∣∣∣∣
= (1 × 14) − (1 × (−16)) + (1 × 5) = 35Hence the inverse of A,
A−1 = 1
35
(14 0 716 −5 −2
5 5 −5
)
(iv) Multiplying each side of (ii) by (iii), andremembering that A × A−1 = I , the unit matrix,gives(
1 0 00 1 00 0 1
)
×(
xyz
)
= 1
35
(14 0 716 −5 −2
5 5 −5
)
×(
433
2
)
(xyz
)
= 1
35
×(
(14 × 4) + (0 × 33) + (7 × 2)(16 × 4) + ((−5) × 33) + ((−2) × 2)(5 × 4) + (5 × 33) + ((−5) × 2)
)
= 1
35
(70
−105175
)
=(
2−3
5
)
(v) By comparing corresponding elements, x = 2,y = −3, z = 5, which can be checked in theoriginal equations.
Ch26-H8152.tex 23/6/2006 15: 9 Page 279
THE SOLUTION OF SIMULTANEOUS EQUATIONS BY MATRICES AND DETERMINANTS 279
F
Now try the following exercise.
Exercise 113 Further problems on solvingsimultaneous equations using matrices
In Problems 1 to 5 use matrices to solve thesimultaneous equations given.
1. 3x + 4y = 0
2x + 5y + 7 = 0 [x = 4, y = −3]
2. 2p + 5q + 14.6 = 0
3.1p + 1.7q + 2.06 = 0
[p = 1.2, q = −3.4]
3. x + 2y + 3z = 5
2x − 3y − z = 3
−3x + 4y + 5z = 3
[x = 1, y = −1, z = 2]
4. 3a + 4b − 3c = 2
−2a + 2b + 2c = 15
7a − 5b + 4c = 26
[a = 2.5, b = 3.5, c = 6.5]
5. p + 2q + 3r + 7.8 = 0
2p + 5q − r − 1.4 = 0
5p − q + 7r − 3.5 = 0
[p = 4.1, q = −1.9, r = −2.7]
6. In two closed loops of an electrical cir-cuit, the currents flowing are given by thesimultaneous equations:
I1 + 2I2 + 4 = 05I1 + 3I2 − 1 = 0
Use matrices to solve for I1 and I2.
[I1 = 2, I2 = −3]
7. The relationship between the displacement,s, velocity, v, and acceleration, a, of a pistonis given by the equations:
s + 2v + 2a = 43s − v + 4a = 253s + 2v − a = −4
Use matrices to determine the values of s, vand a.
[s = 2, v = −3, a = 4]
8. In a mechanical system, acceleration x,velocity x and distance x are related by thesimultaneous equations:
3.4x + 7.0x − 13.2x = −11.39
−6.0x + 4.0x + 3.5x = 4.98
2.7x + 6.0x + 7.1x = 15.91
Use matrices to find the values of x, x and x.
[x = 0.5, x = 0.77, x = 1.4]
26.2 Solution of simultaneousequations by determinants
(a) When solving linear simultaneous equations intwo unknowns using determinants:
(i) write the equations in the form
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
and then
(ii) the solution is given by
x
Dx= −y
Dy= 1
D
where Dx =∣∣∣∣b1 c1
b2 c2
∣∣∣∣
i.e. the determinant of the coefficients leftwhen the x-column is covered up,
Dy =∣∣∣∣a1 c1
a2 c2
∣∣∣∣
i.e. the determinant of the coefficients leftwhen the y-column is covered up,
and D =∣∣∣∣a1 b1
a2 b2
∣∣∣∣
i.e. the determinant of the coefficients leftwhen the constants-column is covered up.
Problem 3. Solve the following simultaneousequations using determinants:
3x − 4y = 12
7x + 5y = 6.5
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280 MATRICES AND DETERMINANTS
Following the above procedure:
(i) 3x − 4y − 12 = 07x + 5y − 6.5 = 0
(ii)x
∣∣∣∣−4 −12
5 −6.5
∣∣∣∣
= −y∣∣∣∣3 −127 −6.5
∣∣∣∣
= 1∣∣∣∣3 −47 5
∣∣∣∣
i.e.x
(−4)(−6.5) − (−12)(5)
= −y
(3)(−6.5) − (−12)(7)
= 1
(3)(5) − (−4)(7)
i.e.x
26 + 60= −y
−19.5 + 84= 1
15 + 28
i.e.x
86= −y
64.5= 1
43
Sincex
86= 1
43then x = 86
43= 2
and since
−y
64.5= 1
43then y = −64.5
43= −1.5
Problem 4. The velocity of a car, acceleratingat uniform acceleration a between two points, isgiven by v = u + at, where u is its velocity whenpassing the first point and t is the time takento pass between the two points. If v = 21 m/swhen t = 3.5 s and v = 33 m/s when t = 6.1 s,use determinants to find the values of u and a,each correct to 4 significant figures.
Substituting the given values in v = u + at gives:
21 = u + 3.5a (1)33 = u + 6.1a (2)
(i) The equations are written in the forma1x + b1y + c1 = 0,
i.e. u + 3.5a − 21 = 0and u + 6.1a − 33 = 0
(ii) The solution is given by
u
Du= −a
Da= 1
D
where Du is the determinant of coefficients leftwhen the u column is covered up,
i.e. Du =∣∣∣∣∣
3.5 −21
6.1 −33
∣∣∣∣∣
= (3.5)(−33) − (−21)(6.1)= 12.6
Similarly, Da =∣∣∣∣1 −211 −33
∣∣∣∣
= (1)(−33) − (−21)(1)= −12
and D =∣∣∣∣1 3.51 6.1
∣∣∣∣
= (1)(6.1) − (3.5)(1) = 2.6
Thusu
12.6= −a
−12= 1
26
i.e. u = 12.6
2.6= 4.846 m/s
and a = 12
2.6= 4.615 m/s2,
each correct to 4 significantfigures
Problem 5. Applying Kirchhoff’s laws to anelectric circuit results in the following equations:
(9 + j12)I1 − (6 + j8)I2 = 5−(6 + j8)I1 + (8 + j3)I2 = (2 + j4)
Solve the equations for I1 and I2
Following the procedure:
(i) (9 + j12)I1 − (6 + j8)I2 − 5 = 0
−(6 + j8)I1 + (8 + j3)I2 − (2 + j4) = 0
(ii)I1∣
∣∣∣−(6 + j8) −5(8 + j3) −(2 + j4)
∣∣∣∣
= −I2∣∣∣∣(9 + j12) −5−(6 + j8) −(2 + j4)
∣∣∣∣
= 1∣∣∣∣(9 + j12) −(6 + j8)−(6 + j8) (8 + j3)
∣∣∣∣
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THE SOLUTION OF SIMULTANEOUS EQUATIONS BY MATRICES AND DETERMINANTS 281
F
I1
(−20 + j40) + (40 + j15)
= −I2
(30 − j60) − (30 + j40)
= 1
(36 + j123) − (−28 + j96)
I1
20 + j55= −I2
−j100
= 1
64 + j27
Hence I1 = 20 + j55
64 + j27
= 58.52∠70.02◦
69.46∠22.87◦ = 0.84∠47.15◦A
and I2 = 100∠90◦
69.46∠22.87◦= 1.44∠67.13◦ A
(b) When solving simultaneous equations in threeunknowns using determinants:
(i) Write the equations in the form
a1x + b1y + c1z + d1 = 0a2x + b2y + c2z + d2 = 0a3x + b3y + c3z + d3 = 0
and then(ii) the solution is given by
x
Dx= −y
Dy= z
Dz= −1
D
where Dx is
∣∣∣∣∣
b1 c1 d1b2 c2 d2b3 c3 d3
∣∣∣∣∣
i.e. the determinant of the coefficientsobtained by covering up the x column.
Dy is
∣∣∣∣∣
a1 c1 d1a2 c2 d2a3 c3 d3
∣∣∣∣∣
i.e., the determinant of the coefficientsobtained by covering up the y column.
Dz is
∣∣∣∣∣
a1 b1 d1a2 b2 d2a3 b3 d3
∣∣∣∣∣
i.e. the determinant of the coefficientsobtained by covering up the z column.
and D is
∣∣∣∣∣
a1 b1 c1a2 b2 c2a3 b3 c3
∣∣∣∣∣
i.e. the determinant of the coefficientsobtained by covering up the constantscolumn.
Problem 6. A d.c. circuit comprises threeclosed loops. Applying Kirchhoff’s laws to theclosed loops gives the following equations forcurrent flow in milliamperes:
2I1 + 3I2 − 4I3 = 26I1 − 5I2 − 3I3 = −87
−7I1 + 2I2 + 6I3 = 12
Use determinants to solve for I1, I2 and I3
(i) Writing the equations in thea1x + b1y + c1z + d1 = 0 form gives:
2I1 + 3I2 − 4I3 − 26 = 0
I1 − 5I2 − 3I3 + 87 = 0
−7I1 + 2I2 + 6I3 − 12 = 0
(ii) the solution is given by
I1
DI1
= −I2
DI2
= I3
DI3
= −1
D
where DI1 is the determinant of coefficientsobtained by covering up the I1 column, i.e.,
DI1 =∣∣∣∣∣
3 −4 −26−5 −3 87
2 6 −12
∣∣∣∣∣
= (3)
∣∣∣∣−3 87
6 −12
∣∣∣∣− (−4)
∣∣∣∣−5 87
2 −12
∣∣∣∣
+ (−26)
∣∣∣∣−5 −3
2 6
∣∣∣∣
= 3(−486) + 4(−114) − 26(−24)
= −1290
DI2 =∣∣∣∣∣
2 −4 −261 −3 87
−7 6 −12
∣∣∣∣∣
= (2)(36 − 522) − (−4)(−12 + 609)
+ (−26)(6 − 21)
= −972 + 2388 + 390
= 1806
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282 MATRICES AND DETERMINANTS
DI3 =∣∣∣∣∣
2 3 −261 −5 87
−7 2 −12
∣∣∣∣∣
= (2)(60 − 174) − (3)(−12 + 609)
+(−26)(2 − 35)
= −228 − 1791 + 858 = −1161
and D =∣∣∣∣∣
2 3 −41 −5 −3
−7 2 6
∣∣∣∣∣
= (2)(−30 + 6) − (3)(6 − 21)
+ (−4)(2 − 35)
= −48 + 45 + 132 = 129
Thus
I1
−1290= −I2
1806= I3
−1161= −1
129
giving
I1 = −1290
−129= 10 mA,
I2 = 1806
129= 14 mA
and I3 = 1161
129= 9 mA
Now try the following exercise.
Exercise 114 Further problems on solvingsimultaneous equations using determinants
In Problems 1 to 5 use determinants to solvethe simultaneous equations given.
1. 3x − 5y = −17.6
7y − 2x − 22 = 0
[x = −1.2, y = 2.8]
2. 2.3m − 4.4n = 6.84
8.5n − 6.7m = 1.23
[m = −6.4, n = −4.9]
3. 3x + 4y + z = 102x − 3y + 5z + 9 = 0x + 2y − z = 6
[x = 1, y = 2, z = −1]
4. 1.2p − 2.3q − 3.1r + 10.1 = 0
4.7p + 3.8q − 5.3r − 21.5 = 0
3.7p − 8.3q + 7.4r + 28.1 = 0
[p = 1.5, q = 4.5, r = 0.5]
5.x
2− y
3+ 2z
5= − 1
20x
4+ 2y
3− z
2= 19
40
x + y − z = 59
60[
x = 7
20, y = 17
40, z = − 5
24
]
6. In a system of forces, the relationshipbetween two forces F1 and F2 is given by:
5F1 + 3F2 + 6 = 0
3F1 + 5F2 + 18 = 0
Use determinants to solve for F1 and F2.
[F1 = 1.5, F2 = −4.5]
7. Applying mesh-current analysis to an a.c.circuit results in the following equations:
(5 − j4)I1 − (−j4)I2 = 100∠0◦
(4 + j3 − j4)I2 − (−j4)I1 = 0
Solve the equations for I1 and I2.[
I1 = 10.77∠19.23◦A,I2 = 10.45∠−56.73◦A
]
8. Kirchhoff’s laws are used to determine thecurrent equations in an electrical networkand show that
i1 + 8i2 + 3i3 = −31
3i1 − 2i2 + i3 = −5
2i1 − 3i2 + 2i3 = 6
Use determinants to find the values of i1, i2and i3. [i1 = −5, i2 = −4, i3 = 2]
9. The forces in three members of a frameworkare F1, F2 and F3. They are related by thesimultaneous equations shown below.
1.4F1 + 2.8F2 + 2.8F3 = 5.6
4.2F1 − 1.4F2 + 5.6F3 = 35.0
4.2F1 + 2.8F2 − 1.4F3 = −5.6
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THE SOLUTION OF SIMULTANEOUS EQUATIONS BY MATRICES AND DETERMINANTS 283
F
Find the values of F1, F2 and F3 usingdeterminants.
[F1 = 2, F2 = −3, F3 = 4]
10. Mesh-current analysis produces the follow-ing three equations:
20∠0◦ = (5 + 3 − j4)I1 − (3 − j4)I2
10∠90◦ = (3 − j4 + 2)I2 − (3 − j4)I1 − 2I3
−15∠0◦ − 10∠90◦ = (12 + 2)I3 − 2I2
Solve the equations for the loop currentsI1, I2 and I3.
[I1 = 3.317∠22.57◦ AI2 = 1.963∠40.97◦ AI3 = 1.010∠−148.32◦ A
]
26.3 Solution of simultaneousequations using Cramers rule
Cramers rule states that if
a11x + a12y + a13z = b1
a21x + a22y + a23z = b2
a31x + a32y + a33z = b3
then x = Dx
D, y = Dy
Dand z = Dz
D
where D =∣∣∣∣∣
a11 a12 a13a21 a22 a23a31 a32 a33
∣∣∣∣∣
Dx =∣∣∣∣∣
b1 a12 a13b2 a22 a23b3 a32 a33
∣∣∣∣∣
i.e. the x-column has been replaced by the R.H.S.b column,
Dy =∣∣∣∣∣
a11 b1 a13a21 b2 a23a31 b3 a33
∣∣∣∣∣
i.e. the y-column has been replaced by the R.H.S.b column,
Dz =∣∣∣∣∣
a11 a12 b1a21 a22 b2a31 a32 b3
∣∣∣∣∣
i.e. the z-column has been replaced by the R.H.S.b column.
Problem 7. Solve the following simultaneousequations using Cramers rule
x + y + z = 42x − 3y + 4z = 333x − 2y − 2z = 2
(This is the same as Problem 2 and a comparisonof methods may be made). Following the abovemethod:
D =∣∣∣∣∣
1 1 12 −3 43 −2 −2
∣∣∣∣∣
= 1(6 − (−8)) − 1((−4) − 12)+1((−4) − (−9)) = 14 + 16 + 5 = 35
Dx =∣∣∣∣∣
4 1 133 −3 4
2 −2 −2
∣∣∣∣∣
= 4(6 − (−8)) − 1((−66) − 8)+1((−66) − (−6)) = 56 + 74 − 60 = 70
Dy =∣∣∣∣∣
1 4 12 33 43 2 −2
∣∣∣∣∣
= 1((−66) − 8) − 4((−4) − 12) + 1(4 − 99)= −74 + 64 − 95 = −105
Dz =∣∣∣∣∣
1 1 42 −3 333 −2 2
∣∣∣∣∣
= 1((−6) − (−66)) − 1(4 − 99)+ 4((−4) − (−9)) = 60 + 95 + 20 = 175
Hence
x = Dx
D= 70
35= 2, y = Dy
D= −105
35= −3
and z = Dz
D= 175
35= 5
Now try the following exercise.
Exercise 115 Further problems on solvingsimultaneous equations using Cramers rule
1. Repeat problems 3, 4, 5, 7 and 8 of Exercise113 on page 279, using Cramers rule.
2. Repeat problems 3, 4, 8 and 9 of Exercise 114on page 282, using Cramers rule.
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284 MATRICES AND DETERMINANTS
26.4 Solution of simultaneousequations using the Gaussianelimination method
Consider the following simultaneous equations:
x + y + z = 4 (1)2x − 3y + 4z = 33 (2)3x − 2y − 2z = 2 (3)
Leaving equation (1) as it is gives:
x + y + z = 4 (1)
Equation (2) − 2 × equation (1) gives:
0 − 5y + 2z = 25 (2′)and equation (3) − 3 × equation (1) gives:
0 − 5y − 5z = −10 (3′)Leaving equations (1) and (2′) as they are gives:
x + y + z = 4 (1)
0 − 5y + 2z = 25 (2′)Equation (3′) − equation (2′) gives:
0 + 0 − 7z = −35 (3′′)By appropriately manipulating the three originalequations we have deliberately obtained zeros in thepositions shown in equations (2′) and (3′′).Working backwards, from equation (3′′),
z = −35
−7= 5,
from equation (2′),−5y + 2(5) = 25,
from which,
y = 25 − 10
−5= −3
and from equation (1),
x + (−3) + 5 = 4,
from which,
x = 4 + 3 − 5 = 2
(This is the same example as Problems 2 and 7,and a comparison of methods can be made). Theabove method is known as the Gaussian eliminationmethod.
We conclude from the above example that ifa11x + a12y + a13z = b1
a21x + a22y + a23z = b2
a31x + a32y + a33z = b3
the three-step procedure to solve simultaneousequations in three unknowns using the Gaussianelimination method is:
1. Equation (2) − a21
a11× equation (1) to form equa-
tion (2′) and equation (3) − a31
a11× equation (1) to
form equation (3′).
2. Equation (3′)− a32
a22× equation (2′) to form equa-
tion (3′′).
3. Determine z from equation (3′′), then y fromequation (2′) and finally, x from equation (1).
Problem 8. A d.c. circuit comprises threeclosed loops. Applying Kirchhoff’s laws to theclosed loops gives the following equations forcurrent flow in milliamperes:
2I1 + 3I2 − 4I3 = 26 (1)I1 − 5I2 − 3I3 = −87 (2)
−7I1 + 2I2 + 6I3 = 12 (3)
Use the Gaussian elimination method to solvefor I1, I2 and I3.
(This is the same example as Problem 6 on page 281,and a comparison of methods may be made)
Following the above procedure:
1. 2I1 + 3I2 − 4I3 = 26 (1)
Equation (2) − 1
2× equation (1) gives:
0 − 6.5I2 − I3 = −100 (2′)
Equation (3) − −7
2× equation (1) gives:
0 + 12.5I2 − 8I3 = 103 (3′)2. 2I1 + 3I2 − 4I3 = 26 (1)
0 − 6.5I2 − I3 = −100 (2′)
Equation (3′) − 12.5
−6.5× equation (2′) gives:
0 + 0 − 9.923I3 = −89.308 (3′′)
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THE SOLUTION OF SIMULTANEOUS EQUATIONS BY MATRICES AND DETERMINANTS 285
F
3. From equation (3′′),
I3 = −89.308
−9.923= 9 mA,
from equation (2′), −6.5I2 − 9 = −100,
from which, I2 = −100 + 9
−6.5= 14 mA
and from equation (1), 2I1 + 3(14) − 4(9) = 26,
from which, I1 = 26 − 42 + 36
2= 20
2= 10 mA
Now try the following exercise.
Exercise 116 Further problems on solv-ing simultaneous equations using Gaussianelimination
1. In a mass-spring-damper system, the acceler-ation x m/s2, velocity x m/s and displacementx m are related by the following simultaneous
equations:
6.2x + 7.9x + 12.6x = 18.07.5x + 4.8x + 4.8x = 6.3913.0x + 3.5x − 13.0x = −17.4
By using Gaussian elimination, determine theacceleration, velocity and displacement forthe system, correct to 2 decimal places.
[x = −0.30, x = 0.60, x = 1.20]
2. The tensions, T1, T2 and T3 in a simpleframework are given by the equations:
5T1 + 5T2 + 5T3 = 7.0T1 + 2T2 + 4T3 = 2.4
4T1 + 2T2 = 4.0
Determine T1, T2 and T3 using Gaussianelimination.
[T1 = 0.8, T2 = 0.4, T3 = 0.2]
3. Repeat problems 3, 4, 5, 7 and 8 of Exer-cise 113 on page 279, using the Gaussianelimination method.
4. Repeat problems 3, 4, 8 and 9 of Exercise 114on page 282, using the Gaussian eliminationmethod.
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Complex numbers and Matrices and Determinants
Assignment 7
This assignment covers the material containedin Chapters 23 to 26.
The marks for each question are shown inbrackets at the end of each question.
1. Solve the quadratic equation x2 − 2x + 5 = 0 andshow the roots on an Argand diagram. (9)
2. If Z1 = 2 + j5, Z2 = 1 − j3 and Z3 = 4 − j deter-mine, in both Cartesian and polar forms, the value
ofZ1Z2
Z1 + Z2+ Z3, correct to 2 decimal places.
(9)
3. Three vectors are represented by A, 4.2∠45◦, B,5.5∠−32◦ and C, 2.8∠75◦. Determine in polarform the resultant D, where D = B + C − A.
(8)
4. Two impedances, Z1 = (2 + j7) ohms andZ2 = (3 − j4) ohms, are connected in series toa supply voltage V of 150∠0◦ V. Determine themagnitude of the current I and its phase anglerelative to the voltage. (6)
5. Determine in both polar and rectangularforms:
(a) [2.37∠35◦]4 (b) [3.2 − j4.8]5
(c)√
[−1 − j3] (15)
In questions 6 to 10, the matrices stated are:
A =(−5 2
7 −8
)
B =(
1 6−3 −4
)
C =(
j3 (1 + j2)(−1 − j4) −j2
)
D =⎛
⎝2 −1 3
−5 1 04 −6 2
⎞
⎠ E =⎛
⎝−1 3 0
4 −9 2−5 7 1
⎞
⎠
6. Determine A × B (4)
7. Calculate the determinant of matrix C (4)
8. Determine the inverse of matrix A (4)
9. Determine E × D (9)
10. Calculate the determinant of matrix D (6)
11. Solve the following simultaneous equations:
4x − 3y = 17x + y + 1 = 0
using matrices. (6)
12. Use determinants to solve the following simul-taneous equations:
4x + 9y + 2z = 21−8x + 6y − 3z = 41
3x + y − 5z = −73 (10)
13. The simultaneous equations representing the cur-rents flowing in an unbalanced, three-phase,star-connected, electrical network are as follows:
2.4I1 + 3.6I2 + 4.8I3 = 1.2−3.9I1 + 1.3I2 − 6.5I3 = 2.61.7I1 + 11.9I2 + 8.5I3 = 0
Using matrices, solve the equations for I1, I2and I3 (10)