Basic Mathematics For Beginners
Basic Mathematics For Beginners
Basic Mathematics For Beginners
25
The theory of matrices and determinants
Similarly, the coefficients of p, q and r in the (a) Adding the corresponding elements gives:
equations
2 −1 −3 0
1.3p − 2.0q + r = 7 −7 4 + 7 −4
3.7p + 4.8q − 7r = 3 2 + (−3) −1 + 0
= −7 + 7 4 + (−4)
4.1p + 3.8q + 12r = −6
−1 −1
=
1.3 −2.0 1 0 0
become 3.7 4.8 −7 in matrix form. (b) Adding the corresponding elements gives:
4.1 3.8 12
The numbers within a matrix are called an array and 3 1 −4 2 7 −5
the coefficients forming the array are called the ele- 4 3 1 + −2 1 0
ments of the matrix. The number of rows in a matrix 1 4 −3 6 3 4
is usually specified by m and the number of columns 3+2 1 + 7 −4 + (−5)
by n and
a matrix referred to as an ‘m by n’ matrix. = 4 + (−2) 3 + 1 1+0
2 3 6 1+6 4 + 3 −3 + 4
Thus, 4 5 7 is a ‘2 by 3’ matrix. Matrices can-
5 8 −9
not be expressed as a single numerical value, but they = 2 4 1
can often be simplified or combined, and unknown 7 7 1
element values can be determined by comparison
methods. Just as there are rules for addition, sub-
traction, multiplication and division of numbers in (ii) Subtraction of matrices
arithmetic, rules for these operations can be applied
to matrices and the rules of matrices are such that If A is a matrix and B is another matrix, then (A − B)
they obey most of those governing the algebra of is a single matrix formed by subtracting the elements
numbers. of B from the corresponding elements of A.
268 MATRICES AND DETERMINANTS
−3 + 2 − 1 0 + ( − 1) − 0
Problem 2. Subtract = 7 + (−7) − (−2) −4 + 4 − (−4)
−3 0 2 −1
(a) 7 −4 from −7 4 and −2 −1
= 2 4 as obtained previously
2 7 −5 3 1 −4
(b) −2 1 0 from 4 3 1
6 3 4 1 4 −3 (iii) Multiplication
2 3 −5 7 3 4 0
Problem 5. If A = 1 −4 and B = −3 4 Problem 7. If A = −2 6 −3 and
find A × B. 7 −4 1
2 −5
B= 5 −6 , find A × B.
C C −1 −7
Let A × B = C where C = C11 C12
21 22
C11 is the sum of the products of the first row ele- The sum of the products of the elements of each row
ments of A and the first column elements of B taken of the first matrix and the elements of each column of
one at a time, the second matrix are taken one at a time. Thus:
i.e. C11 = (2 × (−5)) + (3 × (−3)) = −19 3 4 0 2 −5
−2 6 −3 × 5 −6
C12 is the sum of the products of the first row ele- 7 −4 1 −1 −7
⎛ ⎞
ments of A and the second column elements of B, [(3 × 2) [(3 × (−5))
taken one at a time, ⎜ + (4 × 5) +(4 × (−6)) ⎟
⎜ + (0 × (−1))] +(0 × (−7))] ⎟
i.e. C12 = (2 × 7) + (3 × 4) = 26 ⎜ ⎟
⎜[(−2 × 2) [(−2 × (−5)) ⎟
⎜ ⎟
C21 is the sum of the products of the second row ⎜
= ⎜ + (6 × 5) +(6 × (−6)) ⎟ ⎟
elements of A and the first column elements of B, ⎜ + (−3 × (−1))] +(−3 × (−7))]⎟
⎜ ⎟
taken one at a time, ⎜[(7 × 2) [(7 × (−5)) ⎟
⎝ + (−4 × 5) +(−4 × (−6)) ⎠
i.e. C21 = (1 × ( − 5)) + (−4 × (−3)) = 7
+ (1 × (−1))] +(1 × (−7))]
Finally, C22 is the sum of the products of the second
26 −39 F
row elements of A and the second column elements = 29 −5
of B, taken one at a time, −7 −18
i.e. C22 = (1 × 7) + ((−4) × 4) = −9
Problem 8. Determine
−19 26
Thus, A × B = 7 −9 1 0 3 2 2 0
2 1 2 × 1 3 2
1 3 1 3 2 0
Problem 6. Simplify
3 4 0 2 The sum of the products of the elements of each row
−2 6 −3 × 5 of the first matrix and the elements of each column of
7 −4 1 −1 the second matrix are taken one at a time. Thus:
1 0 3 2 2 0
2 1 2 × 1 3 2
The sum of the products of the elements of each 1 3 1 3 2 0
row of the first matrix and the elements of the second ⎛ ⎞
matrix, (called a column matrix), are taken one at a [(1 × 2) [(1 × 2) [(1 × 0)
time. Thus: ⎜ + (0 × 1) + (0 × 3) + (0 × 2) ⎟
⎜ + (3 × 3)] + (3 × 2)] + (3 × 0)]⎟
⎜ ⎟
3 4 0 2 ⎜[(2 × 2) [(2 × 2) [(2 × 0) ⎟
−2 6 −3 × ⎜ ⎟
5 =⎜ ⎜ + (1 × 1) + (1 × 3) + (1 × 2) ⎟
7 −4 1 −1 ⎟
⎜ + (2 × 3)] + (2 × 2)] + (2 × 0)]⎟
(3 × 2) + (4 × 5) + (0 × (−1)) ⎜ ⎟
⎜[(1 × 2) [(1 × 2) [(1 × 0) ⎟
= (−2 × 2) + (6 × 5) + (−3 × (−1)) ⎝ + (3 × 1) + (3 × 3) + (3 × 2) ⎠
(7 × 2) + (−4 × 5) + (1 × (−1)) + (1 × 3)] + (1 × 2)] + (1 × 0)]
26 11 8 0
= 29 = 11 11 2
−7 8 13 6
270 MATRICES AND DETERMINANTS
⎛ 3⎞
In algebra, the commutative law of multiplication
states that a × b = b × a. For matrices, this law is 3.1 2.4 6.4
⎜ ⎟
only true in a few special cases, and in general A × B F = −1.6 3.8 −1.9 G = ⎝ 4⎠
5.3 3.4 −4.8 2
is not equal to B × A. 1
5
4 1 0
2 3 −2
H= −11 K = 0 1
Problem 9. If A = 1 0 and 5 J=
7 1 0
2 3
B = 0 1 show that A × B = B × A. Addition, subtraction and multiplication
In Problems 1 to 12, perform the matrix opera-
2 3 2 3 tion stated.
A×B= 1 0 × 0 1 ⎡⎛ 1
1 ⎞⎤
3 −
[(2 × 2) + (3 × 0)] [(2 × 3) + (3 × 1)] ⎢⎜ 2 3 ⎟⎥
= [(1 × 2) + (0 × 0)] [(1 × 3) + (0 × 1)] 1. A + B ⎣⎝ 1 2 ⎠⎦
−4 6
4 9 3 5
= 2 3 ⎡⎛ ⎞⎤
1
7 −1 6
2 3 2 3 ⎢⎜ 2⎟ ⎥
B×A= 0 1 × 1 0 ⎢⎜ 1 ⎟⎥
2. D + E ⎢⎜ 3 3 7⎟ ⎥
⎢⎜ ⎟⎥
⎣⎝ 3
2 ⎦
⎠
[(2 × 2) + (3 × 1)] [(2 × 3) + (3 × 0)]
= [(0 × 2) + (1 × 1)] [(0 × 3) + (1 × 0)] 4 7 −3
5
⎡⎛ 1
7 6 2 ⎞⎤
= 1 0 2 −1
⎢⎜ 2 3 ⎟⎥
3. A − B ⎣⎝ 2
4 9 7 6 3 ⎠⎦
Since 2 3 = 1 0 , then A × B = B × A −3 7
3 5
4.8 −7.73̇
4. A + B − C
Now try the following exercise. −6.83̇ 10.3
Exercise 108 Further problems on addition, 18.0 −1.0
5. 5A + 6B −22.0 31.4
subtraction and multiplication of matrices
In Problems 1 to 13, the matrices A to K are: 6. 2D + 3E − 4F
⎛ 1
⎟ 135
⎝ 3
3⎠ −52
10. D × J
−1 0 −85
5
THE THEORY OF MATRICES AND DETERMINANTS 271
⎡⎛ ⎞⎤
1
⎢⎜ 2
3 6
⎟⎥ (1 + j) j2
⎢⎜ 2 ⎟⎥ Problem 11. Evaluate − j3 (1 − j4)
11. E × K ⎢⎜ 12 − ⎟⎥
⎢⎜ 3⎟ ⎥
⎣⎝ 2 ⎠⎦
−
0 (1 + j) j2
5
− j3 (1 − j4) = (1 + j)(1 − j4) − ( j2)(− j3)
55.4 3.4 10.1
12. D × F −12.6 10.4 −20.4 = 1 − j4 + j − j2 4 + j2 6
−16.9 25.0 37.9 = 1 − j4 + j − (−4) + (−6)
13. Show that A⎡× C = C ×
A ⎤ since from Chapter 23, j2 = −1
−6.4 26.1 = 1 − j4 + j + 4 − 6
⎢A × C = 22.7 −56.9 ⎥
⎢ ⎥ = −1 − j3
⎢ −33.5 −53.1 ⎥
⎢C × A = ⎥
⎣ 23.1 −29.8 ⎦
5∠30◦ 2∠−60◦
Hence they are not equal Problem 12. Evaluate 3∠60◦ 4∠−90◦
2 3
showing that c = 1, i.e. c = and −2d = 1, i.e. Exercise 110 Further problems on the
3 2 inverse of 2 by 2 matrices
1
d =− 3 −1
2 1. Determine the inverse of −4
4 7
Since b = −2d, b = 1 and since a = − c, a = −2. ⎡⎛ ⎞⎤
3 7 1
⎢⎜ 17 17 ⎟⎥
1 2 a b ⎢⎜ ⎟⎥
Thus the inverse of matrix 3 4 is c d that ⎣⎝ 4 3 ⎠⎦
−2 1 17 17
is, 3 1 ⎛ ⎞
− 1 2
2 2 ⎜ 2 3⎟
There is, however, a quicker method of obtaining 2. Determine the inverse of ⎜
⎝ 1
⎟
3⎠
the inverse of a 2 by
2 matrix.
− −
p q ⎡⎛ 3 5 ⎞⎤
For any matrix r s the inverse may be 5 4
⎢⎜ 7 7 8 ⎟⎥
7 ⎟⎥
obtained by: ⎢⎜
⎣⎝ 2 3 ⎠⎦
(i) interchanging the positions of p and s, −4 −6
(ii) changing the signs of q and r, and 7 7
THE THEORY OF MATRICES AND DETERMINANTS 273
−1.3 7.4 The value of this determinant is the sum of the prod-
3. Determine the inverse of 2.5 −3.9 ucts of the elements and their cofactors, of any row
⎡ ⎤ or of any column. If the second row or second col-
0.290 0.551 umn is selected, the element 0 will make the product
⎣ 0.186 0.097 ⎦ of the element and its cofactor zero and reduce the
correct to 3 dec. places amount 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 the
25.6 The determinant of a 3 by 3 2 (i.e. the second row and the first column),
is cov-
4 −1
matrix ered up. Thus the cofactor of element 2 is −3 −2
(i) The minor of an element of a 3 by 3 matrix is i.e. −11. The sign of element 2 is minus, (see (ii)
the value of the 2 by 2 determinant obtained by above), hence the cofactor of element 2, (the signed-
covering up the row and column containing that minor)
+11. Similarly the minor of element 7 is
is
3 4
element.
1 2 3
1 −3 i.e. −13, and its cofactor is +13. Hence the
Thus for the matrix 4 5 6 the minor of value of the sum of the products of the elements and
7 8 9 their cofactors is 2 × 11 + 7 × 13, i.e.,
element 4 is obtained bycovering the row 3 4 −1
1 2 0 7 = 2(11) + 0 + 7(13) = 113
(4 5 6) and the column 4 , leaving the 2 by 1 −3 −2
7 The same result will be obtained whichever row or
2 3
2 determinant 8 9, i.e. the minor of element column is selected. For example, the third column
F
expansion
is
4 is (2 × 9) − (3 × 8) = −6. 2 0 3 4 3 4
(−1) 1 −3 − 7 1 −3 + (−2) 2 0
(ii) The sign of a minor depends on its posi-
tion within
the matrix, the sign pattern = 6 + 91 + 16 = 113, as obtained previously.
+ − +
being − + − . Thus the signed-minor
1 4 −3
+ − +
Problem 15. Evaluate −5 2 6
1 2 3 −1 −4 2
of element 4 in the matrix 4 5 6 is
7 8 9
2 3 1 4 −3
− 8 9 = −(−6) = 6.
Using the first row: −5 2 6
−1 −4 2
The signed-minor of an element is called the
2 6 −5 6 −5 2
cofactor of the element. = 1 −4 2 − 4 −1 2 + (−3) −1 −4
(iii) The value of a 3 by 3 determinant is the = (4 + 24) − 4(−10 + 6) − 3(20 + 2)
sum of the products of the elements and their
cofactors of any row or any column of the = 28 + 16 − 66 = −22
corresponding 3 by 3 matrix.
1 4 −3
There are thus six different ways of evaluating a 3×3 Using the second column: −5 2 6
determinant—and all should give the same value. −1 −4 2
−5 6 1 −3 1 −3
= −4 −1 2 + 2 −1
Problem 14. Find the value of
2 −(−4) −5 6
3 4 −1
= −4(−10 + 6) + 2(2 − 3) + 4(6 − 15)
2 0 7
1 −3 −2 = 16 − 2 − 36 = −22
274 MATRICES AND DETERMINANTS
8 −2 −10
Problem 16. Determine the value of
4. Evaluate 2 −3 −2 [−328]
6 3 8
j2 (1 + j) 3
(1 − j) 1 j 5. Calculate the determinant of
0 5
j4
3.1 2.4 6.4
−1.6 3.8 −1.9 [−242.83]
Using the first column, the value of the determinant is: 5.3 3.4 −4.8
j2 2 j
1 j (1 + j) 3 6. Evaluate (1 + j) 1 −3 [−2 − j]
(j2) j4 5 − (1 − j)
j4 5
5 −j4 0
(1 + j) 3 3∠60◦ j2 1
+ (0) 1
j 7. Evaluate 0 (1 + j) 2∠30◦
0 2 j5
= j2(5 − j2 4) − (1 − j)(5 + j5 − j12) + 0
26.94∠−139.52◦ or
= j2(9) − (1 − j)(5 − j7) (−20.49 − j17.49)
adjoint
Problem 17. Determine the inverse of the Inverse =
⎛ ⎞ determinant
3 4 −1
matrix ⎝2 0 7⎠ −17 9 15
The matrix of cofactors is 23 −13 −21
1 −3 −2 18 −10 −16
The transpose
of the matrix of
cofactors (i.e. the
adj A −17
The inverse of matrix A, A−1 = 23 18
|A| adjoint) is 9 −13 −10
15 −21 −16
The adjoint of A is found by:
1 5 −2
(i) obtaining the matrix of the cofactors of the The determinant of 3 −1 4
elements, and −3 6 −7
4 −2 5
21 11 28 −7 4 7
11 −5 −23 6 0 −4
−6 13 −8 1 21 11 28
or 11 −5 −23
113 113 −6 13 −8 2. Write down the transpose of
⎛ ⎞
3 6 21
Problem 18. Find the inverse of ⎝ 5 − 2 7⎠
3
−1 0 35 ⎡⎛ ⎞⎤
1 5 −2
3 5 −1
3 −1 4
−3 6 −7 ⎣⎝ 6 − 23 0⎠⎦
1 3
2 7 5
276 MATRICES AND DETERMINANTS
−16 14 −24
26
The solution of simultaneous
equations by matrices and
determinants
(ii) The matrix equation is
26.1 Solution of simultaneous
equations by matrices 3 5 x 7
4 −3 × y = 19
(a) The procedure for solving linear simultaneous
3 5
equations in two unknowns using matrices is: (iii) The inverse of matrix 4 −3 is
(i) write the equations in the form
a1 x + b1 y = c1
1 −3 −5
3 × (−3) − 5 × 4 −4 3
a2 x + b2 y = c2 ⎛3 ⎞
5
(ii) write the matrix equation corresponding to
these equations, ⎜
i.e. ⎝ 29 29 ⎟
⎠
F
4 −3
a b x c
i.e. a1 b1 × y = c1 29 29
2 2 2
a 1 b1 (iv) Multiplying each side of (ii) by (iii) and remem-
(iii) determine the inverse matrix of a b bering that A × A−1 = I, the unit matrix, gives:
2 2 ⎛3 5⎞
1 b2 −b1 1 0 x ⎜ 29 29 ⎟ 7
i.e.
a1 b2 − b1 a2 −a2 a1 0 1 y = ⎝ 4 −3 ⎠ × 19
(from Chapter 25) 29 29
⎛ 21 95 ⎞
(iv) multiply each side of (ii) by the inverse +
matrix, and x ⎜ 29 29 ⎟
Thus y = ⎝ ⎠
(v) solve for x and y by equating corresponding 28 57
elements. −
29 29
x 4
Problem 1. Use matrices to solve the simulta-
i.e. y = −1
neous equations: (v) By comparing corresponding elements:
3x + 5y − 7 = 0 (1) x=4 and y = −1
4x − 3y − 19 = 0 (2) Checking:
equation (1),
(i) Writing the equations in the a1 x+b1 y = c form 3 × 4 + 5 × (−1) − 7 = 0 = RHS
gives:
equation (2),
3x + 5y = 7
4 × 4 − 3 × (−1) − 19 = 0 = RHS
4x − 3y = 19
278 MATRICES AND DETERMINANTS
(b) The procedure for solving linear simulta- The adjoint of A is the transpose of the matrix of
neous equations in three unknowns using the cofactors of the elements (see Chapter 25).
matrices is: The matrix of cofactors is
(i) write the equations in the form 14 16 5
a1 x + b1 y + c1 z = d1 0 −5 5
7 −2 −5
a2 x + b2 y + c2 z = d2
a3 x + b3 y + c3 z = d3 and the transpose of this matrix gives
(ii) write the matrix equation corresponding 14 0 7
to these equations, i.e. adj A = 16 −5 −2
5 5 −5
a1 b1 c1 x d1
a2 b2 c2 × y = d2 The determinant of A, i.e. the sum of the prod-
a3 b3 c3 z d3 ucts of elements and their cofactors, using a first
row expansion is
(iii) determine the inverse matrix of
−3 4 2 4 2 −3
a1 b1 c1
1−2 −2 − 1 3 −2 + 1 3 −2
a2 b2 c2 (see Chapter 25)
a3 b3 c3
= (1 × 14) − (1 × (−16)) + (1 × 5) = 35
(iv) multiply each side of (ii) by the inverse
Hence the inverse of A,
matrix, and
(v) solve for x, y and z by equating the −1 1 14 0 7
A = 16 −5 −2
corresponding elements. 35 5 5 −5
(iv) Multiplying each side of (ii) by (iii), and
Problem 2. Use matrices to solve the simulta- remembering that A × A−1 = I, the unit matrix,
neous equations: gives
x+y+z−4=0 (1)
1 0 0 x
2x − 3y + 4z − 33 = 0 (2) 0 1 0 × y
3x − 2y − 2z − 2 = 0 (3) 0 0 1 z
1 14 0 7 4
(i) Writing the equations in the a1 x + b1 y + c1 z = = 16 −5 −2 × 33
35 5 5 −5 2
d1 form gives:
x+y+z =4 x 1
y =
2x − 3y + 4z = 33 z 35
3x − 2y − 2z = 2
(14 × 4) + (0 × 33) + (7 × 2)
(ii) The matrix equation is × (16 × 4) + ((−5) × 33) + ((−2) × 2)
(5 × 4) + (5 × 33) + ((−5) × 2)
1 1 1 x 4
2 −3 4 × y = 33
1 70
3 −2 −2 z 2 = −105
35 175
(iii) The inverse matrix of
2
1 1 1
A = 2 −3 4 = −3
3 −2 −2 5
Exercise 114 Further problems on solving 8. Kirchhoff’s laws are used to determine the
simultaneous equations using determinants current equations in an electrical network
and show that
In Problems 1 to 5 use determinants to solve
the simultaneous equations given. i1 + 8i2 + 3i3 = −31
1. 3x − 5y = −17.6 3i1 − 2i2 + i3 = −5
7y − 2x − 22 = 0 2i1 − 3i2 + 2i3 = 6
[x = −1.2, y = 2.8] Use determinants to find the values of i1 , i2
and i3 . [i1 = −5, i2 = −4, i3 = 2]
2. 2.3m − 4.4n = 6.84
8.5n − 6.7m = 1.23
9. The forces in three members of a framework
[m = −6.4, n = −4.9] are F1 , F2 and F3 . They are related by the
3. 3x + 4y + z = 10 simultaneous equations shown below.
2x − 3y + 5z + 9 = 0 1.4F1 + 2.8F2 + 2.8F3 = 5.6
x + 2y − z = 6 4.2F1 − 1.4F2 + 5.6F3 = 35.0
[x = 1, y = 2, z = −1] 4.2F1 + 2.8F2 − 1.4F3 = −5.6
THE SOLUTION OF SIMULTANEOUS EQUATIONS BY MATRICES AND DETERMINANTS 283