5 Determinants
UNIT 1 DETERMINANTS
Structure 1.0 Introduction 1.1 Objectives1.2 Determinants of Order 2 and 3 1.3 Determinants of Order 3
1.4 Properties of Determinants 1.5 Application of Determinants 1.6 Answers to Check Your Progress 1.7 Summary
1.0 INTRODUCTION
In this unit, we shall learn about determinants. Determinant is a square array of numbers symbolizing the sum of certain products of these numbers. Many complicated expressions can be easily handled, if they are expressed as ‘determinants’. A determinant of order n has n rows and n columns. In this unit, we shall study determinants of order 2 and 3 only. We shall also study many properties of determinants which help in evaluation of determinants.
Determinants usually arise in connection with linear equations. For example, if the equations a1x + b1 = 0, and a2x + b2 = 0 are satisfied by the same value of x,
then a1b2 – a2 b1 = 0. The expression a1b2 a2b1 is a called determinant of
second order, and is denoted by
There are many application of determinants. For example, we may use determinants to solve a system of linear equations by a method known as Cramer’s rule that we shall discuss in coordinate geometry. For example, in finding are of triangle whose three vertices are given.
1.1 OBJECTIVES
After studying this unit, you should be able to : define the term determinant;
evaluate determinants of order 2 and 3;
use the properties of determinants for evaluation of determinants; use determinants to find area of a triangle;
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Algebra - I
1.2
DETERMINANTS OF ORDER 2 AND 3
We begin by defining the value of determinant of order 2.
Definition : A determinant of order 2 is written as where a,b, c, d are complex numbers. It denotes the complex number ad – bc. In other words,
Example 1 : Compute the following determinants :
(a) (b) (c) (d) Solutions : (a) = 18 − (–10) = 28 (b) = – =0 (c) = + ( (a + ib) (a − ib) = )
7 Determinants
1.3
DETERMINANTS OF ORDER 3
Consider the system of Linear Equations : x + y + z = ……….. (1) x + y + z = ..……….. (2) x + y + z = ……….. (3)
Where aij C ( 1 ≤ i, j ≤ 3) and , , , C Eliminating x and y from these
equation we obtain
We can get the value of z if the expression
– – –
The expression on the L.H.S. is denoted by
and is called a determinant of order 3, it has 3 rows, 3 columns and is a complex number.
Definition : A determinant of order 3 is written as
where aij C(1 ≤ i, j ≤ 3).
It denotes the complex number +
Note that we can write
= +
= – –
8
Algebra - I Where is written in the last form, we say that it has been expanded along the first row. Similarly, the expansion of along the second row is,
= + –
and the expansion of along the third row is,
= + −
We now define a determinant of order 1.
Definition : Let a A determinant of order 1 is denoted by |a| and its value is a.
Example 2 : Evaluate the following determinants by expanding along the first
row. (a) (b) (c) (d) Solutions: (a) = 2 5 +( 3) = 2(–2 = 2( = 64 + 35 – = 2 0 +1 (c) = x y + z = x (8 12) – y (6 6) + z(4 6) = 6x 2z (d) = 1 a + bc = a – a = a – = ab (b a) + bc (c b) + ca (a c)
9 Determinants
Check Your Progress – 1
1. Compute the following determinants : (a) (b)
4. Evaluate the following determinants :
(a) (b)
5. Show that = abc + 2fgh – a – b – c
1.4
PROPERTIES OF DETERMINANTS
Before studying some properties of determinants, we first introduce the concept of minors and cofactors in evaluating determinants.
Minors and Cofactors
Definition : If is a determinant, then the minor Mij of the element aij is the
determinant obtained by deleting ith row and jth column of . For instance, if
= then
10 Algebra - I M32 = Recall that = – = –
Similarly, the expansion of along second and third rows can be written as = +
and = –
respectively.
Definition : The cofactor Cij of the element aij in the determinant is defined to
be (–1)i+j Mij, where Mij is the minor of the elementaij.
That is, Cij = (–1)i+j Mij
Note that, if = then
= +
= +
= +
We can similarly write expansion of along the three columns :
= +
= +
= +
Thus, the sum of the elements of any row or column of multiplied by their corresponding cofactors is equal to .
Example 3 : Write down the minor and cofactors of each element of the
determinant
Solution: Hence, =
M11 = |5| = 5 M12 = |2| = 2
11 Determinants C11 + (–1)1+1 M11 = (–1)25 = 5 C12 + (–1)1+2 M12 = –2 C21 + (–1)2+1 M21 = (–1)3(–1) = 1 C22 + (–1)2+2 M22 = (–1)4(3) = 3 Properties of Determinants
The properties of determinants that we will introduce in this section will help us to simplify their evaluation.
1. Reflection Property
The determinant remains unaltered if its rows are changed into columns and the columns into rows.
2. All Zero Property
If all the elements of a row(column) are zero. Then the determinant is zero. 3. Proportionality (Repetition) Property
If the elements of a row(column) are proportional (identical) to the element of the some other row (column), then the determinant is zero.
4. Switching Property
The interchange of any two rows (columns) of the determinant changes its sign.
5. Scalar Multiple Property
If all the elements of a row (column) of a determinant are multiplied by a non-zero constant, then the determinant gets multiplied by the same constant. 6. Sum Property
= +
7. Property of Invariance
=
This is, a determinant remains unaltered by adding to a row(column) k times some different row (column).
8. Triangle Property
If all the elements of a determinant above or below the main diagonal consists of zerox, then the determinant is equal to the product of diagonal elements.
12
Algebra - I That is,
= =
Note that from now onwards we shall denote the ith row of a determinant by Ri and its ith column by Ci.
Example 4 : Evaluate the determinant
Solution : Applying R3 → R3 – R2 , and R2 → R2 – R1 , we obtain
Applying R2 → R2 R2 , we obtain
Expanding along C1, we get
( 1) 3+1(1) = 39 6 = 45.
Example 5 : Show that
=(b a) (c a)(c b)
Solution : By applying R2 → R2 – R1 , and R3 → R3 – R1 we get,
=
Taking (b–a) common from R2 and (c–a) common from R3, we get
(b – a) (c – a)
Expanding along C1, we get
(b – a) (c – a)
= (b – a) (c –a)[(c + a) – (b + a)] = (b – a)(c – a)(c – b)
13 Determinants
Example 6 : Evaluate the determinant
where ω is a cube root of unity.
Solution :
1 + +
= (By C1 → C1 C2 + C3)
= ( = 0)
= 0 [C1 consists of all zero entries].
Example 7 : Show that
= 2
Solution : Denote the determinant on the L.H.S. by . Then applying
C1 → C1 + C2 + C3 we get
=
Taking 2 common from C1 and applying C2 → C2 – C1, and C3 → C3 – C1, we
get = 2 – – –
Applying C1 → C2 + C2 C3 and taking (–1) common from bothC2 and C2, we
get
14
Algebra - I Example 8 Show that
= = ( 2 Solution : = ( By applying C1 → C1 + C2 C3) = = (By applying R2→ R2 R1, R3→ R3 R1) = (Expanding along C1)
= (taking (1– a) common from
C1 and C2)
=
= (
Example 9 : Show that
= = 4
Solution : Taking a, b, and c common four C1, C2 and C3 respectively, we get
= abc
=
Applying C1 → C1 + C2 and C1 → C2 + C3 , we get
=
Expanding along C1, we get =
Taking a, b and c common from R1, R2, R3 respectively, we get.
15 Determinants
Example 10 : Show that
=
Solution : We shall first change the form of this determinant by multiplying
R1, R2 and R3 by a, b and c respectively.
Then
Taking a, b and c common from C1, C2 andC3 respectively, we get
Taking 2 common from R1 and applying R1 → R2 – R1 and R3 → R3 – R1 , we
get
2
Applying R1 → R1 +R2 + R3 we get
2
Expanding the determinant along R1 we get
2 + 2
= 2 ) +2
= 4
Check Your Progress – 2
1. Show that = ( b– a)(c – a)(c – b) (a + b + c)
2. Show that
16
Algebra - I 3. Show that
= 4 abc
4. Show that
=
abc (1
+ +
)
1.5 APPLICATION OF DETERMINANTS
We first study application of determinants in finding area of a triangle.
Area of Triangle
We begin by recalling that the area of the triangle with vertices A (x1 y1),B (x2 y2),
and C(x3 y3), is given by the expression
| x
1(y
2– y
3) + x
2(y
3– y
1) + x
3(y
1–
y
2) |
The expression within the modulus sign is nothing but the determinant
Thus, the area of triangle with vertices A( , ), B( , ), and C( , ) is given by
Corollary : The three points A( lie on a straight
line if and only if = 0
Example 11 : Using determinants, find the area of the triangle whose vertices are
(a) A(1, 4) , B(2,3) and C(–5,–3) (b) A(–2,4), B(2,–6) and C(5,4)
Solution :
17 Determinants Area of ABC = | = |70| = square units
Example 12 : Show that the points (a, b+c), (b, c+a) and (c, a + b) are collinear.
denote the area of the triangle formed by the given points.
1 2 1 1 | 2 1 2 4 0 | 2 5 6 4 0 k k k k k 2 2 1 1 1 1 (using C C + C ) 2 1 a a b c b a c a c a a b
= 0. (C1 and C2 are proportional)
the given points are collinear.
Cramer’s Rule for Solving System of Linear Equation’s
Consider a system of 3 linear equations in 2 unknowns : + + =
+ + = ……(1) + + =
A Solution of this system is a set of values of x, y, z which make each of three equations true. A system of equations that has one or more solutions is called
consistent. A system of equation that has no solution is called inconsistent.
If = 0 in (1), the system is said to be homogeneous system of
equations. If atleast one of the system is said to be
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Algebra - I
Let =
Consider x Using the scalar multiple property we can absorb x in the first column of , that is,
x
Applying C1→ C1 yC2+ z C3, we get
x = = =
Note that the determinant can be obtained from
on the R.H.S. of the system of linear equations that is, by .
If ≠ 0, then x = x. Similarly, we can show that if ≠ 0, then y = y and Z = z, when
Where
y = and z = .
This method of solving a system of linear equation is known as Cramer’s Rule. It must be noted that if = 0 and one of x= y= z= 0, then the system has
infinite number of solutions and if = 0 and one of x, y, z is non-zero, the
system has no solution i.e., it is inconsistent.
Example 13 : Solve the following system of linear equation using Cramer’s rule
x + 2 y + 3 z = 6 2x + 4 y + z = 7 3x + 2 y + 9z = 14
Solution : We first evaluate , where
19 Determinants Applying R2→ R2 R1, and R3→ R3 3R1, we get
= = – 20 (expanding along C1)
As ≠ 0, the given system of linear equations has a unique solution. Next we evaluate x, y and z We have
=
Applying R2→ R2 2R1, and R3→ R3 R1, we get
= = – 20 (expanding along C2)
=
[Applying R2→ R2 2R1, and R3→ R3 3R1]
= – 20 [expanding along C1]
and = = Applying R2→ R2 2R1, and R3→ R3 3R1]
= – 20 [expanding along C1]
Applying Cramer’s rule, we get
Remark : If in (1), then
then the only solution of the system of linear homogeneous equations. = 0
= 0
= 0 ……(2)
is x = 0, y = 0, z = 0. This is called the trivial solution of the system of equation (2). If the system (2) has infinite number of solutions.
20
Algebra - I Example 14 : Solve the system of linear homogeneous equation : 2x – y + 3z = 0,
x + 5y – 7z = 0, x – 6y + 10 z = 0
Solution : We first evaluate
= – 20 (expanding along C1)
= 0
(because R1 and R2 are proportional)
Therefore, the given system of linear homogeneous equations has an infinite number of solutions. Let us find these solutions. We can rewrite the first two equations as :
2x – y = 3z
x + 5y = 7z …… (1) Now, we have = = 10 – (–1) = 11.
As the system of equation in (1) has a unique solution. We have = = –15z –(–7z) = – 8z and
= = 14z – (–3z) = 17z
By Cramer’s Rule, x = = =
z and y
= = =z.
We now check that this solution satisfies the last equation. We have x 6y + 10z = = 6
21 Determinants
Check Your Progress – 3
1. Using determinants find the area of the triangle whose vertices are : (a) (1,2), ( 2,3) and ( 3, 4)
(b) ( 3, 5), (3, 6) and (7,2)
2. Using determinants show that ( 1,1), ( 3, 2) and ( 5, 5) are collinear. 3. Find the area of the triangle with vertices at ( k + 1, 2k), (k, 2 2k) and
( 4 k, 6 2k). For what values of k these points are collinear ? 4. Solve the following system of linear equations using Cramer’s rule.
(a) x + 2 y – z = 1, 3x + 8y + 2 z = 28, 4x + 9y + z = 14 (b) x + y = 0, y + z = 1, z + x = 3
5. Solve the following system of homogeneous linear equations : 2x – y + z = 0, 3x + 2y – z = 0, x + 4y +3z = 0.
1.6
ANSWERS TO CHECK YOUR PROGRESS
Check Your Progress – 1
(b) = ab – (c + id) (c id) = ab – c d2 (c) = (n + 1) (n 1) – n2 = n2 1 – n2 = 1 = (a (c = ad – ad – ad = ad ( ) – bc ( ) = (ad – bc) ( )
22 Algebra - I 4. (a) = 2 – (– 1) = 2 (0 – 1) – (8 – 1) + 5(4 – 0) = – 2+ 7 + 20 = 25 (b) = 5 – 3 = 5 (0 – 2) – 3(6 – 1) + 8(4 – 0) = – 10 – 15 + 32 = 7 5. = a + g = a(bc – = abc – a = abc + 2fgh – a
Check Your Progress 2
1. = (Applying C2→ C2 C1 and C3→ C3 –C1) = (b– a) (c– a) = (b– a)(c– a) = (b– a) (c– a)( – – = (b– a) (c– a)[( = (b– a)(c– a) [(c– b)(c+b)+ (c–b)a]
(taking (b – a common form
23 Determinants 2. Taking x, y and z common from C1, C2 andC3 respectively, we get
(Applying C2→ C2 C1, and C3→ C3 –C1) we get
Taking (y – x) common from C2 and (z– x) from C2, we get and (z– x) from C3, we get
= xyz (y – x) (z – x)
Expanding along R1, we get
= xyz (y – x)(z – x) = xyz (y – x) (z–x) (z+ x– y– ) = xyz (y – x) (z – x) (z – y) 3. Let = Applying R1→ R1 R2– R3, we get = = – 2
(by the scalar multiple property)
Applying R2→ R2 R1, and , R3→ R3 R1, we get
=
Expanding along the first column, we get
= – 2
24
Algebra - I 4. We take a, b and c common from C1, C2 and C3 respectively, to obtain
Applying C1→ C1 + C2+ C3, we get
Expanding along C1, we get
Check Your Progress - 3
1. (a) =
– 1
= 1 (By applying R2→ R2 – R1 and R3→ R3 – R1
= (Expanding along C3)
25 Determinants =
– –
= (By applying R2→ R2– R1 and R3→ R3– R1 )
=
= × 92 =46 square units
2. =
= (By applying R2→ R2–R1 and R3→ R3– R1 )
= 12–12 = 0
the given points are collinear.
1 2 1 1 3. Area of triangle | 2 2 1 | 2 4 6 2 1 k k k k k k 1 2 1 1 | 2 1 2 4 0 | 2 5 6 4 0 k k k k k 1 1 = |4k2 + 2k –2| These points are collinear if
i.e., if |4k2 + 2k –2| = 0 i.e., if 2(2k – 1)(k + 1) = 0 i.e., if k = –1,
26
Algebra - I 4. (a) We first evaluate
= [Applying C2→ C2 – 2C1 and C3→ C3 + C1 ]
= 10 – 5 = 5 (expanding along R1)
As 0, the given system of equation has a unique solution. We shall now evaluate , . We have
= = = – (expanding along C3) = –130 = = = – (expanding along R1) = 65 = = (Applying C2→ C2 – 2C1 and C3→ C3 + C1) = 5 (expanding along R1)
Hence by Cramer’s Rule
(b) Here,
=
= [Applying C2→ C2 – C1]
= 2 (Expanding along R1)
Since , the given system has unique solution,
(By applying R2→ R2 – 2R1 and R3→ R3 + R1 we get) (By applying C2→ C2 –2C1 and C3→ C3 + C1 ) a n d C 3 → C 3 + C 1
27 Determinants
Now, = = 2
= = – 2
= = 4
Hence by Cramer’s Rule
5. Here, =
= (Applying R2→ R2 + R1 and R3→ R3 – 3R1)
= 35 + 5 (expanding along C3)
= 40
Since 0, the given system has a unique solution, and the trivial solution x = y = z = 0 is the only solution. In fact,
x = y= z = 0.
= = – (1 – 3) = 2
= = (1 – 3) =– 2
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Algebra - I
1.7 SUMMARY
In this unit, first of all, the definitions and the notations for determinants of order 2 and 3 are given. In sections 1.2 and 1.3 respectively, a number of examples for finding the value of a determinant, are included. Next, properties of determinants are stated. In section 1.4, a number of examples illustrate how evaluation of a determinant can be simplified using these properties. Finally, in section 1.5, applications of determinants in finding areas of triangles and in solving system of linear equations are explained.
Answers/Solutions to questions/problems/exercises given in various sections of the unit are available in section 1.6.
29 Matrices - I
UNIT 2 MATRICES - I
Structure 2.0 Introduction 2.1 Objectives 2.2 Matrices 2.3 Operation on Matrices 2.4 Invertible Matrices2.5 Systems of Linear Equations 2.6 Answers to Check Your Progress 2.7 Summary
2.0
INTRODUCTION
In this Unit, we shall learn about Matrices. Matrices play central role in
mathematics in general, and algebra in particular. A matrix is a rectangular array of numbers. There are many situations in mathematics and science which deal with rectangular arrays of numbers. For example, the following table gives vitamin contents of three food items in conveniently chosen units.
Vitamin A Vitamin C Vitamin D
Food I 0.4 0.5 0.1
Food II 0.3 0.2 0.5
Food III 0.2 0.5 0
The above information can be expressed as a rectangular array having three rows and three columns.
0.4 0.5 0.1
0.3 0.2 0.5
0.2 0.5 0
The above arrangement of numbers is a matrix of order 3 3. Matrices have become an important an powerful tool in mathematics and have found
applications to a very large number of disciplines such as Economics, Physics, Chemistry and Engineering.
In this Unit, we shall see how Matrices can be combined thought the arithematic operations of addition, subtraction, and multiplication. The use of Matrices in solving a system of linear equations will also be studied. In Unit 1 we have already studied determinant. It must be noted that a matrix is an arrangement of numbers whereas determinant is number itself. However, we can associate a determinant to every square matrix i.e., to a matrix in which number of rows is equal to the number of columns.
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Algebra - I
2.1
OBJECTIVES
After studying this Unit, you should be able to : define the term matrix;
add two or more Matrices; multiply a matrix by a scalar; multiply two Matrices;
find the inverse of a square matrix (if it exists); and
use the inverse of a square matrix in solving a system of linear equations.
2.2
MATRICES
We define a matrix as follows :
Def : A m n matrix A is a rectangular array of m n real (or complex numbers) arranged in m horizontal rows an n vertical columns :
a11 a12 ……… a1j ……… a1n a21 a22 ……… a2j ……… a2n : : : : : : : : A = ai1 ai2 ……… aij ……… ain ith row : : : …(1) am1 am2 ……… amj ……… amn jth column
As it is clear from the above definition, the ith row of A is (aij ai2 … ain)
(1 ≤ i ≤ m) and the jth column is a1j
a2j
: (1 ≤ j ≤ n) :
amj
We also note that each element aij of the matrix has two indices : the row
index i and the column index j. aij is called the (i,j )th element of the matrix.
For convenience, the Matrices will henceforward be denoted by capital letters and the elements (also called entries) will be denoted by the corresponding lower case letters.
31 Matrices - I The matrix in (1) is often written in one of the following forms :
A = [aij]; A = (aij), A = (aij)m n or A = (aij)m n
With i = 1, 2, ………….., m and j = 1, 2, …………, n
The dimension or order of a matrix A is determined by the number of rows and columns of the matrix. If a matrix A has m rows and n columns we denote its dimension or order by m n read “m by n”.
For example, A = is a 2 2 matrix and B = is a 2 3 order matrix.
Note a that an m n matrix has mn elements.
Type of Matrices
1. Square Matrix : A square matrix is one in which the number of rows is equal to the number of columns. For instance,
A = , B = , C =
are square Matrices.
If a square matrix has n rows (and thus n columns), then A is said to be a square matrix of order n.
2. Diagonal Matrix : A square matrix A[ ] n n for which
diagonal matrix. For instance, 10 0 0 0 4 0 0 0 0 0 0 D= 0 2 0 and E = 0 0 3 0 0 0 6 0 0 0 5
are diagonal Matrices.
If A = [ ] n n is a square matrix of order n, then the numbers a11, a22, ,
… ann are called diagonal elements, and are said to form the main
diagonal of A. Thus, a square matrix for which every term off the main
32
Algebra - I 3. Scalar Matrix : A diagonal matrix A = [a
ij ] n n for which all the terms
on the main diagonal are equal, that is aij = k for i= j and aij = 0 for i ≠ j
is called a scalar matrix. For instance
H =
are scalar Matrices.
4. Unit or Identity Matrix : A square matrix A = [aij] n n is said to be the unit matrix or identity matrix if
aij = 0 if i ≠j 1 if i = j
Note that a unit matrix is a scalar matrix with is on the main diagonal. We denote the unit matrix having n rows (and n columns) by In. For example,
5. Row Matrix or Column Matrix : A matrix with just one row of
elements is called a row matrix or row vector. While a matrix with just one column of elements is called a column matrix or column vector.
For instance, A = [2 5 −15] is a row matrix whereas B is a column matrix.
6. Zero matrix or Null matrix : An m n matrix is called a zero matrix or null matrix if each of its elements is zero.
We usually denote the zero matrix by Om n
and are example of zero matrices.
Equality of Matrices
Let A = [aij]m×n and B =[ bij]r×s be two Matrices. We say that A and B
are equals if
33 Matrices - I 2. n = s, i.e., the number of columns in A equals the number of
columns in B.
3. aij = bij for I = 1, 2, ……. m and j = 1, 2, …….., n.
We then write A = B, read as “matrix A is equal to B” In other words, two Matrices are equal if their order are equal and their corresponding elements are equal.
Solution: Both the Matrices are of order 2 1. Therefore, by the definition
of equality of two Matrices, we have x + 2 = 4 − y and 3y − 7 = x − 3. That is, x + y = 2 and x − 3y = − 4. Solving these two equations. We get x = 1/2 and y = 3/2. We can check this solution by substitution in A and B.
Transpose of a Matrix
Definition: Let A = [aij] m n, be a matrix. The transpose of A, denoted
by A´, is the matrix A´ = [aij] m n, where bij = aji for each i and j.
The transpose of a matrix A is by definition, that matrix which is obtained from A by interchanging its rows and columns.
So, if A = , then
its transpose is the matrix
A´ = .
Symmetric and Skew Symmetric Matrices
Definition : A square matrix A = [aij] n n, is said to be symmetric if
34
Algebra - I
For example A = is symmetric and B = is a skew – symmetric matrix.
Check Your Progress 1
1. Construct a 2 2 matrix A = [aij] 2 2 where elements are given by
(a) aij = (i + 2j)2 (b) aij = (i −j)2
2. Find x, y when = 3. a, b, c and d such that
.
4. Find the transpose of following Matrices and find whether the matrix is symmetric or skew symmetric.
(a) A=
(c)
2.3 OPERATION ON MATRICES
Addition
Let A = [aij]m×n and B= [bij]r×s be two Matrices. We say that A and B
are comparable for addition if m = r and n = s. That is, A and B are comparable for addition if they have same order.
We define addition of Matrices as follows :
Definition : Let A =[ aij]m×n and B = [bij ]m×n be two Matrices. The sum
of A and B is the m n matrix C =[cij] such that Cij = aij + bij (1
That is, C is obtained by adding the corresponding elements of A an d B. We usually denote C by A + B.
Note that
35 Matrices - I
For example, if A = and B = , then
A + B = +
=
=
It must be noted that Matrices of different orders cannot be added. For instance,
A = Cannot be added.
The following properties of matrix addition can easily be verified. 1. Matrix addition is commutative. That is, if A and B are two m n
matrices, then
A + B = B + A.
2. Matrix addition is associative. That is, if A, B and C are three m n matrices, then
(A + B ) + C = A + (B + C)
3. If A [aij] = is an m×n matrix, then
A + O m×n = O m×n + A = A,
. where O m×n is the m×n null matrix.
4. If A is an m n matrix, then we can find an m n matrix B such that A + B = B + A = O m×n
The matrix B in above property is called „additive inverse‟ or „negative‟ of A and is denoted by − A.
Infact, if A = [aij] m×n then –A = [–aij] m×n
Thus, property 4 can be written as A + ( −A) = ( − A) + A = O m×n
We can now define difference of two Matrices.
Definition : Let A = [aij] m×n and B = [bij] m×n two matrices. We define the difference A − B to be the m n matrix A + (−B).
36
Algebra - I
For example, if A = and B =
then A – B = =
Scalar Multiplication
Definition : Let A =[ ] m×n be a matrix and let K be a complex number.
The scalar multiplication KA of the matrix A and the number K (called the scalar) is the m×n matrix KA [kaij] m×n
For example, let A =
If K = 4, then kA = 4A =
Note that if k = −1, then (−1)A = − A.
This is one of the properties of scalar multiplication. We list some of these properties without proof.
Properties of Scalar Multiplication
1. Let A = [ ] m×n be a matrix and let k1 and k2 be two scalars. Then
(i) (k1 + k2 ) A = k1A + k2 A, and
(ii) k1 (k2 A ) = (k1 k2)A.
2. Let A = [ ] m×n and B = [ ] m×n be two matrices and let k be a scalar.
Then
k (A + B ) = kA + k B.
Multiplication of two Matrices
Let A = [aij] m×n and B = [bij] r×s be two matrices. We say that A and B are comparable for the product AB if n = r, that is, if the number of columns
37 Matrices - I
Definition : Let A = [ ] m×n and B = [ ] n×p be two matrices. Their
product AB is the matrix C = [ ] m×p such that = ai1 bij + ai2 b2j + ai3 b3j +………..+ ainbnj for i ≤ i ≤ m, 1 ≤ j ≤ p. Note that the order of AB is m p.
Example 2 : Let A = and B =
Obtain the product AB.
Solution : Since A is of order 2 and B is or order 2, therefore, the
product AB is defined. Order of AB is 2
AB =
=
=
=
Properties of Matrix Multiplication
Some of the properties satisfied by matrix multiplication are stated below without proof.
1. (Associative Law) : If A = [ ] m×n , B = [ ] n×p and C = [ ] p×q are
three matrices, then
(AB) C = A (BC).
2. (Distributive Law): If A = [ ] m×n , B = [ ] n×p and C = [ ] n×p are
three matrices, then
A (B + C) = AB + AC.
3. If A = [ ] m×n and B = [ ] n×p are two matrices, and k is a complex
number, then
(kA ) B = A(kB) = k(AB).
4. If A = [ ] m×n is an m n matrix, then
Im A = AIn = A,
Where Im and In are unit matrices of order m and n respectively.
Example 3: Let A = and B = [ ]
38
Algebra - I Solution : Since A is 3×1 matrix and B is a 1 3 matrix, therefore, AB is defined and its order is 3 3.
If the number of columns of A is equal to the number of rows of B.
=
Also, BA is defined and a is 1× 1 matrix
BA =
This example illustrates that the matrix multiplication is not
commutative. Infact, it may happen that the product AB is defined but BA is not, as in the following case :
A = and B =
We now point out two more matrix properties which run counter to our experience to number systems.
1. It is possible that for two non-zero matrices and A and B, the product AB is a zero matrix.
2. It is possible that for a non-zero matrix A, and two unequal matrices B and C, we have, AB = AC. That is AB = AC, A ≠ 0 may not imply B = C. In other words, cancellation during multiplication does not hold.
These properties can be seen in the following example.
39 Matrices - I Solution : We have AB= and = O 2 2 AC= = O 2 2
Therefore, AB = AC. We see however, that A ≠ O2 2 and
B ≠ C. Thus, cancellation during multiplication does not hold.
Exponent of a Square Matrix
We now introduce the notion of the exponent of a square matrix. To begin with, we define Am for any square matrix and for any positive integer m.
Let A be a square matrix and m a positive integer. We define. Am = AAA …….A
m times
More formally, the two equations A' = A and Am +1 = Am A define Am recursively by defining it first for m = 1 and then m+1 after it has been defined for m, for all m ≥ 1.
We also define A° = In, where A is a non−zero square matrix of order n.
The usual rules of exponent‟s namely
Am An = Am+n and (Am)n = Amn do hold for matrices if m and n are non-negative integers.
Example 5 : Let A = and f(x) = – 4x + 7. Show that f(A) = O2 2. Use this result to find A5.
Solution : First, we note that by f(A) we mean A2 −4A + 7I2. That is, we
replace x by A and multiply the constant term by I, the unit matrix. Therefore,
f(A) = A2 – 4A + 7I2
=
40
Algebra - I
= – +
=
= = O2 2.
Hence, A2 – 4A – 7 I2, from which we get
A3 = A2A = (4A – 7I2)A
= 4A2 – 7I2A = 4(4A – 7I2 ) − 7A [ I2 A = A]
= 9A − 28 I2
A5 = A2 A3 = (4A − 7I2) (9A − 28I2)
= 36A2 –63I2A –112AI2+ 196 I2 I2 (Distributive Law)
= 36 (4A − 7I2) − 63A − 112A +196 I2
= 144A – 252 I2 −175 A +196 I2
= −31A −56 I2
= –31
= –
=
Check Your Progress – 2
1. If P = , find matrix R such that 5P + 3Q + 2R is a null matrix. 2. If A = , B = and 3. If A = and B = 4. Let f (x) = A = . 5. If A = , show that
6. If A and B are square matrices of the same order, explain why the following may not hold good in general.
(a) (A + B ) (A– B) = (b)
41 Matrices - I
2.4 INVERTIBLE MATRICES
In this section, we restrict our attention to square matrices and formulate the notion of multiplicative inverse of a matrix.
Definition : An n n matrix A is said to be invertible or non-singular if
these exists an n n matrix or non singular if there exists an n n matrix such that AB = BA =
The matrix B is called an inverse of A. If there exists no such matrix B, then A is called non-invertible or singular.
Example 6 : Find whether A is invertible or not where
(a) A = (b) A = .
Solution : (a) We are asked whether we can find a matrix B = such
that AB = I2 . What we require is
= AB =
This would imply that c = 0, d = 1, a = 1 and b = –1, so that matrix
B =
does satisfy AB = I2. Moreover, it also satisfies the equation
BA = I2. This can be verified as follows :
BA = = = = I2.
This implies that A is invertible and B = is an inverse of A.
(b) Again we ask whether we can find a matrix B = such that AB = I2 . What is required in this case is
= AB = .
This would imply that a=1, b = 0 and the absurdity that 1=0. So no such B exists for this particular A. Hence, A is non invertible. We will not show that if A is invertible, then B in the above definition is unique.
42
Algebra - I Theorem : If a matrix has an inverse, then inverse is unique.
Proof : Let B and C be inverses of a matrix A. Then by definition.
AB = BA = In ..(1)
and AC = CA = In .. (2)
Now, B = BIn [property of identify matrix]
= B(AC) [(using 2)] = (BA)C [associative law] = InC [ using (1)]
= C. [property of identity matrix] This means that we will always get the same inverse irrespective of the method employed. We will write the inverse of A, if it exists, as A−1. Thus
AA−1 + A−1 A = In. Definition For example, if A= then A11= (–1)1+1 = 5 A12= (–1)1+2 = 14 and A13= (–1)1+3 = 3. Similarly, A21 = 3, A22= 2 A23 = –7, A31 = 10, A32 = 2 and A33 = 6.
Thus, the cofactor matrix of A is given by C = .
Definition
Let A = (aij)n×n be a square matrix of dimension n×n. The cofactors matrix of
A is defined to be the matrix C = (aij)n×n where Aij denotes the cofactor of the
element aij in the matrix A.
The adjoint of square matrix A = ( ) n×n is defined to be the transpose of the
43 Matrices - I
For example A = , then adj A =
The following theorem will enable us to calculate the inverse of a square matrix. We state the theorem (without proof) for 3×3 matrices only, but it is true for all square matrices of order n× n, where n ≥ 2.
Theorem : If A is a square matrix of order 3×3, then
A(adj)A = (adjA) A = |A|I3.
In view of this theorem, we note that if |A| ≠ 0, then
A = A = I3.
Since, the inverse of a square matrix is unique, we see that if |A| ≠ 0, then A acts as the inverse of A. That is,
Also, a square matrix is invertible (non-singular) if and only if |A| ≠ 0.
Example 7 : Find the inverse of A =
Solution :
We have A11 = (–1)1+1 |4| = 4 and A12 = ( 1)1+2 |2| = 2.
We know that |A| = a11A11 + a12A12 = (–3)(4) + 5(–2) = –22.
Since |A| ≠ 0 the matrix A is invertible, Also,
A21 = (–1)2+1 |5| = –5 and A22 = (–1)2+2 |–3| = –3. Therefore,
adj A = =
Hence A−1 = adj A = =
Example 8 : If A = and B =
verify that (AB)–1 = B–1A–1. = (adjA)
44
Algebra - I Solution : Since |A| = 8 A is invertible.
Similarly, |B| = 20 – 10 = 10 0, B is also invertible. Let Aij denote the cofactor of aij – the (i,j)th element of A. Then
A11 = 0, A12= –4, A21 = –2 and A22 = 3.
Similarly, if Bij is cofactor of (i,j)th element of B, then
B11 = 5, B12= –2, B21 = –5 and B22 = 4
adj A = and adj B =
A–1 = adj A = = and B–1 = adj B = = Let C = AB = = = We have C11 = 20, C12= –16, C21 = –25 and C22 = 16 Also, |C| = 80 C is invertible. Also, adj C = Hence, B–1 A–1 = = = C−1 = (AB) −1
Example 9 : Find the inverse of A =
45 Matrices - I
Solution : Evaluating the cofactors of the elements in the first row of A, we get
A11 = (−1)1+1 = 2, A12 = (−1)1+2 = 3,
and A13 = (–1)1+3 = ,
|A| = a11 A11+ a12 A12+ a13 A13
= (1)(2) + (2)(−3) + (5)(5) = 21 Since |A| ≠ 0, A is invertible. Also,
A21 = (–1)2+1 A22 = (–1)2+2 A23 = (–1)2+3 A31 = (–1)3+1 A32 = (–1)3+2 A33 = (–1)3+3 adj A = = A–1 = adjA = =
To verify that this is the inverse of A, we have
A–1A =
46
Algebra - I Check Your Progress – 3
1. Find the adjoint of each of the following Matrices :
(i) (ii)
2. For A = , verify that
A (adj A) = (adj.A) A = |A| I3.
3. Find the inverse of A =
4. Let A = and B = , Verify that (AB) –1= B–1A–1.
5. If A = , show that A2 = A–1.
What is Adj A ?
6. Let A = Prove that A2 – 4A – 5I3 = 0.
Hence, obtain A–1
7. Find the condition under which
A = is invertible. Also obtain the inverse of A.
2.5 SYSTEMS OF LINEAR EQUATIONS
We can use matrices to solve a system of linear equations. Let us consider the following m linear equations in n unknowns :
a11x1 + a12x2 +……….a1nxn = b1 a21x1 + a22x2 +……….a2nxn = b2 . . . . (1) am1x1 + am2x2 +……….+ amnxn = bm
47 Matrices - I
The m× n matrix is called the coefficient matrix of
the system of linear equations. Using it, we can now write these equations as follows :
=
We can abbreviate the above matrix equation to AX = B, where
A =
and X and B are the n×1 column vectors.
X =
Recall that by a solution of (1) we mean a set of values x1.x2……….xn which
satisfy all the equations in (1) simultaneously.
For example, x1 = 2, x2=–1is a solution of the system of linear equations.
3x1 – 5x2 = 11
2x1 +3x2 = 1
because 3(2) – 5(–1) = 11 and 2(2) +3(–1) = 1.
Also, recall that the system of linear equations (1) is said to be consistent if it has at least one solution; it is inconsistent if it has no solution.
For example, the system of linear equations 3x +2y = 5
6x + 4y = 10 (2)
is consistent. In fact, x = k, y = (5–2k) ( k C) satisfies (2) for all values of k C. However, the system of linear equations.
3x + 2y = 5
6x + 4 y = 11 (3)
is inconsistent. If this system has a solution x = x0, y = y0, then 3x0 + 2 y0= 5
and 6x0 + 4 y0= 11. Multiplying the first equation by 2 and subtracting from
the second equation, we get 0 =1, which is not possible. Thus, the system in (3) has no solution and hence is an consistent.
48
Algebra - I Solution of AX= B (A non-singular)
Let us consider the system of linear equations AX = B, where A is an n matrix. Suppose that A is non-singular. Then A–1 exists and we can pre multiply AX = B by A–1 on both sides to obtain
A–1 (AX) = A–1(B)
(A–1 A)X =A–1B [associative law] InX =A–1B [property of law]
X= A–1 B [property of identity matrix]
Moreover, we have
A(A–1B) = ( A A–1) B [associative law] = InB [property of inverse]
= B.
That is A–1B is a solution of AX = B. Thus, if A is non-singular, the system of equations AX = B has a unique solution. This unique solution is given by X = A–1 B.
Example 10 : Solve the following system of equations by the matrix
inverse method :
x+ 2y = 4 , 2x + 5 y = 9.
Solution : We can put the given system of equations into matrix notation as
follows :
=
Here the coefficient matrix is give by A = .
To check if exists, we note that A11 = ( 1)1+1 |5| = 5 and
A12= ( 1)1+2|2| = 2.
|A| = a11 A11 +a12 A12 = (1)(5)+ (2)( ) = 1 ≠ 0.
Since |A| ≠ 0 A is non- singular (invertible). We also have A21= ( 1)2+1|2|
= 2. A22 = ( 1)2+2 |1| = 1. Therefore, the adjoint of A is
adj A =
X = = =
49 Matrices - I
Example 11 : Solve the following system of equations by using matrix
inverse :
3x + 4 y +7z = 14, 2x – y + 3z = 4, 2x + 2y – 3z = 0
Solution : We can put the given system of equations into the single matrix
equation AX = B, where
, X = and B =
The cofactors of |A| are
A11 = ( 1)1+1 = 3 A12 = ( 1)1+2 =
and A13 = ( 1)1+3 =
|A| = a11 A11 + a12 A12 + a13 A13 = (3)(–3) + 4(9) + 7(5) = 62.
Since |A| ≠ 0, A is non−singular (invertible). Its remaining cofactors are A21 = ( 1)2+1 = A22 = ( 1)2+2 = 16,
A23 = ( 1)2+3 = 2, A31 = ( 1)3+1 =
A32 = ( 1)3+2 = A33 = ( 1)3+3 = 11.
50
Algebra - I
Hence x =1, y = 1, z = 1 is the required solution.
Example 12 : If
are two square matrices, verify that AB = BA = 6I3. Hence, solve the system
of linear equations : x – y =3, 2x+3y+4z = 17, y +2z = 7.
Solution : AB = = = 6 = 6I3 and BA = = = 6 = 6I3 Thus, AB =BA = 6I3
51 Matrices - I or AX = C, where X = and C =
Thus, x = 2, y = 1, z = 4 is the required solution.
Solution of a system of Homogeneous Linear Equations :
These are equations of the type AX = O. Let us consider the system of n homogeneous equations in n unknowns
a11 x1 +a12 x2 +………+a1nxn = 0 a21 x1 +a22 x2 +………+a2nxn = 0 :
an1 x1 +an2 x2 +………+ann xn = 0
We can write this system as follows
We now abbreviate the above matrix equation to AX = 0, where
A=
52
Algebra - I If A is non-singular, then pre multiplying AX = O by A-1, we get A–1(AX) = A–1O
(A–1A)X = O [associative law] InX = O [property of inverse]
X = O [property of identity matrix] x1 = 0, x2 = 0, ………. xn =0.
Also, note that x1 =0, x2 =0, ………. , xn = 0 clearly satisfy the given system of
homogeneous equations.
Thus, when A is non-singular AX = O has the unique solution. x1 =0, x2 =0,
………. xn =0. This is called the trivial solution. Important Result
We now state the following results without proof :
1. If A is singular, then AX = O has an infinite number of solutions.
2. Conversely, if AX = 0 has an infinite number of solution, then A is a singular matrix.
Example 13 : Solve the following system of homogeneous linear equations by the
matrix method :
2x – y + z = 0, 3x+2y – z =0 , x + 4y + 3 z = 0
Solution :
We can rewrite the above system of equations as the single matrix equation AX =0, where
A =
The cofactors of |A| are
|A| = Since |A|
53 Matrices - I
Example 14 : Solve the following system of homogeneous linear equation by the
matrix method :
2x – y + 2z = 0, 5x + 3y – z = 0, x +5y –5 z = 0
Solution :
We can rewrite the above system of equations as the single matrix equation AX =0, where
A =
The cofactors of |A| are
. |A| =
Therefore, A is singular matrix. We can rewrite the first two equation as follows: 2x – y = –2z, 5x + 3y = z or in the matrix form as
Now, we have |3| = 3 and |5| = –5.
|2| = 2. Therefore, the adjoint of A is given by
Therefore, from X =
54
Algebra - I Thus, all the equation are satisfied by the values
Where z is any complex number. Hence, the given system of equation has an infinite number of solutions.
Solutions of AX = B (A Singular)
We state the following result without proof : If A is a singular, that is |A| = 0, and
1. (adjA) B = 0, then AX = B has an infinite number of solutions (consistent).
2. (adj A) B
Example 15 : Solve the following system of linear equation by the matrix
method :
2x – y + 3z = 5, 3x + 2y – z =7 , 4x + 5y − 5 z = 9
Solution :
We can rewrite the above system of equations as the single matrix equation AX =0, where
A =
Here, |A| = 0
adj A =
Thus, AX = B has an infinite number of solutions. To find these solutions, we write 2x – y = 5 –3z, 3x + 2y = 7 + z or as a single matrix equation
Here, |A| = 7 ≠ 0
Since |A| ≠ 0, A is an invertible matrix Now, adj A =
55 Matrices - I Therefore, from X =
Let us check that these values satisfy the third equation. We have
Thus, the values
Satisfy, the given system which therefore has an infinite number of solutions.
In the end, we summarize the results of this section for a square matrix A in the form of a tree diagram.
|A| AX = B X = B (adj A)B = 0 |A| = 0 (adj A)B ≠ 0 Unique Solution Infinite Number of Solutions No Solution Consistent Inconsistent Consistent
56
Algebra - I Check Your Progress 4
1. Solve the following equations by matrix inverse method : 4x – 3y = 5, 3x – 5y = 1
2. Use the matrix inverse to solve the following system of equations : (a) x + y – z = 3, 2x + 3y + z = 10, 3x – y – 7z = 1
(b) 8x + 4y + 3z = 18, 2x + y + z = 5, x + 2y + z = 5
3. Solve the following system of homogeneous linear equations by the matrix method :
3x – y + 2z = 0, 4x + 3y + 3z = 0, 5 x + 7y + 4z = 0
4. Solve the following system of linear equations by the matrix method : 3x + y – 2z = 7
5x + 2y + 3 z = 8 8x + 3y + 8z = 11
2.6
ANSWERS TO CHECK YOUR PROGRESS
Check Your Progress – 1
1. Note that a 2×2 matrix is given by A =
From the formulas given the elements, we have (a) A =
(b) A =
2. From equality of matrices, we have, x = 2x + y, y = x – y
Solving we get x = 0 , y = 0 3. We have
a – b = 5 2c + d = 3 2a – b = 12 2a + d = 15
Solving we get a = 7, b = 2, c = 1 and d = 1. 4. (a) A’ = = A
(b) A’ = = – A
57 Matrices - I
(c) A’ = = –A
Check Your Progress - 2
1. Since P and Q are matrices of order 2× 2, 5P + 3Q is a matrix of order 2×2 and therefore R must be a matrix or order 2 × 2.
Let R = . Then 5 P + 3Q + 2 R = 5 + 3 + 2 = + + = Since 5P + 3Q + 2 R = , we get 48 + 2a = 0, 20 + 2b = 0, 56 + 2c = 0, 76 + 2d = 0 Thus, R = 2. We have
= (A+ B ) A + ( A + B ) B (Distributive Law) = A A + BA + AB + BB
= Therefore,
Thus, we must find a and b such that BA + AB = 0.
We have BA = =
and AB = =
Therefore,
58 Algebra - I = But BA + AB = 0 2a – b + 2 = 0 , – a + 1= 0, 2 a – 2 = 0, – b + 4 = 0 a = 1, b = 4
3. In view of discussion in solution (2), it is sufficient to show that BA + AB = 0 We have BA =
and AB =
Thus, BA + AB = + 4. First, we note that by f(A) we mean
Therefore, = 5. We have A = Therefore, and =
6. In general matrix multiplication is not commulative. Therefore, AB may not be equal to BA, even though both of them exist.
Check Your Progress – 3
59 Matrices - I
adj A = .
(ii) Let A =
The cofactors of the elements of A are
adj A =
2. For the given matrix A, we have
adj A =
Now, A (adj A) =
= = –26
60
Algebra - I
= –26 Also, |A| = – 26
So, A (adj A) = (adj A) A = |A|
3. Here, |A| = (2) (–1) + (–1) (–2) + (3) (1) = 3
and adj A = (see solution 1(ii))
4. We have |A| = –4 and |B| = 20. So, A and B are both invertible. Also, adj A = and adj B =
Let C = AB = =
So, |C| = –80 and adj C =
Hence, =
61 Matrices - I 5. We have To show that . We have 6. We have = Therefore, – 4 = = 0
Also, |A| = 5 ≠ 0. Therefore, A is invertible pre− multiplying 0 by , we get
0 0
=
7. We have |A| = ad−bc. Recall that A is invertible if and only if |A| ≠ 0. That is A = is invertible if and only if ad– bc ≠ 0.
Also, adj A =
62
Algebra - I Check Your Progress – 4
1. We can put the given system of equations into the single matrix equation. = .
Here the coefficient matrix is given by A =
Cofactors of |A| are and
|A| =
Since |A| ≠ 0. A is non-singular (invertible). Also = 4.
Hence x = 2, y = 1 is the required solution.
2. (a) We can put the above system of equation into the single matrix equation AX = B, where
A = , X = and B =
The cofactors of |A| are
and
+
Since |A| ≠ 0, A is non- singular (invertible). The remaining cofactors are
63 Matrices - I
Thus, x = 3, y = 1, z = 1 is the required solution.
2. (b) We can put the above system of equation into the single matrix equation AX = B, where
A = , X = and B = .
The cofactors of |A| are
and
+
Since |A| ≠ 0, A is non−singular (invertible). The remaining cofactors are
64
Algebra - I
Thus, x = 1, y = 1, z = 2 is the required solution.
3. We can put the above system of equation into the single matrix equation AX = 0, where
A = , X = and B =
The cofactors of |A| are
and
+
Therefore, A is a singular matrix. We can rewrite the first two equations as follows :
3x – y = – 2z, 4x + 3y = –3z
or in the matrix form as Now, we have
+
Thus, A is non- singular (invertible). Also,
65 Matrices - I Let us check if these values satisfy the third equation. We have
Thus, all the equations are satisfied by the values
4. We can write the given system of linear equation as the single matrix equation.
AX = B, Where
A = , X = and B =
Here, |A| = 0
Therefore, A is a singular matrix.
Now adj A =
Since (adj A) B ≠ 0, the given system of equations has no solution (inconsistent).
2.7
SUMMARY
In this unit, first of all, definition and notation of an m x n matrix, are given in
section 2.2. Next, in this section, special types of matrices, viz., square matrix,
diagonal matrix, scalar matrix, unit or identity matrix, row or column matrix and zero or null matrix are also defined. Then, equality of two matrices, transpose of a matrix, symmetric and skew matrices are defined. Each of the above concepts is explained with a suitable example. In section 2.3, operations like addition, subtraction, multiplication of two matrices and multiplication of a matrix with a scalar are defined. Further, properties of these matrix operations are stated without proof. Each of these operations is explained with a suitable example. In
section 2.4, the concepts of an invertible matrix, cofactors of a matrix, adjoint of a
square matrix are defined and explained with suitable examples. Finally, in
section 2.5, method of solving linear equations in n variables using matrices, is
given and illustrated with a number of suitable examples. Answers/Solutions to questions/problems/exercises given in various sections of the unit are available in
66 Algebra - I
UNIT 3 MATRICES - II
Structure 3.0 Introduction 3.1 Objectives3.2 Elementary Row Operations 3.3 Rank of a Matrix
3.4 Inverse of a Matrix using Elementary Row Operations 3.5 Answers to Check Your Progress
3.6 Summary
3.0
INTRODUCTION
In Unit 2, we have introduced Matrices. In this Unit, we shall study elementary operation on Matrices. There are basically three elementary operations. Scaling, Interchange and Replacement. These operations are called elementary row operations or elementary column operations according as they are performed on rows and columns of the matrix respectively. Elementary operations play important role in reducing Matrices to simpler forms, namely, triangular form or normal form. These forms are very helpful in finding rank of a matrix, inverse of a matrix or in solution of system of linear equations. Rank of a matrix is a very important concept and will be introduced in this unit. We shall see that rank of a matrix remains unaltered under elementary row operations. This provides us with a useful tool for determining the rank of a givne matrix. We have already defined inverse of a square matrix in Unit 2 and discussed a method of finding inverse using adjoint of a matrix. In this unit, we shall discuss a method of finding inverse of a square matrix using elementary row operations only.
3.1
OBJECTIVES
After studying this Unit, you should be able to : define elementary row operations;
reduce a matrix to triangular form using elementary row operations; reduce a matrix to normal form using elementary operations;
define a rank of a matrix;
find rank of a matrix using elementary operations;
67 Matrices - II
3.2
ELEMENTARY ROW OPERATIONS
Consider the matrices of A = , B = , C = and
D=
Matrices B, C and D are related to the matrix A as follows :
Matrix B can be obtained from A by multiplying the first row of A by 2; Matrix C can be obtained from A by interchanging the first and second rows; Matrix D can be obtained from A by adding twice the second row the first
row.
Such operations on the rows of a matrix are called elementary operations.
Definitions : An elementary row operations is an operation of any one of the
following three types :
1. Scaling : Multiplication of a row by a non zero constant. 2. Interchange : Interchange of two rows.
3. Replacement : Adding one row to a multiple of another row.
We denote scaling by Ri kRi,interchange by Ri Rj and replacement by
Ri Ri + kRj.
Thus, the matrices B, C and D are obtained from matrix A by applying
elementary row operations R1 2R1 R1 R2 and R1 R1 + 2 R2 respectively. Definiton : Two matrices A and B are said to be row equivalent, denoted by
A ~ B, if one can be obtained from the other by a finite sequence of elementary row operations.
Clearly, matrices B, C and D discussed above are row equivalent to the matrices A and also to each other by the following remark.
Remark : If A, B and C are three matrices, then the following is obvious.
1. A ~ A
2. If A ~ B, then B ~ A
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Algebra - I
Example 1 : Show that matrix A = is row equivalent to the matrix.
B =
Solution : We have A =
Applying , we have
A ~
Applying to the matrix on R. H. S. we get.
A ~
Now Applying we have
A ~ = B
The matrix B in above example is a triangular matrix.
Definition : A matrix A = [ ] is called a triangular matrix if aij= 0 whenver i > j.
In the above example, we reduced matrix A to the triangular matrix B by elementary row operations. This can be done for any given matrix by the following theorem that we state without proof.
Theorem : Every matrix can be reduced to a triangular matrix by elementary row
operations.
Example 2 : Reduce the matrix
A =
to triangular form.
69 Matrices - II
~ (by applying R1 R3)
~ (by applying R3 R3– 5R1)
~
which is triangular matrix.
Example 3 : Show that A = is row equivalent to I3.
Solution : A = ~ (R1 R2) ~ (by R2 R2 3R1 and R3 R3 R1) ~ (by R1 R1 R2 ) ~ (by R3 R3) ~ (by R1 R1 5R3 and R2 ) = I3
In above example, we have reduced the square matrix A to identity matrix by elementary row operations. Can every square matrix be reduced to identity matrix by elementary row opearations.
The answer, in general, is no, however, if A is a square matrix with |A| ≠ 0, then A can be reduced to identity matrix by elementary row operations. This we state below without proof.
70
Algebra - I Theorem : Every non-singular matrix is row equivalent to a unit matrix. Below we given an algorithm to reduce a non-singular matrix to identity matrix. 1. Make the first element of first column unity by scaling. If the first element is
zero the first make use of interchange.
2. Make all elements of first column below the first element zero by using replacement.
3. Now make the second element of second column unity and all other elements zero.
4. Continue the process column by column to get an identity matrix. The following example illustrate the process.
Example 4 : Reduce the matrix to I3.
Solution : ~ (by R1 R2) ~ (by (R1 R1) ~ (by R3 R3 R1) ~ (by R2 R2) ~ (by R1 R1 + 2R2 and R3 R3 – R2) ~ (by and R3 R3) ~ (by R1 R1 2R3 and R2 R2 R3) Check Your Progress – 1
1. Write the Matrices obtained by applying the following elementary row operations on
71 Matrices - II
A =
(i) R1 R3 (ii) R2 R2 + 3R1
(iii) R2 R3, then R2 R2 and then R3 R3+ 2R1
2. Reduce the matrix A = to triangular form.
3. Show that is row equivalent to I3.
4. Is the matrix row equivalent to I3.
5. Which of the following is row equivalent to I3.
(a) (b)
3.3 RANK OF A MATRIX
Suppose A is an m × n matrix. We can obtain square sub matrices of order r ( 0 < r least of m and n) from A by selecting the elements in any r rows and r columns of A. We define rank of matrix as follows :
Defintion : Let A be an m × n matrix. The order of the largest square submatrix
of A whose determinant has a non-zero value is called the ‘rank’ of the matrix A. The rank of the zero matrix is defiend to be zero.
It is clear from the definition that the rank of a square matrix is r if and only if A has a square submatrix of order r with nonzero determinant, and all square sub matrices of large size have determinant zero.
Example 5 : Find the rank of the matrix.
