MATRICES AND DETERMINANTS
1.1 THE CONCEPT OF A MATRIX
A matrix is a rectangular array of numbers, parameters or variables enclosed by parenthesis ( ) or bracket [ ] or double vertical bars .
Example:
( )
ijmn mn
mn m
m
n n
a a a
a a
a a
a
a a
a
a a
a
=
=
=
....
...
....
...
....
....
....
11 ..
11
2 1
2 22
21
1 12
11
Where i = 1,2,3…m and j = 1,2,3, ….n. The numbers a11, a12… amn are known as elements of the matrix. The elements in horizontal lines are rows and the elements in vertical lines are columns. The number of rows (r) and the columns (C) represents the dimensions or orders of the matrix (rXc), which is read “r by c”. The above matrix is called an m x n matrix or a matrix of order (dimension) m x n. For the element aij, i is the row subscript and j is the column subscript.
1.2 TYPES OF MATRICES
(a) Vectors: A single row matrix is called a row vector and a single column matrix is called a column vector.
(i) Row Vector: A matrix having only one row is called a row vector or row matrix
Example: (i) A = (a11 a12 a13…. a1n)
ONE
(ii) B = (1 2 3)
(ii) Column Vector: A matrix having only one column is called a column vector or column matrix
Example (i):
=
1 31 21 11
am
a a a A
M
(ii)
= 9 13
6 B
(1) Null Matrix or Zero Matrix: An m x n matrix each of whose entries (elements) is the number 0 is called the m x n zero matrix and denoted as Omxn or Simply O.
Example:
=
=
0 0
0 O 0
0 , 0 0
0 0
O 0
(2) Square Matrix: A matrix having the same number of rows and columns (m = n) is called a square matrix. That is an, m x n.
matrix is a square matrix iff m = n.
Example:
=
=
9 8 7
6 5 4
3 2 1 B ,
33 32 31
23 22 21
13 12 11
a a a
a a a
a a a
A
Note: In a square matrix of order n, the entries a11, a22, a33,… ann which lie on the diagonal extending from the upper left corner to the lower right corner are called the main diagonal entries or simply the main diagonal or the principal diagonal.
(3) Diagonal Matrix: A square matrix whose elements above and below the principal diagonal are all Zero is called diagonal matrix.
Example:
=
=
5 0 0
0 4 0
0 0 3 B , 3 0 0
0 2 0
0 0 1 A
(4) Unit Matrix or Identity Matrix: A diagonal matrix whose elements are all equal to one (1) is called a unit matrix and is denoted by I;
Example:
=
=
1 0 0
0 1 0
0 0 1 I 1 , 0
0
I 1
(5) Scalar Matrix: A diagonal matrix whose all elements are equal is called scalar matrix.
Example:
=
=
5 0 0
0 5 0
0 0 5 B 2 , 0
0 A 2
(6) Triangular Matrix: A Square matrix whose elements above or below the principal diagonal are all zero is called a triangular matrix;
Example: lower triangular
6 5 4
0 3 2
0 0 1
⇒
=
A
ar triangul u
6 0 0
5 4 0
3 2 1
B ⇒ pper
=
(7) Transpose of a Matrix: The transpose of an m x n matrix A is written as n x m matrix A and denoted by AT or AI .
Example:
=
= 2 4 6
5 3 A 1
then 6
5 4 3
2 1
T
A
Note: (i) (AT)T = A
(ii) (A + B)T = AT + BT (iii) (AB)T = BT AT
(8) Symmetric Matrix: A real square matrix A is called a symmetric if A = AT i.e., aij = aji.
Note:
(i) A square matrix A is called a skew symmetric (anti symmetric) if A = -AT.
(ii) For anti-symmetric matrix the elements in the principal diagonal are zero.
Example: (i) For Symmetric, if
=
c f g A
f g
b h
h a
then
=
c f g AT
f g
b h
h a
⇒ A = AT or A is Symmetric Matrix
(ii) If B = then
4 2 - 5
2 0 1
5 1 3 -
−
Matrix.
ymmetric is
B or
4 2 - 5
2 0 1
5 1 3 - BT
S B
B= T
⇒
−
=
For Anti Symmetric:
0 5 - 1
5 0 4 -
1 4 0 A then 0
5 1 -
5 0 4
1 4 - 0 A
T
−
=
−
= If
−
=
⇒
0 5 1 -
5 0 4
1 4 - 0 A - T
(9) Idempotent Matrix: A Symmetric matrix that reproduces itself when multiplied by itself is called an idempotent matrix.
Example: If AA = A2 = A
⇒ A is an idempotent matrix.
If A =
1 0
0 1
Then AA = A2 = A
(10) Singular and Non-Singular Matrices: A square matrix A is a Singular matrix if the determinant of A is zero.
Example: If A = 12 12 0
6 3
4 A 2
then 6 3
4
2 = = − =
A square matrix A is non-singular if the determinant of A is non-zero.
Example: If A = 4 6 2 4
3 2 A 1
then 4 3
2
1 = = − =−
≠0
(11) Equal Matrices: Two matrices A= (aij) and B = (bij) are equal if and only if the corresponding elements are equal. That is, A = B ⇔⇔⇔ ⇔ aij = bij.
Example: From
C A 2 9 3 D and 2 3 8
3 6 5 5 C , 6 5
4 3
2 1 B 4 , 3
2
1 ∴ =
=
=
=
= A
(12) Sub matrix: A matrix obtained by deleting some rows and column’s of a matrix A is called a sub matrix.
Example:
=
6 5 4
3 2
A 1
Sub matrices of A are,
( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
11, 22, , 23, 34, , 153, 66 4 , 6 5 , 5 4
6 3 5 , 2 4 , 1
6 5 4 , 3 2 1
6 5
3 2 6 , 4
3 1 5 , 4
2 1
1.3 OPERATIONS ON MATRICES:
Addition and Substraction of Matrices:
If two matrices are the same order (that is, both have the same number of rows and columns), then addition and substraction operations between two are possible. For addition (substraction) each element of one matrix is added to (subtracted from) the corresponding element of the other matrix.
Example .
8 7
6 B 5
nd 4 a 3
2
A 1
=
=
(i) 10 12 8 6
8 4 7 3
6 2 5 B 1
A
=
+ +
+
= + + (ii)
12 10
8 6
4 8 3 7
2 6 1 A 5
B
=
+ +
+
= + + (iii)
4 4
4 4
8 4 7 3
6 2 5 B 1
A
−
−
−
= −
−
−
−
= −
− (iv)
4 4
4 4 4 8 3 7
2 6 1 A 5
B
=
−
−
−
= −
−
Note: Properties of Matrix Addition: If A, B, C and O are comparable matrices, then the following properties hold for matrix addition:
(i) A+B = B+A Commutative property
(ii) A+(B+C) = (A+B)+C Associative property (iii) A+O = O+A=A Identity property Multiplication of a Matrix by a Scalar:
An ordinary number such as 9, -3, 2.5, 0.09 is called a scalar in matrix algebra. To multiply a matrix by a scalar, each element in the matrix is multiplied by the scalar or number. The process is called scalar multiplication, because it scales the matrix up or down according to the size of the number.
Example: If
4 3
2 1
=
A and it is required to find 4A, then
=
=
=
16 12
8 4 ) 4 ( 4 ) 3 ( 4
) 2 ( 4 ) 1 ( 4 4 3
2 4 1 4 A
Note: If A and B are comparable matrices and λλλλ and µµµµ are scalars then (i) λ
(
A+B)
=λA+λB(ii) (λλλλµµµµ)A = λλλλ(µµµµA) (iii) (λλλλ+µµµµ)A = λλλλA + Aµµµµ Exercise: If λλλλ = -2 and µµµµ = 9
And
=
=
1 8
7 B 6
5 nd 4
3
2 a
A
Then prove that
(1) λ
(
A+B)
=λA+λB(2) (λλλλµµµµ)A = λλλλ(µµµµA) (3) (λλλλ+µµµµ)A = λλλλA + Aµµµµ MATRIX MULTIPLICATION:
The basic rule to multiply a matrix (A) by another matrix (B) is that the number of columns in the first matrix (A) must equal the number of rows in the second matrix (B).
If A is an m x n matrix and B is an n x p matrix, then the product AB is the m x p matrix C whose entry Cij in the ith row and jth column is obtained with the following procedure: Sum the products formed by multiplying, in order, each element (that is, first. Second, etc.), in the ith row of A by the corresponding element (that is, first, second etc.), in the jth column of B.
Example (1): If
[ ]
=
=
7 6 5 B and 4 3 2 A
Then since
1 3
3 1
B A
×
×
i.e., AB =
[ ] [
2(5) 3(6) 4(7)]
7 6 5 4 3
2 = + +
= [10+18+28] = (56) = C1x1
Example (2) If and B
[
2 3]
5 4 3
=
=
A
Then since
2 1
1 3
×
×
B A (AB) = C1x1
(AB) = C3x2
i.e., AB =
[ ]
3215 10
12 8
9 6 5(3) ) 2 ( 5
4(3) ) 2 ( 4
3(3) ) 2 ( 3 3 2 5 4 3
= ×
=
=
C
Example (3) If
=
=
2 1
1 2
2 1 B and 6 5 4
3 2 A 1
Then since
2 3
3 2
×
×
B A
i.e.,
=
2 1
1 2
2 1 6 5 4
3 2 AB 1
2
25 2
20 10 8 12 5 8 6 10 4
6 2 2 3 4 1
) 2 ( 6 ) 1 ( 5 ) 2 ( 4 ) 1 ( 6 ) 2 ( 5 ) 1 ( 4
) 2 ( 3 ) 1 ( 2 ) 2 ( 1 ) 1 ( 3 ) 2 ( 2 ) 1 ( 1
= ×
=
+ + + +
+ + +
= +
+ + +
+
+ + +
+
C
Note: If all the products and sums are defined for the following three matrices A, B and C then the following properties hold for matrix multiplication.
(i) A (BC) = (AB) C Associative Law (ii) A (B+C) = AB + AC Left Distributive Law (iii) (A+B) C = AC + BC Right Distributive Law
(iv) In general AB≠≠≠≠ BA Matrix multiplication is not commutative
(v) In general we have ab = 0 iff a = 0 or b = 0. But if A and B are matrices than AB = 0 does not mean that A = 0 or B = 0.
Example: 1. If
−
=
−
= −
2 1 1
1 0 B 3
4 , 3
2
A 1
and
= 1 1
2 0
0 1 C
(AB) = C2x2
then
− −
=
−
= −
19 6
9 4 4
3 1 - 2 4 3
2 ) 1
(BC A
− −
=
−
−
= −
19 6
9 4 1 1
2 0
0 1 11 4 5
5 2 ) 1
( ABC and
∴∴∴
∴ A(BC) = (AB)C Proved
Example (2) If
=
−
= −
=
7 2
1 C 3
and 3 1
0 B 2
, 2 0
1 A 5
Then A (B+C) =
=
8 6
9 8 4 3
1 1 2 0
1 5
−
−
= −
6 2
3
AB 9 AC =
14 4
12 17
AB + AC =
=
+
−
−
−
8 6
9 8 14 4
12 17 6 2
3 9
∴
∴
∴
∴A (B+C) = AB + AC Proved
Example (3). If A =
=
0 1
1 B 0
and 0 0
0 1
then AB =
=
0 1
0 BA 0 and 0 0
1 0 i.e., AB ≠≠≠≠ BA.
Exam ple (4). In exception some matrices are commutative
If A =
=
4 3
2 B 1
and 1 0
0 1
Then AB =
=
4 3
2 BA 1 and 4 3
2 1 i.e., AB = BA
Example (5) If A =
−
= −
1 1
1 B 1
and 2 2
1 1
then AB = 0 0
0
0
i.e., AB = 0 does not apply that A = 0 or B = 0 Example (6) If A =
2 1 2
6 3 5
2 1 10 B and 3 4 1
6 2 3
5 1 2
=
then AB =
=
+ + + + + +
+ + + + +
+
+ + +
+ +
+
32 16 36
30 15 52
20 10 35 6 24 2 3 12 1 6 20 10
12 12 6 6 6 3 12 10 30
10 6 4 5 3 2 10 5 20
EXERCISES:
(1) From A =
2 0
0 B 3
and 6 0
0
2
=
find AB and BA.
(2) From A =
= −
0 1
0 B 1
and 1 1
1
1 find AB.
(3) From A =
−
−
− 4 4 1
3 3 1
3 4 0
find AA = A2 1.4 DETERMINANTS:
The determinant of a square matrix A is denoted by | A | (that is, employing vertical bars).
A →→→ | A | →
Square matrix Real Number = Determinant of A.
DETERMINANTS ARE DEFINED ONLY FOR SQUARE MATRICES
(i) The determinant of a 2 x 2 matrix is given by:
21 12 22 11 22 21
12
11 a a a a
a a
a
A = a = −
Example: If A =
4 3
2 1
Then 1[4] 2[3] 4 6 2. 4
3 2
1 = − = − −=−
= A
(ii) The determinant of a 3 x 3 matrix is given by
32 31
22 21 13 33 31
23 21 12 33 32
23 22 11 33 32 31
23 22 21
13 12 11
a a
a a a
a a
a a a
a a
a a a
a a a
a a a
a a a
A= = − +
= a11(a22 a33 – a23 a32) – a12(a21 a33 – a23 a31) + a13(a21 a32 – a22 a31) Example: If
6 0 1
3 3 2
4 1 5 A then 6
0 1
3 3 2
4 1 5
−
−
=
−
−
= A
0 1
3 4 2 6 1
3 1 2
6 0
3 53
+ −
−
− −
= −
= 5(18-0) –1 (12-3) + 4(0+3) = 90-9 + 12 = 93 = a Scalar 1.5 BASIC PROPERTIES OF DETERMINANTS
The evaluation of determinants is often simplified by the use of various properties. The following basic properties may be used to find the value of determinant.
Property (1) If in a determinant (a) one row is a multiple of another row (b) one column is a multiple of another column (c) any two rows are equal (d) any two columns are equal (e) each of the elements in a row is zero (f) each of the elements in a column is zero, then the value of the determinant is zero.
Example:
(a) 0
6 5 4
6 4 2
3 2 1
6 5 4
2 3 2 2 2 1
3 2 1
=
=
×
×
×
(b) 0
2 2 1
0 6 3
1 4 2
2 2 1 1
0 2 3 3
1 2 2 2
=
=
×
×
×
(c) 0
4 3 2
6 5 1
4 3 2
=
(d) 0 1 4 4
0 3 3
1 2 2
=
(e) 0
6 5 4
0 0 0
4 3 1
=
(f) 0
2 1 0
3 2 0
1 5 0
=
Property (2) The interchange of rows and columns does not affect the value of a determinant.
Example: 32
1 1 2
5 0 3
3 1 2
=
−
−
= A
and 32
1 5 3
1 0 1
2 3 2
=
−
−
= AT
Property (3) The interchange of two rows (or two columns) alters the sign, but not the numerical vale, of the determinant.
Example: -2
4 3
2
1 =
= A
2
2 1
4
3 =
= B
2
3 4
1
2 =
= C
Property (4) The multiplication of any one row or one column by a scalar K changes the value of the determinant K fold;
Example: 5 12 7
5 4
3
1 = − =−
= A
42 72 5 30
24 3 6 5 6 4
3 6 1
14 24 5 10
4 6 2 5 4
2 3 2 1
−
=
−
=
× =
= ×
−
=
−
=
× =
= ×
C B
Property (5) If all entries below (or above) the main diagonal of a matrix A are zero, then the value of the determinant of A is equal to the product of the main diagonal entries.
30 (3)(5) (2) 5 4 7
0 3 9
0 0 2
30 (5)(6) (1) 6 0 0
4 5 0
3 2 1
=
=
=
=
=
=
B A
Property (6) If a multiple of any row (or column) is added to or subtracted from another row (or column) then the value of the determinant is remain unchanged.
Example: 2
) 4 2 ( 4 ) 4 1 ( 3
2 1
4 3
2
1 =−
×
±
×
= ±
=
A
0
2 ) 2 2 ( 4
1 ) 1 2 ( 2 2 4
1
2 =
×
±
×
= ±
= B
1.6 RANK OF A MATRIX
The rank ρ of a matrix is defined as the maximum number of linearly independent rows or columns in the matrix.
Examples:(a) A 98
2 8 4
4 5 1
2 6 3
A ⇒ =−
−
−
=
With | A| ≠≠≠≠ 0. A is nonsingular and the three rows and columns are linearly independent. Henceρ (A) = 3.
(b) B 0 6
18 3
4 12 2
3 9 5
B ⇒ =
−
−
−
−
=
With |B| = 0, B is singular and the three rows and columns are not linearly independent. Hence ρ (B) ≠≠≠≠ 3. Now testing to see if any two rows or columns are independent
. 0 12 78
2 9
5 − = ≠
∴ρ(B) =2.There are only two linearly independent rows and columns in B
(c) 0
18 6 24
5 . 7 5 . 2 10
6 2 8
=
⇒
−
−
−
−
= C
C
with |C| = 0, ρ (C) ≠≠≠≠ 3. Trying various 2 x 2 Sub matrices . 18 0 6 -
7.5 2.5 - 6 0
- 24
2.5 - 10 7.5 0 2.5 -
6 - 2
5 0 . 2 10
2
8 = = = =
−
− with
all the determinants equals to zero, no two rows or columns of C are linearly independent. So ρ (C) ≠≠≠≠ 2 and ρ (C) = 1.
(d)
= 1 3
11 7
5 2 D
Since the maximum number of linearly independent rows (columns) must equal the maximum number of linearly independent columns (rows), the rank of D cannot exceed 2. Testing a Sub matrix :
2 (D) 0
-13 11 7
5
2 ==== ≠≠≠≠ ∴∴∴∴ρρρρ ====
1.7 MATRIX INVERSION
If A is a square (Non-Singular) Matrix then the inverse of A (denoted by A-1) is defined by the relation
A A−1 = adjA
Where adjA is the adjoint of A and | A | is determinant of A.
Minors and Cofactors:
Minors: The minor of an element in a determinant is known as the determinant of sub matrix that is obtained by deleting the row and column in which the given element occurs. Thus, a minor | Mij | is the determinant of the sub matrix obtained by striking out the ith row and jth column of the matrix.
Cofactors: The cofactor of the elements of A is defined by the relation:
Aij = (-1) i+j.Mij
Where Mij is minor of A.
Note: For every element in a square matrix there is a minor as well as cofactor.
Example: Find the minors and cofactors of the matrix;
1 2 1
4 1 2
3 0 1
= A Solution:
Minors Cofactors
7 8 1 1
2 4
M11 ==== 1 ==== −−−− ====−−−− A11 = (-1)2(-7) = -7
2 4 1 2
1 4
M12 ==== 2 ==== −−−− ====−−−− A12 = (-1)3(-2) = 2
3 1 2 4
1 1
M13 ==== 2 ==== −−−− ==== A13 = (-1)4(3) = 3
6 6 1 0
2 3
M21 ==== 0 ==== −−−− ====−−−− A21 = (-1)3(-6) = 6
2 3 1 1
1 3
M22 ==== 1 ==== −−−− ====−−−− A22 = (-1)4(-2) = -2
2 0 2 2
1 0
M23 ====1 ==== −−−− ==== A23 = (-1)5(2) = -2
3 3 4 0
1 3
M31 ==== 0 ==== −−−− ====−−−− A31 = (-1)4(-3) = -3
2 6 4 4
2 3
M32 ==== 1 ==== −−−− ====−−−− A32 = (-1)5(-2) = 2
1 0 1 1
2 0
M33 ==== 1 ==== −−−− ==== A33 = (-1)6(1) = 1
Therefore,
Co-factors Matrix =
−
−
−
−
=
1 2 3
2 2 6
3 2 7
33 32 31
23 22 21
13 12 11
A A A
A A A
A A A
Adjoint of a Matrix A: An adjoint matrix is the transpose of a cofactor matrix, therefore
T
nn n2
n1
2n 22
21
1n 12
11
A A
A
A A
A
A A
A adjA
=
K M M
K M
K
Where A11, A12 … Ann are the co-factors of the elements a11, a12 … ann. Methods for finding the inverse of a Matrix
A of inverse the find then , 2 1 2
0 1 3
1 2 1 A
= If Solution:
Step 1: det A =
2 1 2
0 1 3
1 2 1
0 9 1 12 2
) 1 ( 1 ) 6 ( 2 ) 2 ( 1 1 2
1 13 2 2
0 23 2 1
0 11
≠
−
= +
−
=
+
−
= +
−
=
Step 2: Minors and Cofactors of A.
Minors Cofactors
2 2 1
0 1
11 = =
M A11 = (-1)2(2) =2
2 6 2
0 3
12 = =
M A12 = (-1)3(6) =-6
1 1 2
1 3
13 = =
M A13 = (-1) 4(1) = 1
2 3 1
1 2
21 = =
M A21 = (-1)3(3) = -3
2 0 2
1 1
22 = =
M A22 = (-1)4(0) = 0
1 3 2
2 1
23 = =−
M A23 = (-1)5(-3) = 3
0 1 1
1 2
31 = =−
M A31 = (-1)4(-1) = -1
0 3 3
1 1
32 = =−
M A32 = (-1)5(-3) = 3
1 5 3
2 1
33 = =−
M A33 = (-1)6(-5) = -5
Step 3: adj A =
T T
A A A
A A A
A A A
−
−
−
−
=
5 3 1
3 0 3
1 6 2
33 32 31
23 22 21
13 12 11
=
∴
5 - 3 1
3 0 6 -
1 - 3 - 2 adjA
Step4: Since A-1 = A adjA
−
−
−
−
=
=
∴
9 5 3 1 9 1
3 1 0 3 2
9 1 9 3 9 2 9
- 5 - 3 1
3 0 6 -
1 - 3 - 2
A-1
PROPERTIES OF INVERSE MATRICES:
(1) AA-1 = A-1A = I (2) (AB)-1 = B-1A-1
(3) (AT)-1 = (A-1)T (4) (A-1)-1 = A
Example (1): Find the inverse of the following matrix:
A =
−
1 0 2
3 3 7
1 2 4
and check the answer.
Solution:
Step 1.
( )
0 2
3 17 1 2
3 2 7 1 0
3 43 1 0 2
3 3 7
1 2 4
+
−
−
=
−
= A
= 4(3-0) + 2(7-6) + 1(0-6)
= 4(3) + 2(1) + 1(-6) = 12 + 2 – 6 = 8 Step 2.
Minors Cofactors
1 3 0
3
M11 = 3 = A11 = (-1)2(3) =3
1 1 2
3 7
12 = =
M A12 = (-1)3(1) = -1
0 6 2
3
M13 = 7 =− A13 = (-1)4(-6) = -6
1 2 0
1 M21 −2 =−
= A21 = (-1)3(-2) = -2
1 2 2
1
M22 = 4 = A22 = (-1)4(2) = 2
0 4 2
2
M23 4 − =
= A23 = (-1)5(4) = -4
3 9 3
1 M31 −2 =−
= A31 = (-1)4(-9) = -9
3 5 7
1
M32 = 4 = A32 = (-1)5(5) = -5
3 26 7
2
M33 4 − =
= A33 = (-1)6(26) = 26
Step 3: adj A =
T T
A A A
A A A
A A A
−
−
−
−
−
=
26 5 9
4 2 2
6 1 3
33 32 31
23 22 21
13 12 11
adj A =
−
−
−
−
− 26 4 6
5 2
1
9 2
3
Step 4:
8 26 4 6
5 2 1
9 2 3
1
−
−
−
−
−
=
=
∴ −
A A adjA
−
−
−
−
−
=
−
−
−
−
−
− =
4 13 2 1 4 3
8 5 4 1 8 1
8 9 4 1 8 3 8 26 8 4 8 6
8 5 8 2 8 1
8 9 8 2 8 3 A 1
Step 5: For Checking results: Since AA-1 = I = A-1A
Therefore, AA-1=
−
−
−
−
−
−
4 13 2 1 4 3
8 5 4 1 8 1
8 9 4 1 8 3 1 0 2
3 3 7
1 2 4
I
1 0 0
0 1 0
0 0 1
=
=
Similarly, A-1A =
−
−
−
−
−
−
1 0 2
3 3 7
1 2 4 4 13 2 1 4 3
8 5 4 1 8 1
8 9 4 1 8 3
I 1 0 0
0 1 0
0 0 1
=
= Therefore, AA-1 = I = A-1A Proved
INVERSE OF A MATRIX OF ORDER 2 X 2
We can use the following four steps to find the inverse of a 2 x 2 matrix.
Step 1: Find the determinant of the matrix.
Step 2: Interchange the positions of the elements on the main diagonal.
Step 3: Negate the elements on the minor diagonal
Step 4: Divide each element by the determinant of the matrix.
Example 1:Find the inverse of
=
3 6
2
A 8 and check the answer.
Solution: Step 1: 24 12 12 3
6 2
8 = − =
=
A
Step 2:
8 6
2
3 Step 3:
−
− 8 6
2 3
Step 4:
−
= −
−
= −
−
3 2 2 1
6 1 4 1 12 8 12 6
12 2 12
1 3 A To check answer:
=
−
−
=
−
1 0
0 1 3 2 2 1
6 1 4 1 3 6
2
1 8 AA
=
−
= −
−
1 0
0 1 3 6
2 8 3 2 2 1
6 1 4
1 1 A A
Therefore, AA-1 = I = A-1A Proved Example 2: Find the inverse of A =
4 3
2
1 and prove that AA-1 = I = A-1A
Solution: Step 1: 4 6 2
4 3
2
1 = − =−
= A
Step 2:
1 3
2
4 Step 3:
−
− 1 3
2 4
Step 4:
−
−
−
−
−
−
−
= −
−
2 1 2 3
1 2 2 1 2 3
2 2 2
A 1 4
To check the answer:
and
AA 1
1 0
0 1 2 3 6 6
1 1 3 2 2 1 2 3
1 2 4 3
2
1 1 =
=
− +
−
− +
= −
−
−
=
−
1 1 0
0 1 2 2 3
3 2
32 3 4 4
4 3
2 1 2 1 2 3
1
1 2 =
=
−
−
+
− +
= −
−
= −
−A A
Therefore, AA-1 = I = A-1A proved.
1.8 SOLUTION OF SIMULTANEOUS EQUATIONS BY
MATRIX ALGEBRA
Matrix algebra can also be used for solving simultaneous equations. In general, a set of n linear equations in n variables X1, X2, X3, … Xn is usually written as follows.
Cn Xn ann n X
a n X a
n C nX a X
a X a
n C nX a X
a X a
= +
+
= +
+ +
= +
+ +
...
2 2 1 1
2 ... 2
2 22 1 21
1 ... 1
2 12 1 11 M
This system can equivalently be written in the matrix equation as follows:
= Cn C C
Xn X a X
M M
L L L
M M
M
L L L
L L L
2 1 2
1
ann an2
an1
a2n a22
a21
a1n a12
11
A X C In a condensed form it can be written as:
AX = C
where A is the n x n matrix of coefficients aij.
X is the column vector of variables (Xj) =
n 2 1
.XXX
and C is column vector
of constant (Ci) =
n 2 1
C. CC
.
Note (1): In order to find a unique solution there must be an equal number of equations and unknowns (or more equations than unknowns) so that a square matrix can be established.
Note (2): A system of two linear equations in two unknowns can be written as follows:
a11 X1 + a12 X2 = C1
a21 X1 + a22 X2 = C2
where X1 and X2 are two unknowns, a11, a12, a22 and a22 are coefficients of unknowns and C1 and C2 are constant.
Similarly, a system of three linear equations in three unknown can be written as follows:
a11 X1 + a12 X2 + a13X3= C1
a21 X1 + a22 X2 + a23 X3 = C2
a31 X1 + a32 X2 +a33X3 = C3
The coefficients matrix A, the vector of variables X, and the vector of constant C can be written as: AX = C
If the inverse of matrix A exists (A-1), then multiplying both sides of above equation by A-1, we get:
A-1AX = A-1C or X = A-1C
Where A-1C represents to the unique vector of solution values.
Example 1: Solve the following system of equations by finding the inverse of the coefficient matrix and also by Cramer’s rule
X1 + 2X2 = 0 4X1 + 9X2 = 1
Solution (a): MATRIX INVERSION METHOD
The given system of equations may be represented in the matrix form as:
AX = C
Where
=
=
=
1 C 0 and X X X 9 , 4
2 1
2
A 1
−
= −
−
1 4
2
1 9 A Since AX = C, then A-1AX = A-1C
∴∴∴∴ X = A-1C
=
−
= −
1 2 - 1 0 1 4
2 9 X
X
2
1 Therefore, X1 = -2 and X2 = 1 (b) CRAMER’S RULE METHOD
Cramer’s rule provides an easy way to find the solution for the unknown variables in a given set of simultaneous equations.
Cramer’s rule may be used to solve the above system of equations:
X1 + 2X2 = 0 4X1 + 9X2 = 1
Solution: Step 1 Express the equation in matrix form as
=
1 0 X X 9 4
2 1
2
1 or AX = C
Step 2 Find the determinant of A: 9 8 1. 9
4 2
1 = − =
=
A
Step 3 In order to solve for X1, replace column1 of matrix A, the coefficients of X1, with the vector of constants C and form a new matrix A1.
= 9 1
2 A1 0
Step 4 Find the determinant of A1 0 2 2 9
2 1 0
1 = = − =−
A
Step 5 In order to solve for X2, replace column 2 of matrix A, the coefficients of X2, with the column vector of constants C and form a new matrix A2.
= 1 4
0 1 A2
Step 6. Find the determinant of A2 1 0 1. 1
4 0 1
2 = = − =
A
Step 7. Using Cramer’s rule we have solution for X1and X2 as follows:
1 X
&
2 1 1
X 1 . 1 2
2
2 1
2 2 1
1 = = − =− = = = ∴X =− =
A and A
A X A
Example 2: Solve the following system of equations by finding inverse of the coefficients matrix.
2X – Y = 2 3Y + 2Z = 16 5X + 3Z = 21 Solution: Since
=
−
21 16 2 3
0 5
2 3 0
0 1 2
Z Y X
or AX = C or X = A-1C
8 10 18 ) 15 ( 0 ) 10 ( 1 ) 9 ( 2 3 0 5
2 3 0
0 1 2
=
−
=
− +
− +
=
−
= A
2 Z and 4 Y 3, X
2 4 3 21 16 2 6 5 15
4 6 10
2 3 9 8 1 Z Y X
C A X
8 6 5 15
4 6 10
2 3 9
6 5 15
4 6 10
2 3 9
1 - 1
=
=
=
∴
=
−
−
−
−
=
∴
=
−
−
−
−
=
=
∴
−
−
−
−
=
−
Q
A A adjA adjA
Example 3. Solve the following system of equations by Cramer’s rule.
2X1 + X2 + X3 = 0 4X1 + 3X2 + 2X3 = 2 2X1 - X2 - 3X3 = 0
Solution.
=
−
− 0
2 0 3
1 2
2 3 4
1 1 2
3 2 1
X X X
Which is of the form AX = C The determinant of the coefficients matrix A is
16 3 0 2
2 2 4
1 0 2
4 3 1 0
2 3 2
1 1 0
8 10 16 14 ) 10 ( 1 ) 16 ( 1 ) 7 ( 2 3 1 2
2 3 4
1 1 2
2 1
−
=
−
=
=
−
−
=
−
=
− +
−
=
− +
−
−
−
=
−
−
=
A A A
8 0 1 2
2 3 4
0 1 2
3 =
−
= A
8 1 8
8 2 16
2 1 8 Therefore, 4
3 3
2 2
1 1
−
− =
=
=
− =
=−
=
−
− =
=
=
A X A
A X A
A X A
∴ The solution is X1 = -1/2 X2 = 2 & X3 = -1 1.9 REVIEW SECTION
MATRIX REPRESENTATION
(i) We assume that the prices (in Rs. Per unit) for commodities X, Y and Z be represented by the prices matrix like:
(
4 5 6)
Z Y X
= P
Quantities of X, Y and Z which are purchased are given by the column matrix like:
= 7 8 9 Q
The cost of the purchases is
( ) [
4 9 5 8 6 7]
(118)7 8 9 6 5
4 = × + × + × =
= PQ
(ii) We assume that a builder has taken contract for 3 A type houses, 9 B type houses and 12 C type houses. This is represented by the row matrix
Q = (3 9 12)
Raw materials required for the construction of houses may be known as Iron, Cement, Bricks, Wood and Labours. The entries in the matrix P given below represent the amounts of each raw material required into each type of house:
P C
B A
=
13 6 20 30 14
4 4 16 10 12
9 2 5 20 10
If the builder wishes to calculate the amount of each raw material required for the construction. Then, this may be calculated as follows:
( )
(
306 510 399 114 219)
13 6 20 30 14
4 4 16 10 12
9 2 5 20 10 12 9 3
=
= QP
It shows that builder should order 306 units of Iron, 510 units of cement, 399 Units of brick, 114 units of wood and 219 units of labour. With the assumption that Iron costs Rs. 9 per unit, Cement costs Rs. 8 per Unit, Brick costs Rs. 6 per unit, Wood costs Rs. 3 per unit and labour costs Rs.
5 per unit. This may be written in the column cost matrix as follows:
Iron Cement Brick Wood Labour
= 5 3 6 8 9
C
Builder can calculate the costs, which he has to pay for the raw materials for each type of house (PC) as follows:
=
=
569 316 331
5 3 6 8 9
13 6 20 30 14
4 4 16 10 12
9 2 5 20 10 PC
Therefore, the cost of raw materials for the A type house is Rs. 331, for B type house Rs. 316 and for the C type house Rs. 569. The total cost of raw materials for all the houses will be:
( )
10665 Rs.
) 10665 ( 569 316 331 12 9 3 )
( =
= PC Q
Exercise 1: The equilibrium conditions for two related markets (Apple & Orange) are given by
18pa - po = 87 -2pa + 36po = 98 Find the equilibrium price for each market.
Solution:
=
−
−
98 87 36
2 1 18
o a
p p
3
&
3 5
5 323
9 323
1
646 1 323
18
18 2
1 A 36
646
1 ⇒ = =
=
∴
=
=
=
−
o
a p
p X
A
Adj A
