Matrices and Determinants for Undergraduates. By Dr. Anju Gupta. Ms. Reena Yadav

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1 Matrices and Determinants for Undergraduates By Dr. Anju Gupta Director, NCWEB, University of Delhi Ms. Reena Yadav Assistant Professor, NCWEB, University of Delhi Matrices A rectangular arrangement consisting of a set of mn numbers( which may be real or imaginary) into m horizontal rows and n vertical columns enclosed by a pair of brackets [ ]. For instance aa 11 aa 12 aa 1nn aa 21 aa 22 aa 2nn aa mm1 aa mm2 aa mmmm The above matrix is of order m n where m is the number of rows and n is the number of columns. It is to be noted here that a matrix is always denoted by capital letters such as A, B, C, D and so on. The above matrix is of order m n i.e. m rows and n columns. Illustration 1 Find the order of each of the following: A= B = C = Solution Matrix A has 2 rows and 2 columns i.e. 2 2 matrix. Matrix B has 2 rows and 3 columns i.e. 2 3 martix. 1

2 Matrix C has 3 rows and 3 columns i.e. 3 3 matrix. TYPES OF MATRICES Rectangular matrix- is the one in which number of rows is not equal to the number of columns. For example , the order of this matrix is Square matrix is the one in which number of rows is equal to the number of columns. For example- 1 2, the order of this matrix is Row matrix is the one which consists of only one row. For example [ ] 1 Column matrix is the one which consists of only one column. For example 2 0 Diagonal matrix is that square matrix in which all entries outside the principal diagonal are zero. For example Scalar matrix is the one in which all elements on principal diagonal are same. For example Identity matrix is the one which has each diagonal element equal to unity. For example Null matrix is the one in which all the elements of a matrix are zero. For example- Triangular matrix it is of two types a) Upper triangular matrix if all the entries below the principal diagonal are zero. For example

3 b) Lower triangular matrix if all entries above the principal diagonal are zero For example ALZEBRA OF MATRICES Addition of matrices If there are two matrices (say A & B) of same order, then addition of matrices denoted by + B is obtained by adding corresponding elements of A & B. For example If A = 1 3 5, B = Then, A+B = = A To generalize we can say that if A= aa 11 aa 12 aa 21 aa 22 & B = bb 11 bb 12 bb 21 bb 22 then, A+B = aa 11 + bb 11 aa 12 + bb 12 aa 21 + bb 21 aa 22 + bb 22 PROPERTIES OF A MATRIX CAN BE SUMMED AS FOLLOWS If there are 3 matrices A, B and C of same order then, i. A+B=B+A ii. (A+B)+C=A+(B+C) iii. k(a+b)= ka+kb iv. A+O=A=O+A v. A+(-A)=A+(-A)=0 vi. A+C=B+C which gives A=B Subtraction of matrices If A = aa 11 aa 12 aa 21 aa 22 & B = bb 11 bb 12 bb 21 bb 22 then, A-B = aa 12 bb 11 aa 12 bb 12 aa 21 bb 21 aa 22 bb 22 Illustration 1 3

4 If A = & B = 2. Find A-B Solution A-B = = Illustration 2 If A = , B = C = , O = Find whether (I) A+B=B+A (II) (A+B)+C=A+(B+C) (III) A+O=O+A Multiplication of matrices We can find the product of two matrices only if they are said to be comfortable for product. In other words, if there are two matrices say A and B then, we can find their product (AB) only if number of columns of matrix A is equal to the number of rows of matrix B. EXAMPLE 4

5 0 2 If A = 1 4, B = then, find A*B matrix SOLUTION The final matrix should be of order 3*4. 0*6+2*8 0*1+2*3 0*-2+2*2 0*4+2*7 A*B = 1*6+(-4)*8 1*1+(-4)*3 1*-2+(-4)*2 1*4+(-4)*7 5*6+3*8 5*1+3*3 5*-2+3*2 5*4+3*7 The final answer comes out to be Properties of matrix multiplication can be summed as follows 1. If there are three matrices A,B and C of order m*n, n*p and n*p respectively then, A(B+C)= AB+AC. This property is called distributive property. 2. If there are three matrices A,B and C of order m*n, n*p and p*s respectively then, (AB)C= A(BC). This property is called associativity property. 3. If A is a square matrix of order n*n and I is unit matrix of the same order i.e. n*n then, AI=A=IA. This property is called existence of multiplicative identity. 4. It does not admits cancellation laws i.e. if AB=AC then, it does not imply that B=C. Illustration 2 8 If A = 3 0 and B = 2 0. Calculate AB. Can you calculate BA. Explain your answer. ( B.A ECO(HONS) DU 1991) Solution 5

6 Matrix A is of order 3*2 and matrix B is of order 2*2 which shows that number of columns of A is equal to number of rows of column B. Therefore, we can find AB and final order of AB matrix will be 3*2. 2*2+8*3 2*0+8*8 AB = 3*2+0*3 3*0+0*8 5*2+1*3 5*0+1* = BA is not possible because for finding BA the order becomes, 2 columns of B and 3 rows of A. Since, number of rows is not equal to number of columns, therefore solution for BA is not possible. DETERMINANTS A determinant is a pure number which is obtained by enclosing the elements of square matrix A within two parallel lines. The determinant of square matrix A is written as AA is equal to aa 11 aa 12.. aa 1nn aa 12 aa 22.. aa 2nn aa 31 aa 32.. aa 3nn aa nn1 aa nn2. aa mmmm Determinant of order one If A= [aa 11 ] i.e.1*1 matrix then, determinant of A is number a 11 itself. For instance, A= [8] 6

7 then, AA = 8. Determinant of order two If A = aa 11 aa 12 aa 21 aa 22 i.e. 2*2 matrix then, AA = aa 11 aa 12 aa 21 aa 22 = a 11 a 22 a 21* a 12. EXAMPLE: If A= 3 2 then, AA 2 = 3 = 3*6 2*5 = = Determinant of order three In order to understand this, we first need to know the concept of minor and co-factor. Minors The minor of any element in AA is a determinant of the sub-matrix obtained from A by deleting the row and column containing that particular element. a 11 a 12 a 13 AA = a 21 a 22 a 23 a 31 a 32 a 33 minor of element a 11 can be denoted by M 11 and can be found by dropping all elements in first row and first column i.e. a 11 a 12 a 13 A = a 21 a 22 a 23 a 31 a 32 a 33 M 11 = aa 22 aa 23 aa 32 aa 33 = a 22 * a 33 - a 23 * a 32 Similarly, if we need to find out minor of a 22,then 7

8 aa 11 aa 12 aa 13 AA = aa 21 aa 31 aa 22 aa 32 aa 23 aa 33 In this case, we will drop 2 nd row and 2 nd column, therefore M 22 = aa 11 aa 13 aa 31 aa 33 = a 11 * a 33 a 31 * a 13 Cofactors The cofactor of any element in AA is the minor of that element in AA with proper sign depending upon the number of row and column in which the element occurs. Cofactor is denoted by Cij. Moreover, C ij = m ij, if i + j is even. = -m ij, if i + j is odd. For instance, aa 11 aa 12 aa 13 AA = aa 21 aa 31 aa 22 aa 32 aa 23 aa 33 Here, to find co-factor C 11, 1 st we need to find minor M 11 M 11 = aa 22 aa 23 aa 32 aa 33 C 11 = M 11 (because i + j is even here). To find cofactor of C 12, we need to first find minor M 12 M 12 = aa 21 aa 23 aa 31 aa 33 C 12 = -M 12 (since, i+j is odd i.e.1+2=3). The assignment of sign is as follows

9 ADJOINT OF A SQUARE MATRIX 1. Adjoint of a square matrix of order 3 aa 11 aa 12 aa 13 A = aa 21 aa 31 aa 22 aa 32 aa 23 is given by aa 33 cc 11 cc 21 cc 31 Adj A = cc 12 cc 13 cc 22 cc 23 cc 32 cc Adjoint of a square matrix of order 2 A = aa 11 aa 12 aa 21 aa is given by simply interchanging the diagonal elements and changing 22 signs of non-diagonal elements. For instance, if A = then, adj A = INVERSE OF A SQUARE MATRIX Inverse of a square matrix can be found by two methods by- a) Adjoint method. b) Elementary row operation. Adjoint method A square matrix A of order n is invertible if there exists a square matrix B of same order such that AB=BA=I. in such a situation, we say that B is inverse of A ( i.e., A -1 = B). We have two types of matrices : a) Singular matrix- a square matrix A is said to be singular if AA = 0. b) Non-singular matrix- a square matrix A is said to be non-singular if AA 0. NOTE:- only if AA 0 then, only A -1 exists which is given by A -1 = 11 AA.adj (A) Illustration Find inverse of matrix A = (D.U. BA Eco(H) 1987) 9

10 Solution AA = 4(3-0)-(-2)(7-6)+ 1(0-6)= 4*3+2*1-6=12+2-6=14-6=8 0 Therefore, A -1 exist and is given by A -1 = 11 AA.adj (A). Minors for above matrix are : M 11 = 3 M 12 =7-6=1 M 13 =-6 M 21 =-2 M 22 =4-2=2 M 23 =4 M 31 =-9 M 32 =12-7=5 M 33 =12+14=26 Co-factors for the above are C 11 = 3 C 12 =-1 C 13 =-6 C 21 =2 C 22 =2 C 23 =-4 C 31 =-9 C 32 =-5 C 33 =26 cc 11 cc 21 cc 31 Therefore, Adj A = cc 12 cc 22 cc 32 cc 13 cc 23 cc 33 = A -1 = adj (A) = AA Elementary row operations.i.e. Gauss elimination method The working rules of this method are:- a) The system of equation need to be written in the form AX= B b) Then, we need to construct augmented matrix i.e. A : B (by introducing the vector B to the right of coefficient matrix A). c) The third step is to use elementary row operation to transform [A:B] matrix into upper triangular matrix (i.e. the one in which entries below principal diagonal are 0). d) Last step is to rewrite the system of equation from reduced matrix and solve it. Illustration Solve the following system of equation by elementary row operation method. X + 2Y +3Z=4 2X +3Y + 8Z=7 10

11 X-Y-9Z=1 (this question has been taken from taxmann s business mathematics book) Solution Step 1 : AX = B xx 4 yy = 7 zz 1 Step 2: presenting it in the form of [AA: BB] matrix Step3: using elementary row operation [AA: BB] = R 2 -> R 2 2R R 3 -> R 3 R R 3 -> R 3 3R Step 4: Writing the above matrix in the equation form x+2y+3z = 4 eq.(1) -y+2z = -1 eq.(2) -18z = 0 eq.(3) From eq.(3), z = 0 Putting value z in eq.(2) 11

12 -y + 2*0 = -1 -y = -1 => 1 Now, putting value of y and z in eq. (1) x+2y+3z = 4 x+2*1+3*0 = 4 x+2 = 4 x = 4-2 => x = 2 therefore, solution is x = 2, y = 1 and z = 0. Application based question Question: In an engineering workshop there are 10 machines for drilling, 8 machines for turning and 7 machines for grinding. Three types of brackets are made. Type I bracket require 0 minute for drilling, 5 minute for turning and 4 minute for grinding. The corresponding times for type II and III brackets are 3, 3, 2 and 3, 2, 2 minutes respectively. How many brackets of each type should be produced per hour so that all the machines remain fully occupied during an hour? Solve by matrix algebra. [B.Com(H),D.U 2008,2010] Solution: The given data can be summarized as : Brackets Drilling Turning Grinding Type I Type II Type III Time(in min) 10*60=600 8*60=480 7*60=420 Let x, y, z denotes the number of brackets produced of each type. Then we have 0x+3y+3z=600 5x+3y+2z=480 4x+2y+2z=420 Presenting the matrix in the form of AX=B xx 600 yy = 480 zz

13 AA = 0(6-4)-3(10-8)+ 3(10-12)= -12 0, therefore, A -1 exist adjoint (A) = then, A -1 = AA.adj (A) = X = A -1 B = = Therefore, 5 brackets of type I should be produced, of type II 55 brackets must be produced and 145 of type III. 13

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