2018-G11-Math-E

31.. MQautaridcreastiacndEqDueatetiromninsants eLearn.Punjab(ii) The matrix form of the system 2x1x1++24x2x2==412 is eLearn.Punjab version: 1.1    1  241 x1  = 4 A 1 2= = x2 12 A 2 2 and = 2 = 4 - 4 = 0, so A-1 does not exist. 4Multiplying the irst equation o f the above system by 2, we have 2x1 + 4x2 = 8 but 2x1 + 4x2 = 12which is impossible. Thus the system has no solution. Exercise 3.1==1. If A 12 53 and B 16 74 , then show that i) 4A - 3A = A ii) 3B - 3A = 3(B - A)2. If A = 1i 0  , show that A4 = I2. -i3. Find x and y if=i) x-+33 3y1- 4 2 1 ii) x +3 1   y 1  -3 2  -3 3y - 4 -3 2x==4. If A -11 2 32 and B 10 3 22 , ind the following matrices; 0 -1 i) 4A - 3B = A ii) A + 3(B - A)5. Find x and y If 12 0 3x + 2 10 x -y1 =14 -62 13 y 2 17

13.. MQautaridcreastiacnEdqDueatetiromnisnants eLearn.Punjab6. If A = [aij]3%3, ind the following matrices; eLearn.Punjab version: 1.1 i) λ (µ A) = (λµ ) A ii) (λ + µ ) A =λ A + µ A ii) λ A - A = (λ -1) A 7. If A = [aij]2%3 and B = [bij]2%3, show that λ( A + B) = λ A + λB.==8. If A a1 b2 and A2 00 00 , ind the values of a and b.==9. If A 1a -b1 and A2 10 10 , ind the values of a and b.==10. If A 10 -1 12 and B 12 3 -01 , then show that (A + B)t = At + Bt. 3 2 1 1 -63311. Find A3 if A =  5 2 -2 -112. Find the matrix X if;. i) X -52 12 1-21 53 ii) -52 12 X = 52 11013. Find the matrix A if, i) 053 -101 A = 703 -027 ii) -21 -21 A = 03 -3 -87 3 r cosφ 0 -sinφ   cosφ 0 sinφ    -r 14. Show that 0 r 0 0 1 0 = rI3. sinφ 0 cosφ sinφ 0 cosφ r r 18

31.. MQautaridcreastiacndEqDueatetiromninsants eLearn.Punjab eLearn.Punjab3.4 FieldA set F is called a ield if the operations of addition ‘+ ’ and multiplication ‘. ’ on F satisfythe following properties written in tabular form as:Addition Multiplication Closurei) For any a,b U F, For any a,b U F, a+bUF a.b U F Commutativityii) For any a,b U F, For any a,b U F,a+b=b+a a.b = b.a Associativityiii) For any a,b,c U F, For any a,b,c U F,(a + b) + c = a + (b + c) (a.b).c = a.(b.c) Existence of Identityiv) For any For anya U F ,\ 0 U F such that a U F ,\ 1 U F such thata+0=0+a=a a.1 = 1.a = a Existence of Inversesv) For any v) For any a U F, a ≠ 0 ∃1 U F such that a U F ,\ -a U F such that a + (-a ) = (-a ) + a = 0 a a=.( 1 ) (=1 ).a 1 Distributivity aavi) For any a,b,c U F, D1 : a(b + c) = ab + ac D2 : (b + c)a = ba + caAll the above mentioned properties hold for Q, _, and C.3.5 Properties of Matrix Addition , Scalar Multiplication and Matrix Multiplication.are true: If A, B and C are n%n matrices and c and d are scalers, the following properties version: 1.1 19

13.. MQautaridcreastiacnEdqDueatetiromnisnants eLearn.Punjab1. Commutative property w.r.t. addition: A + B = B + A eLearn.Punjab version: 1.1Note: w.r.t. is used for “with respect to”.2. Associative property w.r.t. addition: (A + B) + C - A + (B + C)3. Associative property of scalar multiplication: (cd)A = c(dA)4. Existance of additive identity: A + O = O + A - A (O is null matrix)5. Existance of multiplicative identity: IA = AI = A (I is unit/identity matrix)6. Distributive property w.r.t scalar multiplication: (a) c(A + B) = cA + cB (b) (c + d)A = cA + dA7. Associative property w.r.t. multiplication: A(BC) = (AB)C8. Left distributive property: A(B + C) = AB + AC9. Right distributive property: (A + B)C = AC + BC10. c(AB) = (cA)B = A(cB)Examp=le 1: Find AB and BA if A -132 0 162 and B= 112 -1 031 4 3 0 -2=Solution : AB -123 0 162 112 -1 031 4 3 0 -2= 132×××111+++400×××222+++261×××111 2 × (-1) + 0 × 3 +1× (-2) 132×××000+++400×××(((---111)))+++261 ××× 333 1× (-1) + 4 × 3 + 2 × (-2) 3× (-1) + 0 × 3 + 6 × (-2) (1)= 1931 -4 1328 7 -15=BA 112 -1 -031 132 0 162 3 4 -2 0 20

31.. MQautaridcreastiacndEqDueatetiromninsants eLearn.Punjab eLearn.Punjab 1× 2 + (-1) ×1 + 0 × 3 1× 0 + (-1) × 4 + 0 × 0 1×1 + (-1) × 2 + 0 × 6  12××22++ 2 × 0 + 3× 4 + (-1) × 0 12××11++ (3-×2)2×+2(-+13) ×× 66= 3×1 + (-1) × 3 1× 0 + (-2) × 4 + 3× 0 (-2) ×1 + 3× 3 (2) = 194 -4 1-251 12 -8Thus from (1) and (2), AB ≠ BANote: Matrix multiplication is not commutative in general  2 -1 3 0 Exam=ple 2: If - 1 0 4 2 ,then ind AAt and (At). -3 5 2 -1Solution : Taking transpose of A, we have  2 1 -3  , At = -31 0 5 so 4 2  0 -2 -1 2 -1 3 -021 -0231 1 --5231=AAt - 1 0 4 0 2 4 -3 5 -2 = -246++-015+++1926+++000 2 + 0 +12 + 0 -6 - 5 + 6 + 120 1 + 0 +16 + 4 -3 + 0 + 8 + -3 + 0 + 8 + 2 9 + 25 + 4 + = 11-445 14 -3795 21 7 version: 1.1 21

13.. MQautaridcreastiacnEdqDueatetiromnisnants eLearn.Punjab eLearn.Punjab  2 1 -3 -123 -1 3 -021 which is A,( )=As A=t -31 04  0 4 5- , so At t 2 2 5  0 -2 -1That is, (At)t = A. (Note that this rule holds for any matrix A.) Exercise 3.21. If A = [aij]3%4, then show thati) I3A = A ii) AI4 = A2. Find the inverses of the following matrices. i) 23 -11 ii) --42 53 iii) 2ii -ii iv) 62 133. Solve the following system of linear equations.i) 2=x1 - 3x2 45 ii) 4=x1 + 3x2 75 iii) 3x1 - 5 y ==1-3 =5x1 + x2 =3x1 - x2 -2x + y==4. If A -131 -1 524 , B -121 1 -141 and C =  1 3 --021 , then ind 2- 3 1 2 2 3 4 0i) A - B ii) B - A iii) (A - B) - C iv) A - (B - C)=5. If A =1i -2ii , B -2ii 1i and C = -2ii -i1 , then show thati) (AB)C = A(BC) ii) (A + B)C = AC + BC6. If A and B are square matrices of the same order, then explain why in general;i) (A + B)2 ≠ A2 + 2AB + B2 ii) (A - B)2 ≠ A2 - 2AB + B2iii) (A + B )(A - B ) ≠ A2 - B2 version: 1.1 22

31.. MQautaridcreastiacndEqDueatetiromninsants eLearn.Punjab eLearn.Punjab  2 -1 3 0 =7. If A - 1 0 4 2 then ind AAt and AtA. -3 5 2 -18. Solve the following matrix equations for X:i) 3=X=- 2A = B if A -21 3 -52 and B 52 -3 -11 , 1 4ii) 2=X=- 3A = B if A -12 -1 52 and B 43 -1 10 4 29. Solve the following matrix equations for A: (i) 42 23 A - -21 -32 =-31 -64 (ii) A43 12 - -31 12 =-21 053.6 Determinants The determinants of square matrices of order n83, can be written by followingthe same pattern as already discussed. For example, if n = 4 A = aaaa13421111 a12 a13 a14  , then the determinat of =A A= a11 a12 a13 a14 a22 a23 a24  a21 a22 a23 a24 a32 a33 a34 a31 a32 a33 a34 a42 a43 a44 a41 a42 a43 a44 Now our aim is to compute the determinants of various orders. But before describinga method for computation o f determinants of order n83, we introduce the followingdeinitions. version: 1.1 23

13.. MQautaridcreastiacnEdqDueatetiromnisnants eLearn.Punjab eLearn.Punjab3.6.1 Minor and Cofactor of an Element of a Matrix or its DeterminantMinor of an Element: Let us consider a square matrix A of order 3 .Then the minor ofan element aij, denoted by .Mij is the determinant of the (3 - 1) % (3 - 1) matrix formed bydeleting the ith row and the jth column of A(or|A|).For example, ifA =  a11 a12 a13  , then the matrix obtained by deleting the irst row and the second a21 a22 a23 a31 a32 a33  ----------------------------------------column of A is aa3211 a23  (see adjoining igure) aa1211 a12 a13  and its determinant is the a33 a31 a22 a23  a32 a33 minor of an, that is,M12 = a21 a23 a31 a33Cofactor of an Element: The cofactor of an element aij denoted by Aij is deined by Aij = (-1)i+j %Mijwhere Mij is the minor of the element aij of A or |A|. For example, A12 = (-1)1+2 M12 = (-1)3 a21 a23 = - a21 a23 a31 a33 a31 a333.6.2 Determinant of a Square Matrix of Order n83: The determinant of a square matrix of order n is the sum of the products of eachelem ent of row (or column) and its cofactor. version: 1.1 24

31.. MQautaridcreastiacndEqDueatetiromninsants eLearn.Punjab eLearn.Punjab aa1211 a12 a13  a1 j  a1n  a22 a23  a2 j  a2n If A =  a31 a32 a33  a3 j  a3n  , then       ai1 ai 2 ai3  aij ain      an1 an2 an3  anj  ann  |A| = ai1Ai1 + ai2Ai2 + ai3Ai3 + .... + ainAin for i = 1, 2, 3,...., n or |A| = a1j A1j + a2j A2j + a3j A3j + .... + anj Anj for j = 1, 2, 3,...., nPutting i = 1, we have|A| = a11 A11 + a12 A12 + a13 A13 + .... + a1n A1n which is called the expansion of |A|by the irstrow.If A is a matrix of order 3, that is, A = aa1211 a12 a13  , then: a22 a23  a31 a32 a33  |A| = ai1Ai1 + ai2Ai2 + ai3Ai3 + .... + ainAin for i = 1, 2, 3 (1) (2)or |A| = a1j A1j + a2j A2j + a3j A3j + .... + anj Anj for j = 1, 2, 3For example, for i = 1, j = 1 and j = 2, we have|A| = a11 A11 + a12 A12 + a13 A13 (i)or |A| = a11 A11 + a21 A21 + a31 A31 (ii)or |A| = a12 A12 + a22 A22 + a32 A32 (iii)(iii) can be written as:|A| = a12(-1)1+2 M12 + a22(-1)2+2 M22 + a32(-1)3+2 M32i.e., |A| = -a12M12 + a22M22 - a32M32 (iv)Similarly (i) can be written as |A| = a11M11 - a12M12 - a13M13 (v)Putting the values of M11.M12: and M13 in (iv). we obtainA = a11 a22 a23 - a12 a21 a23 + a13 a21 a22 a32 a33 a31 a33 a31 a32or |A| = a11(a22a33 - a23a32) - a12(a21a33 - a23a31) + a13(a21a32 - a22a31) (vi)or |A| = a11a22a33 + a12a23a31 + a13a21a32 - a11a23a32- a12a21a33- a13a22a31 (vi)’ version: 1.1 25

13.. MQautaridcreastiacnEdqDueatetiromnisnants eLearn.Punjab eLearn.Punjab The second scripts of positive terms are in circular order ofanti-clockwise direction i.e., these are as 123, 231, 312 (adjoining igure)while the second scripts of negative terms are such as 132, 213, 321.An alternative way to remember the expansion of the determinant |A| given in (vi)’ isshown in the igure below.Example 1: Evaluate the determinant of A= -142 -2 123 3 -3Solution : 1 -2 3 A = -2 3 1 4 -3 2Using the result (v) of the Art.3.6.2, that is, |A| = a11 M11 - a12M12 + a13M13 , we get,=A 3 1 - (-2) -2 1 + 3 -2 3 1 -3 24 24 -3 = 1[6 - 1(-3)] + 2[(-2).2 -1.4] + 3[(-2)(-3) -12] = (6 + 3) + 2(-4 - 4) + 3(6 -12) = 9 -16 -18 = -25 version: 1.1 26

31.. MQautaridcreastiacndEqDueatetiromninsants eLearn.Punjab eLearn.Punjab  1 -2 3Example 2: Find the cofactors A12, A22 and A32 if A= -2 3 1 and ind |A|.  4 -3 2Solution : We irst ind M12, M22 and M32, M12 = -2 1 =-4-4=-8; M 22 = 1 3 = 2 -1 2 = -10 and M 32 = 1 3 = 1- (-6) =7 4 2 4 2 -2 1Thus A12 = (-1)1+2 M12 = (-1)(-8 ) = 8 ; A22 = (-1)2+2 M22 = 1(-10)= -10 A32 = (-1)3+2 M32 = (-1)(7 ) = -7 ;and |A| = a12A12 + a22 A22 + a32 A32 = (-2)8 + 3(-10) + (-3)(-7) = -16 - 30 + 21 = -25Note that a11A12 + a21 A22 + a31 A32 = 1(8) + (-2)(-10) + 4(-7) = 8 + 20 - 28 = 0and a13A12 + a23 A22 + a33 A32 = 3(8) + 1(-10) + 2(-7) = 24 - 10 - 14 = 0Similarly we can show that a11A13 + a21 A23 + a31 A33 = 0; a11A21 + a12 A22 + a13 A23 = 0; and a11A31 + a12 A32 + a13 A33 = 0;3.7 Properties of Determinants which Help in their Evaluation1. For a square matrix A, |A| =IAtI2. If in a square matrix A, two rows or two columns are interchanged, the determinant of the resulting matrix is -|A|.3. If a square matrix A has two identical rows or two identical columns, then |A| = 0 .4. If all the entries of a row (or a column) of a square matrix A are zero, then |A| = 0 .5. If the entries of a row (or a column) in a square matrix A are multiplied by a number k U_, then the determinant of the resulting matrix is k |A|.6. If each entry of a row (or a column) of a square matrix consists of two terms, then itsdeterminant can be written as the sum of two determinants, i.e., if aa2111 +  =B + b11 a12 a13 , then b21 a22 a23 a32 a31 + b31 a33  version: 1.1 27

13.. MQautaridcreastiacnEdqDueatetiromnisnants eLearn.Punjab eLearn.Punjab a11 + b11 a12 a13 a11 a12 a13 b11 a12 a13 + B =a21 b21 =a22 a23 a21 +a22 a23 b21 a22 a23 a33 b31 a32 a33 a31 + b31 a32 a33 a31 a327. If to each entry of a row (or a column) of a square matrix A is added a non-zero multiple of the corresponding entry of another row (or column), then the determinant of the resulting matrix is |A|.8. If a matrix is in triangular form, then the value of its determinant is the product of the entries on its main diagonal. Now we prove the above mentioned properties of determinants.Proporty 1: If the rows and columns of a determinant are interchanged, then the valueof the determinant does not change. For example. a11 a12 = a11a22 - a12a21 = a11a22 - a21a12 = a11 a21 (rows and columns are interchanged) a21 a22 a12 a22Property 2: The value of a determinant changes sign if any two rows (columns) areinterchanged. For example, a11 a12 = a11a22 - a12a21 a21 a22and a12 a11 = a12a21 - a11a22 = -(a11a22 - a12a21) (columns are interchanged) a22 a21Property 3: If all the entries in any row (column) are zero, the value of the determinant iszero. For example,0 a12 a13 = 0 a22 a23 - 0 a12 a13 + 0 a12 a13 = 0 (expanding by C )0 a22 a23 a32 a33 a32 a33 a22 a230 a32 a33Property 4: If any two rows (columns) of a determinant are identical, the value of thedeterminant is zero. For example, version: 1.1 28

31.. MQautaridcreastiacndEqDueatetiromninsants eLearn.Punjab eLearn.Punjaba bca b c = 0, (it can be proved by expanding the determinant)x yzProperty 5: If any row (column) of a determinant is multiplied by a non-zero number k, thevalue of the new determinant becomes equal to k times the value of original determinant.For example,A = a11 a12 , multiplying irst row by a non-zero number k, we get a21 a22ka11 ka12 = ka11a22 - ka12a21 = k (a11a22 - a12a21) = k a11 a12a21 a22 a21 a22Property 6: If any row (column) of a determinant consists of two terms, it can be written asthe sum of two determinants as given below: a11 + b11 a12 a13 a11 a12 a13 b11 a12 a13 a21 + b21 a22 a23 = a21 a22 a23 + b21 a22 a23 (proof is left for the reader) a31 + b31 a32 a33 a31 a32 a33 b31 a32 a33Property 7: If any row (column) o f a determinant is multiplied by a non-zero number k andthe result is added to the corresponding entries of another row (column), the value of thedeterminant does not change. For example, a11 a12 = a11 a12 + ka11 (k multiple of C1 is added to C2) a21 a22 a21 a22 + ka21It can be proved by expanding both the sides. Proof is left for the reader.Example 3: If A = -1235 -2 3 -0241 , evaluate |A| 5 1 1 0 -3 -1 version: 1.1 29

13.. MQautaridcreastiacnEdqDueatetiromnisnants eLearn.Punjab eLearn.PunjabSolution : 2 -2 3 4 A = 3 1 5 -1 -5 -3 1 0 1 -1 0 2 0030 =0 4 5 -7 By R1 + (-2 )R4, R2 + (-3 )R4 and R3 + 5R4 0 -8 1 10 1 -1 0 2Expanding by irst column, we have|A| = 0.A11 + 0.A21 + 0.A31 + 1.A41 030 030 =(-1)4+1 × 4 5 -7 =(-1) 4 5 -7 -8 1 10 -8 1 10 = ( -1)(- 3)[4 % 10 - (-7)(-8)] = 3(40 - 56) = - 48 x a+x b+cExample 4: Without expansion, show that x b + x c + a =0 x c+x a+bSolution : Multiplying each entry of C1 by -1 and adding to the corresponding entry of C2 i.e.,by C2 + (-1)C1, we getx a + x b + c x a + x + (-1)x b + cx b + x c +=a x b + x + (-1)x c + ax c + x a + b x c + x + (-1)x a + b = x a b+c 1 a b+c  by property 5 or  x b c +=a x1 b c+a taking x common x c a+b c a+b 1 from C1 = 1 a + (b + c) b+c  adding the entries of C3 to the  x1 b + (c + a) c+a, corresponding entries of C2 c + (a + b) a+b 1 version: 1.1 30

31.. MQautaridcreastiacndEqDueatetiromninsants eLearn.Punjab 1 1 b+c eLearn.Punjab = x(a + b + c) 1 1 c + a , (by property 5) version: 1.1 1 1 a+b = x(a + b + c).0 (a C1 and C2 are identical or by property 3) =0 x 011Example 5: Solve the equation 0 1 -1 -1 = 0 1 -2 3 4 -2 x 1 -1Solution: By C3 + C2 and C4 + C2, we have x0 1 1 01 0 0 =0 1 -2 1 2-2 x x +1 x -1 x1 1Expanding by R2, we get 1 1 2 = 0 (a (-1)2+2 = 1) -2 x +1 x -1 x1 1By R3 + 2R2, we get 1 1 2 =0 0 x+3 x+3 x11or (x + 3) 1 1 2 =0 (by taking x + 3 common from R3) 011 x11⇒ x + 3 = 0 or 1 1 2 = 0 011⇒ x = -3 or x = 0 (a R1 and R3 are identical if x = 0)Thus the solution set is {-3, 0}. 31

13.. MQautaridcreastiacnEdqDueatetiromnisnants eLearn.Punjab eLearn.Punjab3.8 Adjoint and Inverse of a Square Matrix of Order n83If A = aa1211 a12 a13  , then the matrix of co-factors of A =  A11 A12 A13  , a22 a23 A21 A22 A23 a31 a32 a33   A31 A32 A33 and adj A=  A11 A21 A31  , A12 A22 A32  A13 A23 A33 Inverse of a Square Matrix of Order n83: Let A be a non singularsquare matrix of order n. If there exists a matrix B such thatAB = BA = In, then B is called the multiplicative inverse of A and is denoted by A-1. It is obviousthat the order of A-1 is n % n.Thus AA-1 = In and A-1A = In. If A is a non singular matrix, thenA-1 = 1 adjA A 1 0 2Example 6: Find A-1 if A = 0 2 1 1 -1 1Solution: We irst ind the cofactors of the elements of A. A11 =(-1)1+1 2 1 =1.(2 +1) =3, A12 =(-1)1+2 0 1 =(-1)(-1) =1 -1 1 1 1 A13 =(-1)1+3 0 2 =1.(0 - 2) =-2, A21 =(-1)2+1 0 2 =(-1)(0 + 2) =-2 1 -1 -1 1 A22 =(-1)2+2 1 2 =1.(1 - 2) =-1, A23 = (-1)2+3 1 0 = (-1)(-1 - 0) =1 1 1 1 -1A31 =(-1)3+1 0 2 =1.(0 - 4) =-4, A32 =(-1)3+2 1 2 =(-1)(1 - 0) =-1 2 1 0 1 version: 1.1 32

31.. MQautaridcreastiacndEqDueatetiromninsants eLearn.Punjab eLearn.Punjab A33 =(-1)3+3 1 0 =1.(2 - 0) =2 0 2Thus [ Aij ]3×3 = AA1211 A12 A13  = 32- 1 -12and adj A=  A31 A22 A23 1- A32 A33  -4 -1 2  -2 --241 [ Ai′j ]3×=3 -132 -1 (a A’ij = Aji for i, j=1, 2, 3 ) 1Since |A|= a11A11 + a12A12 + a13A13So = 1(3) + 0(1) + 2(-2) = 3 + 0 - 4 = -1,  3 -2 -4 -3 2 4  A-1 =1 =-11 -12 -21 =-21 -12 A adj A -1 1 1 -1Example 7: If A = -121 -241 and  1 13 then verify that (AB)t = BtAt 2Solution: so AB =-121 -241 =-12 13 =-211-+-824 -363+-+142 =--475 -751 , ( AB )t = --51 -7 54 7and Bt At = 13 -12 -21 1 -21  At = -21 1 -21 and Bt = 13 -12  4 4== --31 - 4 1-8 2+ 12 --51 -7 54 + 2 3+4 6- 7 Thus (AB)t = Bt At version: 1.1 33

13.. MQautaridcreastiacnEdqDueatetiromnisnants eLearn.Punjab Exercise 3.3 eLearn.PunjabEvaluate the following determinants. = version: 1.11. i) 5 -2 -4 5 2 -3 1 2 -3 3 -1 -3 ii) 3 -1 1 iii) -1 3 4 -2 1 2 -2 1 -2 -2 5 6 a+l a-l a 1 2 -2 2a a aiv) a a + l a - l v) -1 1 -3 vi) b 2b b a-l a a+l 2 4 -1 c c 2c2. Without expansion show that 678 2 3 -1 123=i=)=3 4 5 0 ii) 1 1 0 0 iii) 4 5 6 0 234 2 -3 5 7893. Show that a11 a12 a13 + a13 a11 a12 a13 a11 a12 a13i) a21 a22 a=23 + a23 a21 a22 a23 + a21 a22 a23 a31 a32 a33 + a33 a31 a32 a33 a31 a32 a33 2 3 0 210 a+l a aii) 3 9 6 =9 1 1 2 iii) a a l+ a l=2 (3a l+) 2 15 1 2 5 1 a a a+l 111 111 b+c a ai=v) x y z x =y +z v) b c a b 4abc yz zx xy x2 y2 z2 c c a+b b -1 a r cosφ 1 - sinφvi) a b 0 =a3 b3 vii) 0 1 0+ r 1ab r sinφ 0 cosφ a b+c a+bviii) b c+a b + c = a3 + b3 + c3 - 3abc a+b c+a c a+λ b c + λ =λ λ +λ 34

φ φ -31.. MQautaridcreastiac+ndEqDueat+etiromninsants eLearn.Punjab + += + + eLearn.Punjab c a+b c+a version: 1.1a+λ b cix) a b + λ c =λ2 (a + b + c + λ)a b c+λ 111x) a b c =(a - b)(b - c)(c - a) a2 b2 c2 b + c a a2xi) c + a b=b2 (a + b + c)(a - b)(b - c)(c - a) a + b c c24. If If A =-102 -222 -103 and B-= -532 -2 -542 ,then f-ind; 1 1i) A12, A22, A32 and |A| ii) B21, B22, B23 and|B|5. Without expansion verify that 1 a2 aa b +g 1 bc 1 2 3xi) b =g + a =1 0 ii=) 2 3 6x 0 iii) 1 b2 b 0 cag a+b 1 3 5 9x 1 c2 c ab a-b b-c c-aiv) b - c c - a a - b =0 c-a a-b b-c bc ca abv) 1 1 1 = 0 ab c ab cmn l l2 1 l2 l3 = 35

13.. MQautaridcreastiacnEdqDueatetirom=nisnants eLearn.Punjab ab c eLearn.Punjab version: 1.1 mn l l 2 1 l 2 l3vi) nl m m2 = 1 m2 m3 lm n n2 1 n2 n3 2a 2b 2cvii) a + b 2b b + c a + c b + c 2c7 2 6 7 2 7 7 2 -1 -a 0 cviii) 6 3 2= 6 3 5 + 6 3 -3 ix) 0 a -b-3 5 1 -3 5 -3 -3 5 4 b -c 06. Find values of x if 1 x -1 3 121 3 1x 1+ x 1=2 0 iii) 2 x 2 0= i) -1 3 -4 =30 i-i) 2 -2 x 36x x 107. Evaluate the following determinants:34 2 7 2 3 1 -1 -3 9 1 125 0 3 40 2 1 0 3 -1 2i) 1 2 -3 5 ii) 5 2 -1 6 iii) 9 7 -1 14 1 -2 6 3 -7 2 -2 -2 0 1 -1 x1118. Show that 1 x 1 1 =(x + 3)(x - 1)3 11x1 111x9. Find |AAt| and |AtA| if==i) A 23 2 -31 ii) A 1223 1143 1 36

31.. MQautaridcreastiacndEqDueatetiromninsants eLearn.Punjab10. If A is a square matrix of order 3, then show that |KA|= k3|A|. eLearn.Punjab11. Find the value of λ if A and B singular. version: 1.1 4 λ 3 B = 8523 1 2 1103 A = 72 3 16 , 5 2 0 3 -1 2 λ12. Which of the following matrices are singular and which of them are non singular? i) 103 0 -431 ii) 122 3 -501 iii) 1132 1 2 --2431 1 2 -1 2 1 3 1 -3 -1 3 2 1 013. Find the inverse of A = 1 1 0 and show that A-1 A = I3 2 -3 514. Verify that (AB)-1 = B-1A-1 if==i) A =-11 02 , B =-43 -11 ii) A 52 12 , B 42 1315. Verify that (AB)t = Bt At and if A 10 -1 12 and B 103 -121 316. If A= 2 -1 v( eA=r-if23y th-1a1t ( A-1 )t =( At )-1 3 1 17. If A and B are non-singular matrices, then show that i) (AB)-1 = B-1A-1 ii) (A-1)-1 = A 37

13.. MQautaridcreastiacnEdqDueatetiromnisnants eLearn.Punjab eLearn.Punjab3.9 Elementary Row and Column Operations on a Matrix Usually a given system of linear equations is reduced to a simple equivalent system byapplying in turn a inite number of elementary operations which are stated as below:1. Interchanging two equations.2. Multiplying an equation by a non-zero number.3. Adding a multiple of one equation to another equation.Note: The systems of linear equations involving the same variables, are equivalent if they have the same solution. Corresponding to these three elementary operations, the following elementary rowoperations are applied to matrices to obtain equivalent matrices.i) Interchanging two rowsii) Multiplying a row by a non-zero numberiii) Adding a multiple of one row to another row.Note: Matrices A and B are equivalent if B can be obtained by applying in turn a inite number of row operations on A.Notations that are used to represent row operations for I to III are given below:Interchanging Ri and Rj is expressed as RiGRj .k times Ri, is denoted by kRi D R’iAdding k times Rj to Ri is expressed as Ri + kRj D R’i (R’i is the new row obtained after applying the row operation).opeFIrfoaArtiRoeqnBustihavenanldeBnntoRmtAa.atiAtorlniscosestifhAAataRnaBrdeaBnu,dsweBedRwfoCrr,itttehhAeenmRA..BR.C. Now we state the elementary columni) Interchanging two columns Ci G Cjii) Multiplying a column by a non-zero number kCi D C’iiii) Adding a multiple of one column to another column Ci + kCj D C’i Consider the system of linear equations; x+ y +=+482zzz-===1123 which can be written in matrix forms as 2x - y3x + 5y version: 1.1 38

31.. MQautaridcreastiacndEqDueatetiromninsants eLearn.Punjab eLearn.Punjab132 1 842  x  =1-123 or [x y z ] 112 2 543= [1 12 -3] -1 y -1 5 z 8that is, AX = B (i) X t At = Bt (ii) 1 1 2 x  1 A =-23 84=, X and=B 1-23where 1 y 5 z A is called the matrix of coeicients. Appending a column of constants on the left of A, we get the augmented matrix of thegiven system, that is,1 1 2  1 32 1-23 -1 8  (Appended column is separated by a dotted line segment) 5 4 Now we explain the application of elementary operations on the system-of linear equationsand the application of elementary row operations on the augmented matrix of the systemwriting them side by side. =32xx x+++5-yyy+++482zzz-==11-23 12 1 2  1  1 8  12 3 5 4  -3Adding -2 times the irst equation to the (By R2+ (-2)R1 DR’2 andsecond and -3 times the irst equation to R3 + (-3)R1D R’3, we get)the third, we get =x-+23yyy -++ 224zzz-==1160- R 10 1 2  1  0 3 4  10 2 -2  -6 version: 1.1 39

13.. MQautaridcreastiacnEdqDueatetiromnisnants eLearn.Punjab eLearn.PunjabInterchanging the second and third equations, we have (By R2 GR3, we get)=-x32+yy+-y 4+2zz2-=z 1=061- R 10 1 2  -16 0 2 -2  3 4  10 Multiplying the second equation by 1 , we get By 1 R2 DR’2, we get. 2 2-x3+yyy+-+4z2z =z==-1310 R 100 1 2 1-103 1 -1  -3 4Adding 3 times the second equation to By R3 + 3R2DR’3, we obtain, the third, we obtain,x + y -+zz2==z 1=-31 10 1 2 -13 y 0 1 1  R 0 -1  1The given system is reduced to the triangular form which is so called because on the leftthe coeicients (of the terms) within the dotted triangle are zero. Putting z = 1 in y - z = -3 , we have y - 1 = - 3 ⇒ y = - 2 Substiliting z = 1, y = -2 in the irst equation, we get x + (-2 ) + 2(1) = 1 ⇒ x = 1 Thus the solution set o f the given system is {(1, -2,1)}. Appending a row of constants below the matrix At, we obtain theaugmented matrix for the matrix equation (ii), that is  1 2 3  1 -1 5 2 8 4 1 -3  12Now we apply elementary column operations to this augmented matrix. version: 1.1 40

31.. MQautaridcreastiacndEqDueatetiromninsants eLearn.Punjab eLearn.Punjab 1 2 3 1 0 0  12  12 -22 -1 5 C -3 By C2 + (-2)C1 D C’2 and 8 4 C3 + (-3)C1 D C’3  4    1 12 -3  1 10 -6 1 0 0 1 0 0  2 -3 C 12 -43 By  1 1 1 2 C 2 -2 4  By C2 ↔ C3 -1 C2 → C'2     1 -6 10  1 -3 101 0 0C 12 10  By C3 + 3C2 → C'3 1 -1  1 -3 1 Thus [x y z] 112 0 100 = [1 -3 1] 1 -1or [x + y + 2z y - z z] = [1 -3 1] x + y -+zz2==z 1=-31 y⇒Upper Triangular Matrix: A square matrix A = [aij] is called upper triangular if all elementsbelow the principal diagonal are zero, that is, aij = 0 for all i > jLower Triangular Matrix: A square matrix A = [aij] is said to be lower triangular if all elementsabove the principal diagonal are zero, that is, aij = 0 for all i < j version: 1.1 41

13.. MQautaridcreastiacnEdqDueatetiromnisnants eLearn.Punjab eLearn.PunjabTriangular Matrix: A square matrix A is named as triangular whether it is upper triangularor lower triangular. For example, the matrices100 2 634 and -1341 0 0 1000 are triangular matrices of order 3 and 4 1 2 0 0 1 5 2 3respectively. The irst matrix is upper triangular while the second is lower triangular.Note: Diagonal matrices are both upper triangular and lower triangular.Symmetric Matrix: A square matrices A = [aij]n x n is called symmetric if At = A.From At = A, it follows that [a’ij]n x n = [aij]n x nwhich implies that a’ij = aji for i, j = 1, 2, 3, ......., n.but by the deinition of transpose, a’ij = aji for i, j = 1, 2, 3, ......., n.Thus aij = aji for i, j = 1, 2, 3, ......., n.and we conclude that a square matrix A = [aij]n x n is symmetric if aij = aji.For example, the matrices 13 23 ,  a h g  and -1231 3 2 --6321 are symmetric. h b f 0 5 g f c 5 1 6 -2Skew Symmetric Matrix : A square matrix A = [aij]n x n is called skew symmetric or anti-symmetric if At = -A. From At = -A, it follows that [a’ij] = for i, j = 1, 2, 3, ......., n which implies that a’ij = -aij for i, j = 1, 2, 3, ......., n but by the deinition of transpose a’ij = aji for i, j = 1, 2, 3,..., n Thus -aij = aji or aij = -aji Alternatively we can say that a square matrix A = [aij]nxn is anti-symmetricif aij = -aji . version: 1.1 42

31.. MQautaridcreastiacndEqDueatetiromninsants eLearn.Punjab eLearn.Punjab For diagonal elements j = i, so aii = -aii or 2aii = 0 ⇒ aii = 0 for i = 1, 2, 3, ......., n For example if B = -041 -4 -103 ,then 0 3 Bt 0 4 -1  0 -4 1  =-4 0 3  =(-1)  4 0 -3 =-B  1 -3 0  -1 3 0  Thus the matrix B is skew-symmetric.Let A = [aij] be an n x m matrix with complex entries, Then the n x m matrix [aij] where aij is thecomplex conjugate of aij for all i, j, is called conjugate of A and is denoted by A . For example, if ==A=32-i i 1-+ii , then A 32-i i -i  3-+2ii 1 i i  1+ i -( )Hermitian Matrix: A square matrix A = [aij]nxn with complex entries, is called hermitian if t A= A.( )From, A t = A it follows that ai′j n×n = [aij]nxn which implies that ai/j = aij for i, j = 1 ,2 ,3,....,n butby the deinition of transpose, a / = aij for i, j = 1, 2, 3,......, n. ij Thus aij = aij for i, j = 1, 2, 3,......, n and we can say that a square matrixA = [aij]nxn is hermitian if aij = a for i, j = 1, 2, 3, .... ,n. ji For diagnal elements, j = i so aii = a which implies that aii is real for i = 1, 2, 3, ..., n iiFor example, if A = 1 1 1 - i  , then + 2 i ( )A = 1 1 1 + i  ⇒ t = 1 1 1 - i  = A - 2 + 2 i A i Thus A is hermitian. version: 1.1 43

13.. MQautaridcreastiacnEdqDueatetiromnisnants eLearn.Punjab eLearn.PunjabSkew Hermitian Matrix: A square matrix A = [aij]nxn with complex entries, is called skew-( )hermitian or anti-hermitian if A t = -A. ( )From A t = -A, it follows that ai/j n×n = [-aij]nxn which implies that ai/j = -aij for i, j = 1, 2, 3, ..., n. but by the deinition of transpose, ai/j = a ji for i, j = 1, 2, 3, ...., n. Thus -aij = a ji or aij = -a ji , for i, j = 1, 2, 3, ...., n.and we can conclude that a square matrix A = [aij]nxn is anti-hermitian if aij = -a ji For diagonal elements j = i, so aii = -aii ⇒ aii + -aii = 0 which holds if aii = 0 or aii = iλ where λ is real because 0 + 0 = 0 or iλ + iλ = iλ - iλ = 0For example, if A = -20+ 3i 2 +03i , thenA = -20+ 3i 2 +03i( )⇒A t = 2 0 -2 + 3i  = (-1) -20- 3i 2 - 3i  = -A + 3i 0 0Thus A is skew-hermitian.3.10 Echelon and Reduced Echelon Forms of Matrices In any non-zero row of a matrix, the irst non-zero entry is called the leading entry ofthat row. The zeros before the leading entry of a row are named as the leading zero entriesof the row.Echelon Form of a Matrix: An m x n matrix A is called in (row) echelon form ifi) In each successive non-zero row, the number of zeros before the leading entry is greaterthan the number of such zeros in the preceding row,ii) Every non-zero row in A precedes every zero row (if any),iii) The irst non-zero entry (or leading entry) in each row is 1. version: 1.1 44

31.. MQautaridcreastiacndEqDueatetiromninsants eLearn.Punjab eLearn.PunjabNote: Some authors do not require the condition (iii)The matrices 00 1 -2 42 and 10 2 -3 42 are in echelon form 0 1 0 1 0 0 0 0 0 0 0 1but the matrices 000 0 1 -021 and 000 1 --421 are not in echelon form. 1 3 0 0 0 0Reduced Echelon Form of a Matrix: An m x n matrix A is said to be in reduced (row)echelon form if it is in (row) echelon form and if the irst non-zero entry (or leading entry) inRi lies in Cj , then all other entries of Cj are zero.The matrices 000 1 0 042 and 100 2 0 100 are in (row) reduced 0 1 0 1 0 0 0 0echelon form.Example 1: Reduce the following matrix to (row) echelon and reduced (row) echelon form, 123 3 -1 -923 -1 2 1 3Solution: 12 3 -1 -93 R 12 -1 2 -93 By R1 ↔ R2 -1 2 3 -1 3 1 3 2  3 1 3 2 R 100 -1 2 11-513 By R2 +( - 2)R1 → R '2 R 100 -1 2 -3  1 5 and R3 +( - 3)R1 → R '3 1 5 -5 -1 3 By R2 → R '2 4 -3 4 -3 111 -1 2 -3  1 0 1 0  R 00  R 00  1 -1 3 By R3 +( - 4)R2 → R '3 1 -1 3 By R1 +1.R2 → R '1 0 -1 0 1 -1 -1 version: 1.1 45

13.. MQautaridcreastiacnEdqDueatetiromnisnants eLearn.Punjab eLearn.PunjabR 100 0 0 -121 By R1+( -1)R3 → R '1 1 0 and R2 + 1.R3 → R '2 0 1Thus 10 -1 2 3  and 10 0 0 1  are (row) echelon and reduced (row) echelon forms 1 -1 3 1 0 2 0 0 1 -1 0 0 1 -1of the given matrix respectively.Let A be a non-singular matrix. If the application of elementary row operations on AI insuccession reduces A to I, then the resulting matrix is I  A-1.Similarly if the application of elementary column operations on.A..in succession reduces I  IA to I, then the resulting matrix is A-1Thus A I R I  A-1 and.IA.. C .I...   A-1Example 2: Find the inverse of the matrix A = 132 5 --221 4 2Solution: A = 123 5 --221 = 2(-8 - 4) - 5(-6 - 2) -1(6 - 4) = -24 + 40 - 2 4 2 = 40 - 26 = 14 As |A| ≠ 0, so A is non-singular. 2 5 -1  1 0 0 version: 1.1Appending I3 on the left of the matrix A, we have 3 4 2  0 1 0 1 2 -2  0 0 1 46

31.. MQautaridcreastiacndEqDueatetiromninsants eLearn.Punjab eLearn.Punjab Interchanging R1 and R3 we get..132 2 -2  0 0 100 R 100 2 -2  0 0 --123 By R2 + ( - 3)R1 → R'2 4 2  0 1 -2 8  0 1 and R3 + ( - 2)R1 → R'3 5 -1  1 0 1  1 0 3 By - 1 R2 → R'2, we get 2 100 2 -2  0 0 -1232 R 100 0 6  0 0 -2  1 -4  0 1 -4  0 3 By R3 + ( -1)R2 → R'3 1 3  1 -1 0 1 -1 2 7  2 2  and R1 + ( - 2)R2 → R'1 0 1 -7 2 2 By 1 R3 → R'3, we have 7100 0 6  0 1 -2  100 0 0 -6 4 1   1 -4  3 1 0 7 7 0 0 -1 R 0 1 4 -3 - 1 By R1 + ( - 6)R3 → R'1 1  1 2 2 7 14 - 2 and R2 + 4R3 → R'2 7 1 -1 1 1 1 7 14 2 14 2Thus the inverse of A is --717764 4 1  7 -3 -1 14 2 1 14 -1 2Appending I3 below the matrices A, we have 2 5 -1 1013 102400-22 0 0 1  version: 1.1 47

13.. MQautaridcreastiacnEdqDueatetiromnisnants eLearn.PunjabInterchanging C1 and C3, we get eLearn.Punjab version: 1.1101023 100524100--221 C 100--221100542 100123 C  1 5 2 -2 4 3 0202 11 By (-1)C1 D C’1  0 1 0 -1 0 0 By C2 + (-5)C1 D C’2 and C3 + (-2)C1 D C’3, we have1 0 0 CC ----0010221211--105044107474 ..10210...-.-.70...33 By 1 C2 →→ C '2--002121051-48102.-.7..3. B 14By C1 + (2)C2 D C’1 and C3 + (-7)C2 D C’3 we have1 0 0   1 0 0  0 1 0   0 1 0 76170-174104 .1...112..  -74760 -074134.1....-.112  By C1 +  - 6  C3 → C '1 and C2 7 C +  4  C3 → C '2 7- 2 5 - 1   1 1 - 1  7 14 2 7 14 2 48

31.. MQautaridcreastiacndEqDueatetiromninsants eLearn.Punjab eLearn.PunjabThus the inverse of A is -771476 4 1  7 -3 -1 14 2 1 14 -1 2Rank of a Matrix: Let A be a non-zero matrix. If r is the number of non-zero rows when it isreduced to the reduced echelon form, then r is called the (row) rank of the matrix A.Example 3: Find the rank of the matrix 132 -1 2 ---1731 07 1 12Solution: 132 -1 2 ---1731 R 100 -1 2 ---231 By R2 + (-2) R1 → R '2 07 2 3 and R3 + (-3) R1 → R '3 6 1 12 4 1 -1 2 -3  By 1 R2 → R '2 1 -1 2 -3  By R3 + (- 4)R2 → R '3R 00 2  2 R 00 1 -1  4 3 -1 0 3 2 2 2 2 6 0 0 -2R 100 0 7 -7  By R1 + 1.R2 → R '1 2 2 1 3 0 2 -1 0 2 0 As the number of non-zero rows is 2 when the given matrix is reduced to the reducedechelon form, therefore, the rank of the given matrix is 2. version: 1.1 49

13.. MQautaridcreastiacnEdqDueatetiromnisnants eLearn.Punjab eLearn.Punjab Exercise 3.4 1. If A = -12 -2 -51 and B =  -3 1 --21 ,then show that A + B is symmetric. 3 1 0  5 -1 0  -2 -1 2 2. If A = -131 2 -021 , show that 2 3 i) A + At is symmetric ii) A - At is skew-symmetric.3. If A is any square matrix of order 3, show that i) A + At is symmetric and ii) A - At is skew-symmetric.4. If the matrices A and B are symmetric and AB = BA, show that AB is symmetric.5. Show that AAt and AtA are symmetric for any matrix of order 2 x 3.6. If A = 1i 1+ i ,show that -i i) A + ( A)t is hermitian ii) A - ( A)t is skew-hermitian.7. If A is symmetric or skew-symmetric, show that A2 is symmetric.8. If A = 1 1 i  , ind A( A)t . +  i 9. Find the inverses of the following matrices. Also ind their inverses by using row andcolumn operations.  1 2 -3  1 2 -1 1 -3 2 -02 ii) 10  02 10 i) -2 0 -1 3 iii) 1 -2 2 0 2 -1 version: 1.1 50

31.. MQautaridcreastiacndEqDueatetiromninsants eLearn.Punjab eLearn.Punjab10. Find the rank of the following matricesi) 123 -1 2 -113 ii) 1132 -4 -1437 iii) 1322 -1 3 0 --5321 -6 5 -5 2 -1 -3 4 -2 3 4 2 5 -7 5 -2 -33.11 System of Linear Equations An equation of the form: ax + by = k (i) where a ≠ 0, b ≠ 0, k ≠ 0is called a non-homogeneous linear equation in two variables x and y. Two linear equations in the same two variables such as: a1x + b1 y ==kk12  (I) a2 x + b2 yis called a system of non-homogeneous linear equations in the two variables x and y ifconstant terms k1, k2 are not both zero. If in the equation (i), k = 0 , that is, ax + by = 0 , then it is called ahomogeneous linear equation in x and y. If in the system (I), k1 = k2 = 0 , then it is said to be a system of homogenous linearequations in x and y. An equation of the form:ax + by + cz = k . . . . (ii)is called a non-homogeneous linear equation in three variables x, y and z if a ≠ 0, b ≠ 0, c ≠ 0and k ≠ 0. Three linear equations in three variables such as: a1x + bbb132yyy+++ccc132zzz===kkk132  (II) a2 x + version: 1.1 a3 x + 51

13.. MQautaridcreastiacnEdqDueatetiromnisnants eLearn.Punjab eLearn.Punjabis called a system of non-homogeneous linear equations in the three variables x, y and z, ifconstant terms k1, k2 and k3 are not all zero. If in the equations (ii) k = 0 that is, ax + by + cz = 0.then it is called a homogeneous linear equation it x, y and z. If in the system (II), k1 = k2 = k3 = 0, then it is said to be a system of homogeneous linearequations in x ; y and z. A system of linear equations is said to be consistent if the system has a uniquesolution or it has ininitely many solutions. A system of linear equations is said to be inconsistent if the system has no solution. The system (II), consists of three equations in three variables so it is called 3 x 3 linearsystem but a system of the form: x- y + 2z ==46 2x + y + 3zis named as 2 x 3 linear system.Now we solve the following three 3 x 3 linear systems to determine the criterion for a systemto be consistent or for a system to be inconsistent. 32xxx+++425yyy+--22zzz-===1513 ....(1), 32xxx+-+5yyy+++724zzz-===1113 (2)and 32xxx+--56yyy+++245zzz-===173 ....(3)The augmented matrix of the system (1) is132 5 -1  1-513 4 2  2 -2  We apply the elementary row operations to the above matrix to reduce it to the equivalentreduced (row) echelon form, that is, version: 1.1 52

31.. MQautaridcreastiacndEqDueatetiromninsants eLearn.Punjab eLearn.Punjab 2 5 -1  5  1 2 -2  -3 13 1-13 R 23 11 BY R 1↔ R 3 4 2  4 2  5  2 -2  5 -1  1 2 -2  -3 1 2 -2  -3 R 02 250 BY R2 + ( - 3)R1 → R2′ R 00  R3′ -2 8  -2 8  20 BY R3 + ( - 2)R1 → 5 -1  1 3  11 BY - 1 R2 → R2′ , we get 2 100 2 -2  -1-1130 R 100 0 6  -121710 By R1+ ( - 2)R2 → R'1 1 -4  1 -4  and R3 + ( -1), R2 → R'3 1  0 7  3R 100 0 6  -13170 BY 1 R3 → R′3 R 100 0 0  -321 By R1+ ( - 6)R3 → R'1 1 -4  7 1 0  and R2 + 4R3, R'2 0 1  0 1  Thus the solution is x = -1, y = 2 and z = 3.The augmented matrix for the system (2) is 123 1 2  1-113 -1 7  5 4  Adding (-2)R1 to R2 and (-3)R1 to R3, we get132 1 2  1-113 R 100 1 2  -196 -1 7  -3  5 4  2 3  -2R 100 1 2  1 R 100 0 3 -043 BY R1 + ( -1) R2 → R1′ 1  --63 1 → R2′ 1 and R3 + ( -2) R2 → R3′ 2 -1  By - 3 R2 0 -1  -2 0 version: 1.1 53

13.. MQautaridcreastiacnEdqDueatetiromnisnants eLearn.Punjab eLearn.PunjabThe system (2) is reduced to equivalent system x + 3z = 4 y-z=-3 0z = 0 The equation 0z = 0 is satisied by any value of z. From the irst and second equations, we get x = -3z + 4 .......(a) and y = z - 3 ........(b) As z is arbitrary, so we can ind ininitely many values of x and y from equation (a) and (b)or the system (2), is satisied by x = 4 - 3t, y = t - 3 and z = t for any real value of t. Thus the system (2) has ininitely many solutions and it is consistent. The augmented matrix of the system (3) is 132 -1 2  -173 -6 5  4  5 Adding (-2)R1, to R2 and (-3)R1, to R3 we have1 -1 2  1  1 -1 2  1 23 -73 R 00 -56 -6 5  -4 1  5 4  8 -2 R 100 -1 2  1  R 100 0 7  -1  By R1 + 1.R2 → R1′ 1  4  8 -1  - 5 By - 1 R2 → R2′ 1 4  and R3 + ( -8) R2 → R3′ 4 4 4 0 -1  -5 -6  4 -2 4 4 0 Thus the system (3) is reduced to the equivalent system x + 7 -z =1 44 y - 1 -z =5 44 0z = 4 version: 1.1 54

31.. MQautaridcreastiacndEqDueatetiromninsants eLearn.Punjab eLearn.PunjabThe third equation 0z = 4 has no solution, so the system as a whole has no solution. Thus thesystem is inconsistent. We see that in the case of the system (1), the (row) rank of the augmented matrix and thecoeicient matrix of the system is the same, that is, 3 which is equal to the number of thevariables in the system (1). Thus a linear system is consistent and has a unique solution if the(row) rank of the coeicient matrix is the same as that of the augmented matrix of thesystem. In the case of the system ( 2), the (row) rank of the coeicient matrix is the same as thatof the augmented matrix of the system but it is 2 which is less than the number of variablesin the system (2). Thus a system is consistent and has ininitely many solutions if the (row) ranks of thecoeicient matrix and the augmented matrix of the system are equal but the rank is lessthan the number of variables in the system. In the case of the system (3), we see that the (row) rank of the coeicient matrix is notequal to the (row) rank of the augmented matrix of the system. Thus we conclude that a system is inconsistent if the (row) ranks of the coeicient matrixand the augmented matrix of the system are diferent.3.11.1 Homogeneous Linear EquationsEach equation of the system of following linear equations: a11x1 + a12 x2 + a13x3 ===000 ...(((iiiiii)))a21x1 + a22 x2 + a23x3a31x1 + a32 x2 + a33x3is always satisied by x1 = 0, x2 = 0 and x3 = 0, so such a system is always consistent. Thesolution (0, 0, 0) of the above homogeneous equations (i), (ii), and (iii) is called the trivialsolution. Any other solution of equations (i), (ii) and (iii) other than the trivial solution iscalled a non-trivialsolution. The above system can be written as AX = O , where O = 000 version: 1.1 55

13.. MQautaridcreastiacnEdqDueatetiromnisnants eLearn.Punjab eLearn.PunjabIf |A| ≠ 0, then A is non-singular and A-1 exists, that is, A-1(AX) = A-1O = Oor (A-1A )X = O ⇒ X = O , i.e.,  x1  = 000 x2 x 3In this case the system of homogeneous equations possesses only the trivial solution.Now we consider the case when the system has a non-trivial solution. Multiplying the equations (i), (ii) and (iii) by A11, A21 and A31 respectively and adding theresulting equations (where A11, A21 and A31 are cofactors of the corresponding elements of A),we have(a11 A11 + a21 A21 + a31 A31)x1 + (a12 A11 + a22 A21 + a32 A31)x2 + (a13 A11 + a23 A21 + a33 A31)x3 = 0, that is,|A|x1 = 0. Similarly, we can get |A|x2 = 0 and |A|x3 = 0 For a non-trivial solution, at least one of x1, x2 and x3 is diferent from zero. Let x1 ≠ 0 , thenfrom |A|x1 = 0, we have |A| = 0. For example, the system x1 + x2 + x3 =0 (((IIII)II)) x1 - x2 + 3x3 =0 x1 + 3x2 - x3 =0has a non-trivial solution because 11 1 10 0 =-2 2 =0 A =1 -1 2 -2 3 =1 -2 -2 2 13 -1 1 2 Solving the irst two equations of the system, we have 2x1 + 4x3 = 0 (adding (I) and (II)) ⇒ x1 = -2x3 and 2x2 - 2x3 = 0 (subtracting (II) from (I)) ⇒ x2 = x3 Putting x1 = -2x3 and x2 = x3 in (III), we see that (-2x3) + 3(x3) - x3= 0, which shows that theequation (I), (II) and (III) are satisied by x1 = -2t, x2 = t and x3 = t for any real value of t. Thus the system consisting of (I), (II) and (III) has ininitely many solutions. But the system version: 1.1 56

31.. MQautaridcreastiacndEqDueatetiromninsants eLearn.Punjab eLearn.Punjabx1 + x2 + x3 ===000 has only the trivial solution,x1 - x2 + 3x3x1 + 3x2 - 2x3because in this caseA= 1 1 1 1 0 0 -2 2 = 6 - 4= 2≠0 1 -1 1 -2 2 -3 1 3 3= 1 2 2= -2 -3 Solving the irst two equations of the above system, we get x1 = -2x3 and x2 = x3. Puttingx1 = -2x3 and x2 = x3 in the expression. x1 + 3x2 - 2x3 , we have - 2x3 + 3(x3) - 2x3 = - x3, that is,the third equation is not satisied by putting x1 = -2x3 and x2 = x3 but it is satisied only if x3 =0. Thus the above system has only the trivial solution.3.11.2 Non-Homogeneous Linear Equations Now we will solve the systems of non-homogeneous linear equations with help of thefollowing methods.i) Using matrices, that is, AX = B ⇒ X = A-1B.ii) Using echelon and reduced echelon formsiii) Using Cramer’s rule.Example 1: Use matrices to solve the system x1 - 2x2 + x3 - ===564 2x1 - 3x2 + 2 x3- 2x1 + 2x2 + x3Solution: The matrix form of the given system is 12 -2 12-  x1  =-64 -3 x2 2 2 1  x3   5  version: 1.1 57

13.. MQautaridcreastiacnEdqDueatetiromnisnants eLearn.Punjab or AX = B (i) eLearn.Punjab where R′ version: 1.1 A-=12 -23 =- 12 ,X xx12= and B -46 As 2 2 1  x3   5  2 1 -2 1 1 -2 1 A= 2 -3 2= 0 1 0 B+y -R2 ( →2 )R1 221 221 =(-1)2+2 1 1 =(1 - 2) =-1 , that is, 21 |A| ≠ 0, so the inverse of A exists and (i) can be written as X = A-1B (ii) Now we ind adj A.Since  Aij 3×=3 --471 2 1-106 ,  A-11 =7=, A12 2=, A13 10=, A21 4 1 -1 A-22 =- 1=,-A23==6, A31 =1, A32 0, A33 0So a=dj AA= 1-207 4 -101 -1 -6 -7 4 -1  7 -4 1  A-1 =1 adjA= 1 120  =-120 -01and A -1 -1- 0 1 -6 1 6Thus  x1  -4 = 7 -4 1  --46 =-288-+62+4 + 5 ,i.e., x2 =A-1 -6 -2 1 0 0  x3   5  -10 6 -1  5   40 - 36 - 5   x1  =  1   x2  2  x3  -1 Hence x1= 1, x2 = 2 and x3 = - 1. 58

31.. MQautaridcreastiacndEqDueatetiromninsants eLearn.Punjab eLearn.PunjabExample 2: Solve the system; x1 + 3x2 + 2x3 ===3432 , 4 x1 + 5x2 - 3x3- 3x1 - 2 x2 + 17x3by reducing its augmented matrix to the echelon form and the reduced echelon form.Solution: The augmented matrix of the given system is 14 3 2  -33 5 -3  3 -2 17  42 We reduce the above matrix by applying elementary row operations, that is, 143 3 2  -4323 R 100 3 2 -33135 By R2 + (-4) R1 → R2′ -3  and R3 + (-3) R1 → R3′ 5 17  -7 -11  -2 -11 11 1 3 2  3 R 00 33  By R2 ↔ R3 -11 11  -15 -7 -11 R 100 3 2 --3135 By  - 1  R2 → R2′ R 100 3 2  3  By R3 + 7R2 → R3′ 11 1  -3 1 -1  0 -1  -7 -11  -18 -36R 100 3 2 -323 By  -1  R3 → R3′ 1 18 0 -1  1 The equivalent system in the (row) echelon form is version: 1.1 59

13.. MQautaridcreastiacnEdqDueatetiromnisnants eLearn.Punjab eLearn.Punjab x1 + 3x2 + 2xx3x3-3===233 x2 -Substituting x3 = 2 in the second equation gives: x2 - 2 = -3 ⇒ x2 = -1Putting x2 = -1 and x3 = 2 in the irst equation, we have x1 + 3 (-1) + 2(2) = 3 ⇒ x1 = 3 + 3 - 4 = 2.Thus the solution is x1 = 2, x2 = -1 and x3 = 2Now we reduce the matrix 100 3 2 -323 to reduced (row) echelon form, i.e., 1 0 -1  -1 100 3 2 -323 R 100 0 5 1-223 By R1 + (-3) R2 → R1′ 1 1 0 -1  0 -1  1 1 R 100 0 0  -221 By R1 + (-5) R3 → R1′ 1 0  0 1  andR2 + 1.R3 → R2′The equivalent system in the reduced (row) echelon form is x1 = 2 x2 = - 1 x3 = 2which is the solution o f the given system.3.12 Cramer’s RuleConsider the system of equations, a11x1 + a12 x2 + a13 x3 ===bbb132  (1) a21x1 + a22 x2 + a23 x3 version: 1.1 a31x1 + a32 x2 + a33 x3 60

31.. MQautaridcreastiacndEqDueatetiromninsants eLearn.Punjab eLearn.PunjabThese are three linear equations in three variables x1, x2, x3 with coeicients and constantterms in the real ield R. We write the above system of equations in matrix formas: AX = B (2)where=A =a=ij 3×3 , X  x1  and B bb12   x2   x3 b3 We know thatthe matrix equation (2) can be written as: X = A-1B (if A-1 exists)Note: A-1(AX) ≠ BA-1We have already proved that A-1 = 1 adj A and A=adj A =Ai′j 3×3  A11 =AA2221 AA3321  ( ) Ai′j Aji A12 A23 A33   A13Thus xx12= 1  A11 A21 A31 =bb12  1  A11b1 + A21b2 + A31b3   x3  A A12 A22 A32 A A+12b1 + A2+2b2 A32b3 A23 A23b2 + A33b3  A13 A33  b3  A13b1  x1  =  A11b1 + A21b2 + A31b3  x2  A12b1 + A + A32b3  A13b1 + + A33b3  x3 A22b2 A A23b2  A ==Hence x1 b1 A11 + b2 A21 + b3 A31 b1 a12 a13 (i) A b2 a22 a23 b3 a32 a33 A version: 1.1 61

13.. MQautaridcreastiacnEdqDueatetiromnisnants eLearn.Punjab eLearn.Punjab==x2 b1 A12 + b2 A22 + b3 A32 a11 b1 a13 (ii) A a21 b2 a23 a31 b3 a33 A==x3 b1 A13 + b2 A23 + b3 A33 a11 a12 b1 (iii) A a21 a22 b2 a31 a32 b3 AThe method of solving the system with the help of results (i), (ii) and (iii) is often referred toas Cramer’s Rule.Example 3: Use Crammer’s rule to solve the system. 3x-1xx1+1++x2x22-x-22-xx33x==3 =--414 3 1 -1Solution: Here A = 1 1 -2 = 3(-1+ 4) -1.(-1- 2) -1.(2 +1) -1 2 -1 =9+3-3=9 -4 1 -1 -4 1 -2 -4(-1+ 4) -1(4 + 2) -1(-8 -1)==So x1 1 2 -1 9 9 = -12 - 6 +=9 -=9 1 99 3 -4 -1 1 -4 -2 3(4 + 2) + 4(-1- 2) -1(1- 4)==x2 -1 1 -1 9 9 = 18 -12 + 3= 9= 1 99 version: 1.1 62

31.. MQautaridcreastiacndEqDueatetiromninsants eLearn.Punjab 3 1 -4 eLearn.Punjab version: 1.1 1 1 -4 3(1+ 8) -1(1- 4) - (2 +1)==x3 -1 2 1 9 9 = 27 + 3 -1=2 1=8 2 99Hence x1 = -1 , x2 = 1, x3 = 2 Exercise 3.51. Solve the following systems of linear equations by Cramer’s rule.i) 325xxx-++22yyy--+32zzz===231 ii) 432xx1x11+--24xxx222++-3xx3x33===538 iii) 2x1 - x2 + x3 ===186 x1 + 2x2 + 2x3 x1 - 2x2 - x32. Use matrices to solve the following systems:i) 3xx-+2yyy-+-2zzz===-141 ii) -32xxx111+-+xxx222-++223xxx333===03-4 iii) 22yxx-+-3yzz===21-13. Solve the following systems by reducing their augmented matrices to the echelon form and the reduced echelon forms.i) x1 - 2x2 - -2+3x-x3x33===111 ii) 22xxx+++2y3y+y+-2zzz===2-91 iii) x1 + 4x2 + 2 x3 ===1922 2x1 + 3x2 2x1 + x2 - 2 x3 5x1 - 4x2 3x1 + 2x2 - 2 x34. Solve the following systems of homogeneous linear equations.i) 52xxx+++24yyy-++258zzz===000 ii) x1 + 4x2 +--324xxx333===000 iii) 2xxx111-+-2xx2x22++-54xxx333===000 2x1 + x2 3x1 + 2x2 63

13.. MQautaridcreastiacnEdqDueatetiromnisnants eLearn.Punjab eLearn.Punjab5. Find the value of λ for which the following systems have non-trivial solutions. Also solve the system for the value of λ .i) x+ y + z zz===000 ii) 3x2x1x1+1++4λxxx222+--λ34xxx333===000 2x + y - λ x + 2y - 26. Find the value of λ for which the following system does not possessa unique solution. Also solve the system for the value of λ . x1 + 4x2 + λ x3 ===11216 2x1 + x2 - 2x3 3x1 + 2x2 - 2x3 version: 1.1 64

CHAPTER version: 1.14 Quadratic Equations Animation 4.1: Completing the square Source & Credit: 1ucasvb

14.. QQuuaaddrraattiiccEEquqautaiotinosns eLearn.Punjab eLearn.Punjab4.1 Introduction A quadratic equation in x is an equation that can be written in the form ax2 + bx + c = 0;where a, b and c are real numbers anda ≠ 0. Another name for a quadratic equation in x is 2nd Degree Polynomial in x. The following equations are the quadratic equations: i) x2 - 7x + 10 = 0; a = 1, b = -7, c = 10 ii) 6x2 + x - 15 = 0; a = 6, b = 1, c = -15 iii) 4x2 + 5x + 3 = 0; a = 4, b = 5, c=3 iv) 3x2 - x = 0; a = 3, b = -1, v) x2 = 4; a = 1, b = 0, c=0 c = -44.1.1 Solution of Quadratic Equations There are three basic techniques for solving a quadratic equation: i) by factorization. ii) by completing squares, extracting square roots. iii) by applying the quadratic formula.By Factorization: It involves factoring the polynomial ax2 + bx + c. It makes use of the fact that if ab = 0, then a = 0 or b = 0.For example, if (x - 2) (x - 4) = 0, then either x - 2 = 0 or x - 4 = 0.Example 1: Solve the equation x2 - 7x + 10 = 0 by factorization.Solution: x2 - 7x + 10 = 0 ⇒ (x - 2) (x - 5) = 0 ∴ either x - 2 = 0 ⇒x=2 or x - 5 = 0 ⇒x=5 ∴ the given equation has two solutions: 2 and 5 ∴ solution set = {2, 5}Note: The solutions of an equation are also called its roots. ∴ 2 and 5 are roots of x2 - 7x + 10 = 0 version: 1.1 2