Determinant and Matrices PDF

4 Determinants and Matrices  1st row Let's Study a b · Definition and Expansion of Dete...

4 Determinants and Matrices  1st row Let's Study a b · Definition and Expansion of Determinants c d 2nd row · Minors and Co-factors of determinants 1st 2nd · Properties of Determinants column column · Applications of Determinants · Introduction and types of Matrices a b The value of the determinant is · Operations on Matrices c d ad – bc. · Properties of related matrices 4.1 Introduction SOLVED EXAMPLES We have learnt to solve simultaneous equations in two variables using determinants. 7 9 cos θ sin θ We will now learn more about the determinants Ex. Evaluate i) ii) −4 3 − sin θ cos θ because they are useful in Engineering applications, and Economics, etc. 4 i log 42 log 42 iii) where i2 = –1 iv) The concept of a determinant was −2i 7 2 4 discussed by the German Mathematician G.W. Leibnitz (1676-1714) and Cramer (1750) Solution : developed the rule for solving linear equations using determinants. 7 9 i) = 7 × 3 − (−4) × 9 = 21 + 36 = 57 −4 3 Let's Recall cos θ sin θ ii) = cos2θ − (−sin2θ) − sin θ cos θ 4.1.1 Value of a Determinant In standard X we have studied a method of = cos2θ + sin2θ = 1 solving simultaneous equations in two unknowns using determinants of order two. In this chapter, 4 i we shall study determinants of order three. iii) = 4 × 7 − (−2i) × i = 28 + 2i2 −2i 7 a b = 28 + 2(−1) [Qi2 = −1] The representation is defined as the = 28 − 2 = 26 c d determinant of order two. Numbers a, b, c, d log 42 log 42 iv) = 4×log4 2 − 2×log42 are called elements of the determinant. In this 2 4 = log424 − log422 arrangement, there are two rows and two columns. 59  = log416 − log44 a11 a12 a13 = 2 log44 − log44 D = a21 a22 a23 = 2 × 1 − 1 = 2 − 1 = 1 a31 a32 a33 The determinant can be expanded as follows: Let's Understand a22 a23 a21 a23 a21 a22 D = a11 − a12 + a13 4.1.2 Determinant of order 3 a32 a33 a31 a33 a31 a32 Definition - A determinant of order 3 is a square arrangement of 9 elements enclosed between two vertical bars. The elements are SOLVED EXAMPLES arranged in 3 rows and 3 columns as given below. a11 a12 a13 R1 R1 are the rows Evaluate: a21 a22 a23 R2 Cj are the column 3 -4 5 a31 a32 a33 R3 i) 1 1 -2 C1 C2 C3 2 3 -1 Here aij represents the element in ith row and sec θ tan θ 0 jth column of the determinant. ii) tan θ sec θ 0 e.g. a31 represents the element in 3rd row and 0 0 1 1st column. In general, we denote determinant by Capital 2-i 3 -1 Letters or by ∆ (delta). iii) 3 2-i 0 where i = -1 We can wirte the rows and columns 2 -1 2-i separately. e.g. here the 2nd row is [a21 a22 a23] and 3rd column is Solution :  a13  a  3 -4 5 1 -2 1 -2  23  i) 1 1 -2 = 3 − (−4)  a33  3 -1 2 -1 2 3 -1 Expansion of Determinant 1 1 +5 2 3 We will find the value or expansion of a 3x3 determinant.We give here the expansion by the = 3(−1+6) + 4(−1+4) + 5(3−2) 1st row of the determinant D. = 3×5+4×3+5×1 There are six ways of expanding a determinant of order 3, corresponding to each = 15+12+5 of three rows (R1, R2, R3) and three columns = 32 (C1, C2, C3). 60  sec θ tan θ 0 Definitions ii) tan θ sec θ 0 The minor of aij - It is defined as the determinant obtained by eliminating the ith row 0 0 1 and jth column of A. That is the row and the column that contain the element aij are omitted. sec θ 0 tan θ 0 We denote the minor of the element aij by Mij = secθ − tan θ 0 1 0 1 In case of above determinant A a22 a23 tan θ sec θ Minor of a11 = M11 = a a = a22. a33 − a32.a23 +0 32 33 0 0 a21 a23 = secθ (secθ – 0) − tanθ (tanθ – 0) + 0 Minor of a12 = M12 = = a21. a33 − a31. a23 a31 a33 = sec2θ − tan2θ a21 a22 =1 Minor of a13 = M13 = = a21. a32 − a31. a22 a31 a32 Similarly we can find minors of other 2−i 3 −1 elements. iii) 3 2−i 0 Cofactor of aij - 2 −1 2−i cofactor of aij = (−1)i+j minor of aij = Cij = (2 − i)[(2 − i) − 0] − 3[3(2 − i) − 0] 2 ∴ Cofactor of element aij = Cij = (−1)i+j Mij − 1[ − 3 − 2(2 − i)] The same definition can also be given for = (2 − i)3 − 9(2 − i) + 3 + 2(2 − i) a b elements in 2×2 determinant. Thus in = 8 − 12i + 6i2 − i3 − 18 + 9i + 3 + 4 − 2i c d The minor of a is d. = 8 − 12i + 6(−1) + i − 18 + 9i + 3 + 4 − 2i (since i2= −1) The minor of b is c. The minor of c is b. = 8 − 6 − 18+ 7 − 12i + 9i − 2i + i The minor of d is a =− 9 − 4i SOLVED EXAMPLES Let's Learn Ex. 1) Find Minors and Cofactors of the 4.1.3 Minors and Cofactors of elements of elements of determinant determinants 2 −3 i) a11 a12 a13 4 7 Let A = a21 a22 a23 be a given determinant. a11 a12 2 −3 Solution : Here = a31 a32 a33 a21 a22 4 7 M11 = 7 C11 = (−1)1+1 M11 = (−1)1+1.7 = 7 61  M12 = 4 1 2 M23 = = −1 −10 = −11 C12 = (−1) 1+2 M12 = (−1) 1+2.4 = −4 5 −1 M21 = −3 C23 = (−1) 2+3 M23 = (−1) 2+3. (−11) = 11 C21 = (−1)2+1 M21 = (−1)2 +1. (−3) = 3 2 −3 M31 = = 8 − 0= 8 M22 = 2 0 4 C22 = (−1)2+2 M22 = (−1)2+2. 2= 2 C31 = (−1) 3+1 M31 = (−1) 3+1. 8 = 8 1 −3 1 2 −3 M32 = = 4 − 6 = −2 −2 4 ii) −2 0 4 5 −1 3 C32 = (−1)2 3+2 M32 = (−1) 3+2. (−2) = 2 Solution : 1 2 M33 = = 0+4 = 4 a11 a12 a13 −2 0 1 2 −3 Here a21 a22 a23 = −2 0 4 C33 = (−1) 3+3 M33 = (−1) 3+3. 4 = 4 a31 a32 a33 5 −1 3 Expansion of determinant by using Minor 0 4 and cofactors of any row/column M11 = =0+4=4 −1 3 a11 a12 a13 C11 = (−1)1+1 M11 = (−1)1+1. 4 = 4 A = a21 a22 a23 a31 a32 a33 −2 4 M12 = = −6 −20 = −26 = a11C11 + a12C12 + a13C13 (By 1st row) 5 3 C12 = (−1)1+2 M12 = (−1)1+2.(−26) = 26 = a12C12 + a22C22 + a32C32 (By 2nd column) Ex. 2) Find value of x if −2 0 M13 = =2−0 =2 x −1 2 5 −1 i) 2 x 1 − 3 = − 10 C13 = (−1)1+3 M13 = (−1)1+3. 2 = 2 3 −4 5 2 −3 ∴ x (5 − 12) − (−1) (10x + 9) + 2(−8x − 3) M21 = =6−3 =3 = −10 −1 3 ∴ x(−7) + (10x + 9) − 16x − 6 = −10 C21 = (−1) 2+1 M21 = (−1) 2+1. 3 = −3 ∴ −7x + 10x − 16x + 9 − 6 + 10 = 0 1 −3 ∴ 10x − 23x +13 =0 M22 = = 3+15 = 18 5 3 ∴ 13x = 13 C22 = (−1) 2+2 M22 = (−1)1+1. 18 = 18 ∴ x=1 62  x 3 2 = 2(0 + 6) − 5(0−2) – 1(3−2) ii) x x 1 = 9 = 2(6) − 5(−2) – 1(1) 1 0 1 = 12 + 10 − 1 ∴ x (x−0) − 3(x−1) + 2(0−x) =9 = 22 – 1 ∴ x2 − 3x + 3 − 2x = 9 = 21 ∴ x2 − 5x +3 =9 Interpretation: From (a) and (b) it is seen that ∴ x2 − 5x − 6 = 0 the expansion of determinant by both ways gives the same value. ∴ (x−6) (x+1) =0 ∴ x−6 = 0 or x+1 = 0 ∴ x = 6 or x = −1 EXERCISE 4.1 1 −1 2 Q.1) Find the value of determinant Ex. 3) Find the value of −2 3 5 by −2 0 − 1 3 −4 5 2 −4 2i 3 i) ii) iii) 1 1 − 2 expanding along a) 2nd row b) 3rd column and 7 − 15 4 −i Interprete the result. 2 3 1 a) Expansion along the 2nd row a h g = a21 c21 + a22 c22 + a23 c23 iv) h b f g f c −1 2 1 2 = –2(−1)2+1 + 3(−1)2+2 −2 − 1 0 −1 Q.2) Find the value of x if 1 −1 + 5(−1)2+3 x −1 2 −2 0 x2 − x + 1 x + 1 i) = 0 ii) 2 x 1 − 3 = 29 x +1 x +1 = 2(+1 −0) + 3(–1+4) − 5(0−2) 3 −4 5 = 2(1) + 3(3) − 5(−2) 4i i 3 2i = 2 + 9 + 10 Q.3 Find x and y if 1 3i 2 4 = x+iy where = 21 i2 = −1 5 −3 i b) Expansion along 3rd coloumn = a13 c13 + a23 c23 + a33 c33 Q.4) Find the minor and cofactor of element of the determinant −2 3 1 −1 = 2(−1)1+3 + 5(−1)2+3 + 2 −1 3 −2 0 −2 0 D = 1 2 − 1 1 −1 − 1(−1)3+3 5 7 2 −2 3 63  2 -3 5 = a1 (b2 c3 – b3c2) – a2(b1c3 + c1b3) + a3(b1c2 – c1b2) Q.5) Evaluate A = 6 0 4 Also find minor 1 5 -7 = a1(b2 c3 – b3c2) – b1(a2c3 – a3c2) + and cofactor of elements in the 2 row of nd c1(a2b3 – b2a3) ---------- (ii) determinant and verify From (i) and (ii) D= D1 a) − a21.M21 + a22.M22 − a23.M23 = value of A Ex. b) a21 C21 + a22 C22 + a23.C23 = value of A 1 2 -1 where M21, M22 , M23 are minor of a21 , a22, a23 Let A = 3 -1 2 and C21, C22, C23 are cofactor of a21 , a22, a23 0 2 1 Q.6) Find the value of determinant expanding -1 2 3 2 3 -1 along third column = 1 –2 –1 2 1 0 1 0 2 -1 1 2 = 1(–1–4) – 2 (3–0) – 1(6–0) -2 3 - 4 = –5 – 6 – 6 -3 4 0 = –17 ………………………….. (i) 4.2 Properties of Determinants by interchanging rows and columns of A we get determinant A1 In the previous section we have learnt how to expand the determinant. Now we will study 1 3 0 -1 2 2 2 some properties of determinants. They will help A1 = 2 - 1 2 = 1 -3 us to evaluate the determinant more easily. 2 1 -1 1 -1 2 1 ∙ Let’s Verify… 2 -1 + 0 Property 1 - The value of determinant -1 2 remains unchanged if its rows are turned into = 1(–1–4) – 3 (2+2) +0 columns and columns are turned into rows. = –5 –12 Verification: = –17 ………………………………...(ii) a1 b1 c1 ∴ A = A1 from (i) and (ii) Let D = a2 b2 c2 Property 2 - If any two rows (or columns) of a3 b3 c3 the determinant are interchanged then the value = a1.(b2 c3 − b3 c2) – b1 (a2c3 – a3c2) + of determinant changes its sign. c1(a2b3 – a3b2 ) The operation Ri ↔ Rj change the sign of the ---------- (i) determinant. a1 a2 a3 Note : We denote the interchange of rows by Let D1 = b1 b2 b3 Ri ↔ Rj and interchange of columns by Ci ↔ Cj. c1 c2 c3 Property 3 - If any two rows (or columns) = a1 (b2 c3 – b3 c2) – a2 (b1c3 – c1b3) + of a determinant are identical then the value of a3(b1c2 − c1b2) determinant is zero. 64 R1 ↔ R2 D = D1 R1 → R1 + kR3 then D1 = −D....... (property 2)...... (I) a1 + ka3 b1 + kb3 c1 + kc3 But R1 = R2 hence D1 = D.......... (II) A1 = a2 b2 c2 ∴ adding I and II a3 b3 c3 2D1 = 0 ⇒ D1 = 0 i.e. D = 0 Simplifying A1, using the previous properties, we get A1 = A. Property 4 - If each element of a row (or a column) of determinant is multiplied by a constant k then the value of the new determinant 1 2 3 is k times the value of given determinant. Ex. : Let B = -1 2 0 = 1(2–0) – 2(–1–0) + The operation Ri → kRi gives multiple of the 1 2 1 determinant by k. 3(–2–2) = 2 + 2 – 12 Remark i) Using this property we can take = 4–12 = –8 ----------(i) out any common factor from any one row (or any 1 2 3 one column) of the given determinant Now, B = -1 2 0 ii) If corresponding elements of any two rows 1 2 1 (or columns) of determinant are proportional (in the same ratio) then the value of the R1 → R1 + 2R2 determinant is zero. 1 + 2(−1) 2 + 2(2) 3 + 2(0) Property 5 - If each element of a row B1 = −1 2 0 (or column) is expressed as the sum of two 1 2 1 numbers then the determinant can be expressed as sum of two determinants -1 6 3 For example, B1 = -1 2 0 = –1(2–0) – 6(–1–0) + 3(–2–2) a1 + x1 b1 + y1 c1 + z1 a1 b1 c1 x1 y1 z1 1 2 1 a2 b2 c2 = a2 b1 c1 + a2 b2 c2 = –2 + 6 – 12 = 6 – 14 = –8 ----(ii) a3 b3 c3 a3 b3 c3 a3 b3 c3 From (i) and (ii) B = B1 Property 6 - If a constant multiple of all Remark : If more than one operation from elements of any row (or column) is added to above are done, make sure that these operations the corresponding elements of any other row are completed one at a time. Else there can be (or column ) then the value of new determinant mistake in calculation. so obtained is the same as that of the original Main diagonal of determinant : The main determinant. The operation Ri ↔ Ri + kRj does diagonal (principal diagonal) of determinant A is not change the value of the determinant. collection of entries aij where i = j Verification OR a1 b1 c1 Main diagonal of determinant : The set of A = a2 b2 c2 elements ( a11 , a22, a33 , ---- ann) is called the main a3 b3 c3 diagonal of the determinant A. 65  a11 a12 a13 1 2 3 e.g. D = a21 a22 a23 here a11, a22 , a33 are = 100 505 606 707 by using property a31 a32 a33 1 2 3 element of main diagonal = 100 × 0 (R1 and R3 are identical) Property 7 - (Triangle property) - If each = 0 element of a determinant above or below the main diagonal is zero then the value of the determinant 312 313 314 is equal to product of its diagonal elements. ii) 315 316 317 = 0 that is 318 319 320 a1 b1 c1 a1 0 0 312 313 314 0 b2 c2 = a2 b2 0 = a 1b 2c 3 L.H.S. = 315 316 317 0 0 c3 a3 b3 c3 318 319 320 Remark : If all elements in any row or any C 2→ C 2 – C 1 column of a determinant are zeros then the value 312 1 314 of the determinant is zero. = 315 1 317 318 1 320 SOLVED EXAMPLES C 3→ C 3 – C 1 312 1 2 Ex. 1) Show that = 315 1 2 101 202 303 318 1 2 i) 505 606 707 = 0 take 2 common from C3 1 2 3 312 1 1 101 202 303 = 2 315 1 1 LHS = 505 606 707 318 1 1 1 2 3 = 2 (0) ( C2 and C3 are identical) R1 → R1 – R3 1 a bc 1 a a2 100 200 300 Ex. 2) Prove that 1 b ca = 1 b b 2 = 505 606 707 1 c ab 1 c c2 1 2 3 1 a bc 100 × 1 100 × 2 100 × 3 L.H.S. = 1 b ca = 505 606 707 1 c ab 1 2 3 R1 →aR1 66  a a 2 abc x y z 1 1 b ca = Ex. 3) If - x y z = k.xyz then find the value a x -y z 1 c ab of k R2 → bR2 Sloution : x y z a a 2 abc 1 1 b b 2 abc L.H.S. = - x y z = × a b x -y z 1 c ab R2→R2 +R1 R3 → cR3 x y z 2 a a abc = 0 2 y 2 z 1 1 1 = 2 × × b b abc x -y z a b c c c 2 abc R3 → R3 + R1 a a2 1 x y z 1 2 = 0 2 y 2 z = × abc b b 1 abc 2x 0 2z c c2 1 x y z (taking abc common from C3) = 2 × 2 0 y z taking (2 common a a2 1 x 0 z = b b2 1 from R2 and R3) 2 c c 1 = 4[x(yz) – y(0 – xz)+z(0 – xy)] C1 ↔C3 = 4[xyz + xyz – xyz] = 4xyz 1 a2 a From given condition = (–1) 1 b 2 b L.H.S. = R.H.S. 1 c2 c 4xyz = k xyz C2 ↔C3 ∴k = 4 1 a a2 EXERCISE 4.2 = (–1)(–1) 1 b b 2 1 c c2 Q.1) Without expanding evaluate the following determinants. 1 a a2 2 1 a b+c 2 3 4 2 7 65 = 1 b b = R.H.S. 8 iii) 3 8 75 i) 1 b c + a ii) 5 6 1 c c2 1 c a+b 6 x 9 x 12 x 5 9 86 67  x+ y y+z z+x 4.3 APPLICATIONS OF DETERMINANTS Q.2) Prove that z + x x + y y + z 4.3.1 Cramer’s Rule y+z z+x x+ y In linear algebra Cramer’s rule is an explicit x y z formula for the solution of a system of linear = 2 z x y equations in many variables. In previous class y z x we studied this with two variables. Our goal here is to expand the application of Cramer’s rule to Q.3) Using properties of determinant show that three equations in three variables (unknowns). Variables are usually denoted by x, y and z. a+b a b a a+c c = 4abc Theorem - Consider the following three linear i) b c b+c equations in variables three x, y, z. a 1x + b 1y + c 1z = d 1 1 log x y log x z a2x + b2y +c2z = d2 ii) log y x 1 log y z = 0 a 3x + b 3y + c 3z = d 3 log z x log z y 1 Here ai , bi , ci and di are constants. Q.5) Solve the following equations. The solution of this system of equations is x + 2 x + 6 x −1 Dx Dy Dz x= ,y= ,z= i) x + 6 x −1 x + 2 = 0 D D D x −1 x + 2 x + 6 provided D≠0 where x -1 x x-2 a1 b1 c1 d1 b1 c1 ii) 0 x -2 x -3 = 0 D = a2 b2 c2 Dx = d 2 b2 c2 0 0 x -3 a3 b3 c3 d3 b3 c3 4+ x 4− x 4− x a1 d1 c1 a1 b1 d1 Q.6) If 4 − x 4 + x 4 − x = 0 then find the Dy = a2 d 2 c2 Dz = a2 b2 d 2 4− x 4− x 4+ x a3 d3 c3 a3 b3 d3 values of x Remark : Q.7) Without expanding determinants show that 1) You will find the proof of the Cramer’s Rule in QR code. 1 3 6 2 3 3 1 2 1 6 1 4 + 4 2 1 2 = 10 3 1 7 2) If D = 0 then there is no unique solution for the given system of equations. 3 7 12 1 7 6 3 2 6 68  Ex. 2) By using Cramer’s rule solve the following SOLVED EXAMPLES linear equations. x +y – z = 1, 8x +3y – 6z = 1, –4x – y + 3z = 1 Ex. 1) Solve the following equation by using Cramer’s rule. Solution : Given equations are x+y+z = 6, x–y+z = 2, x+2y–z = 2 x+y–z=1 8x + 3y – 6z = 1 Solution : Given equations are –4x – y + 3z =1 x+y+z = 6 x–y+z = 2 x+2y–z = 2 1 1 1 1 1 −1 D = 8 3 −6 D = 1 −1 1 −4 − 1 3 1 2 −1 = 1(9–6) – 1(24–24) –1(–8+12) = 1(1–2) – 1(–1–1) + 1(2+1) = 3+0–4 = –1 + 2 + 3 = –1 = –1 + 5 1 1 −1 = 4 Dx = 1 3 −6 6 1 1 1 −1 3 Dx = 2 − 1 1 = 1(9–6) – 1(3+6) –1 (–1–3) 2 2 −1 = 3–9+4 = 6(1–2) – 1(–2–2) + 1(4+2) = –2 = –6 + 4 + 6 1 1 −1 = 4 Dy = 8 1 −6 −4 1 3 1 6 1 Dy = 1 2 1 = 1(3+6) – 1(24–24) – 1(8+4) 1 2 −1 = 9 – 0 – 12 = –3 = 1(–2–2) – 6(–1–1) + 1(2–2) 1 1 1 = –4 + 12 + 0 Dz = 8 3 1 = 8 −4 − 1 1 1 1 6 Dz = 1 − 1 2 = 1(3+1) – 1(8+4) + 1(–8+12) 1 2 2 = 4 – 12 + 4 = 8 – 12 = 1(–2–4) – 1(2–2) + 6(2+1) = –4 = –6 + 0 + 18 = 12 Dx −2 Dy −3 ∴ x = = = 2, y = = = 3 and Dx 4 Dy D −1 D −1 8 Dz ∴x= = = 1, y = = =2 and z = Dz −4 D 4 D 4 D ∴z= = =4 D −1 12 = 3 are solutions of given equation. ∴x =2, y = 3, z = 4 are the solutions of the given 4 equations. 69 Ex. 3) Solve the following equations by using Dp -5 5 determinant ∴ p = = = , D -3 3 1 1 1 1 2 1 2 1 3 Dq 5 -5 + + = –2 , − + =3 , − + = –1 x y z x y z x y 2 q = = = , q -3 3 Solution : Dr 6 r = = = –2 1 1 1 D -3 Put =p =q =r x y z 1 5 3 ∴ =p= ∴ x = , ∴ Equations are x 3 5 p + q + r = –2 -5 1 -3 p – 2q + r = 3 ∴ =q= ∴y= , y 3 5 2p – q + 3r = –1 1 -1 ∴ = r = –2 ∴ z = 1 1 1 z 2 D = 1 -2 1 3 -3 -1 2 -1 3 ∴x= ,y= ,z= are the solutions of 5 5 2 = 1(–6+1) – 1(3–2) + 1(–1+4) the equations. = –5 – 1 + 3 = –3 Ex. 4) The cost of 2 books, 6 notebooks and 3 pens is Rs.120. The cost of 3 books, -2 1 1 4 notebooks and 2 pens is Rs.105. while the Dp = 3 -2 1 cost of 5 books, 7 notebooks and 4 pens is -1 - 1 3 Rs.183. Using this information find the cost of 1 book, 1 notebook and 1 pen. = –2(–6+1) – 1(9+1) + 1(–3–2) Solution : Let Rs. x, Rs. y and Rs. z be the cost of = 10 – 10 – 5 one book, one notebook and one pen respectively. = –5 Then by given information we have, 1 -2 1 2x + 6y + 3z = 120 Dq = 1 3 1 3x + 4y + 2z = 105 2 -1 3 5x + 7y + 4z = 183 = 1(9+1) + 2(3-2) + 1(–1–6) 2 6 3 = 10 + 2 – 7 = 5 D = 3 4 2 5 7 4 1 1 -2 Dr = 1 -2 3 = 2(16–14) – 6(12–10) + 3(21–20) 2 -1 -1 = 2(2) – 6(2) + 3(1) = 1(2+3) – 1(–1–6) – 2(–1+4) = 4 – 12 + 3 = 5 + 7 –6 = 7 – 12 = 6 = –5 70  120 6 3 4.3.2 Consistency of three equations in two 40 6 3 105 4 2 = 3 35 4 2 variables Dx = 183 7 4 61 7 4 Consider the system of three linear equations in two variables x and y = 3 [40(16–14) – 6(140–122) + 3(245–244)] = 3[40(2) – 6(18) + 3(1)] a1x+ b1y +c1 = 0 (I) a2x+ b2y +c2 = 0 (II) = 3[80 –108 +3] a3x+ b3y +c3 = 0 (III) = 3[ 83 – 108] = 3[– 25] = –75 These three equations are said to be consistent if they have a common solution. 2 120 3 2 40 3 Dy = 3 3 105 2 = 3 35 2 Theorem : The necessary condition for the 5 183 4 5 61 4 equation a1x + b1y + c1 = 0 , a2x + b2y + c2 = 0, a3x + b3y + c3 = 0 to be consistent is = 3[2(140–122) – 40(12–10) + 3(183–175)] = 3[ 2(18) – 40(2) + 3(8)] a1 b1 c1 a2 b2 c2 = 0 = 3[ 36–80+24] a3 b3 c3 = 3[ 60–80] = 3[–20] Proof : Consider the system of three linear equations in two variables x and y. = –60 2 6 40 2 6 120 a 1x + b 1y + c 1 = 0 Dz = 3 3 4 35 = 3 4 105 a 2x + b 2y + c 2 = 0 (I) 5 7 61 5 7 183 a 3x + b 3y + c 3 = 0 = 3[2(244–245) – 6(183–175) + 40(21–20)] We shall now obtain the necessary condition for the system (I) be consistent. = 3[2(–1) – 6(8) + 40(1)] Consider the solution of the equations = 3[–2 – 48 + 40 ] a2x+b2y = – c2 = 3[–50+40] a3x+b3y = – c3 = 3[–10] a2 b2 = –30 If ≠ 0 then by Cramer’s Rule the system a3 b3 Dx −75 Dy −60 of two unknowns, we have ∴x= = = 15, y = = = 12, D −5 D −5 −c2 b2 a2 −c2 Dz −30 z= = =6 −c3 b3 a3 −c3 D −5 x = , y= put these a2 b2 a2 b2 ∴ Rs.15, Rs. 12, Rs. 6 are the costs of one book, a3 b3 a3 b3 one notebook and one pen respectively. values in equation a1x + b1y + c1 = 0 then 71  -c2 b2 x + y − 2 = 0, 2x +3y − 5 = 0, 3x − 2y − 1 = 0 a2 -c2 -c3 b3 a3 -c3 a1 b1 c1 1 1 -2 a1 + b1 + c1 = 0 a2 b2 a2 b2 a2 b2 c2 = 2 3 -5 a3 b3 a3 b3 a3 b3 c3 3 -2 -1 -c2 b2 a -c2 a b = 1(–3–10) –1(–2+15) –2(–4–9) i.e a1 + b1 2 + c1 2 2 = 0 -c3 b3 a3 -c3 a3 b3 = –13 –13 + 26 = 0 c2 b2 a c2 a b ∴ Given equations are consistent. i.e − a1 − b1 2 + c1 2 2 = 0 c3 b3 a3 c3 a3 b3 ii) x + 2y − 3 = 0, 7x + 4y − 11 = 0, b2 c2 a c2 a b 2x +3y + 1 = 0 i.e a1 − b1 2 + c1 2 2 = 0 b3 c3 a3 c3 a3 b3 a1 b1 c1 1 2 -3 a2 b2 c2 = 7 4 -11 a1 b1 c1 a3 b3 c3 2 3 1 i.e a2 b2 c2 = 0 a3 b3 c3 = 1(4+33) –2(7+22) –3(21–8) = 37 –58 39 = 37 –97 Note : This is a necessary condition that the = −60 ≠ 0 equations are consistant. The above ∴Given system of equations is not consistent. condition of consistency in general is not sufficient. iii) x + y = 1, 2x + 2y = 2, 3x + 3y = 5 Solution : x + y = 1, 2x + 2y = 2, 3x + 3y = 5 are SOLVED EXAMPLES given equations. Check the condition of consistency. Ex. 1) Verify the consistency of following equations 1 1 -1 1 1 -1 2x+2y = −2, x + y = −1, 3x + 3y = −5 2 2 -2 = 2 1 1 -1 Solution : By condition of consistency consider 3 3 -5 3 3 -5 a1 b1 c1 2 2 2 = 2(0) = 0 (R1 and R2 are identical) = 1 1 1 a2 b2 c2 Let us examine further. a3 b3 c3 3 3 5 Note that lines given by the equations = 2(5−3) − 2(5−3) + 2(3−3) = 4 − 4 + 0 = 0 x + y = 1 and 3x + 3y = 5 are parallel to each But the equations have no common Solution. other. They do not have a common solution, so (why?) equations are not consistent. Ex. 3) Find the value of k if the following Ex. 2) Examine the consistency of following equations are consistent. equations. 7x – ky = 4 , 2x + 5y = 9 and 2x + y = 8 i) x + y = 2, 2x +3y = 5, 3x − 2y = 1 Solution : Given equations are Solution : Write the given equation in standard form. 7x – ky – 4 = 0, 2x + 5y –9 = 0, 72  2x + y – 8 = 0 are consistent 1 Area of triangle ABC = PQ.[AP+CQ] 7 −k 4 2 ∴ 2 5 −9 = 0 1 1 + QR.[QC+BR]– PR.[AP+BR] 2 1 −8 2 2 1 1 ∴ 7(–40+9) + k(–16+18) −4(2−10) = 0 = (y1+y3) (x3–x1) + (y2+y3) (x2– x3) 2 2 ∴ 7(–31) +2k –4(–8) = 0 1 – (y1+y2) (x2–x1) ∴ –217 + 2k + 32 = 0 –185 + 2k = 0 2 185 1 ∴ 2k = 185 ∴ k = = [y1.x3 –x1y1 + x3y3–x1y3+x2y2−x3y2 +x2y3 2 2 –x3y3−x2y1+x1y1–x2y2+x1y1] 4.3.3 Area of triangle and Collinearity of three points. 1 = [ y1x3–x1y3–x3y2+x2y3–x2y1+x1y2] 2 Theorem : If A(x1 , y1) , B(x2 , y2) and C(x3 , y3) 1 = [x1(y2–y3) – x2(y1–y3) + x3(y1–y2)] 2 are vertices of triangle ABC then the area of x1 x2 x3 x1 y1 1 triangle is 1 y y y 1 x y 1 = 1 2 3 = 2 2 2 2 x y 1 x1 y1 1 1 1 1 3 3 1 x2 y2 1 Remark: 2 x y3 1 3 i) Area is a positive quantity. Hence we Proof : Consider a triangle ABC in Cartesian always take the absolute value of a determinant. coordinate system. Draw AP, CQ and BR ii) If area is given, consider both positive and perpendicular to the X axis negative values of the determinant for calculation of unknown co-ordinates. iii) If area of a triangle is zero then the given three points are collinear. SOLVED EXAMPLES Ex. 1) Find the area of the triangle whose vertices are A(−2, −3) , B(3, 2) and C(−1, −8) Solution : Given (x1, y1) = (−2, −3), (x2, y2) = (3, 2), and (x3, y3) = (−1 , −8) Fig. 4.1 We know that area of triangle From the figure, x1 y1 1 −2 −3 1 Area of ∆ABC = Area of trapezium PACQ +Area 1 x y2 1 = 1 3 2 1 = 2 of trapezium QCBR –Area of trapezium PABR 2 x y3 1 2 3 −1 −8 1 73  1 Solution : Given (x1, y1) ≡ (3, 7), (x2, y2) ≡ (4, –3) = [–2(2+8)+3(3+1)+1(–24+2)] 2 and (x3, y3) ≡ (5, –13) 1 x1 y1 1 3 7 1 = [−20+12−22] 1 1 2 Area of ∆ = x2 y2 1 = 4 −3 1 2 2 1 1 x3 y3 1 5 −13 1 = [−42+12] = [–30] = –15 2 2 1 Area is positive. = [3(–3+13) – 7(4–5) + 1(−52+15)] 2 ∴ Area of triangle = 15 square unit This gives the area of the triangle ABC 1 1 = [30+7–37] = [37–37] = 0 in that order of the vertices. If we consider the 2 2 same triangle as ACB, then triangle is considered A(∆ABC) = 0 ∴ A, B, C are collinear points in opposite orientation. The area then is 15 sq. units. This also agrees with the rule that Ex. 4) Show that the following points are collinear interchanging 2nd and 3rd rows changes the sign by determinant method. of the determinant. A(2, 5), B(5, 7), C(8, 9) Ex. 2) If the area of triangle with vertices Solution : Given A ≡ (x1, y1) = (2, 5) , P(−3, 0), Q(3, 0) and R(0, K) is 9 square B ≡ (x2, y2) ≡ (5, 7), C ≡ (x3, y3) ≡ (8,9) unit then find the value of k. x1 y1 1 Solution : Given (x1, y1) ≡ (–3, 0), (x2, y2) If x2 y2 1 = 0 then, A, B, C are collinear ≡ (3, 0) and (x3, y3) ≡ (0, k) and area of ∆ is 9 sq. x3 y3 1 unit. 2 5 1 x1 y1 1 1 ∴ 5 7 1 = 2(7–9) –5(5−8) +1(45–56) We know that area of ∆ = x2 y2 1 2 8 9 1 x3 y3 1 = –4 +15 − 11 = –15 +15 = 0 −3 0 1 1 ∴ A, B, C are collinear. ∴ ±9 = 3 0 1 ( Area is positive but the 2 0 k 1 determinant can be of either sign) EXERCISE 4.3 1 Q.1) Solve the following linear equations by ∴ ±9 = [–3(0 – k) + 1(3k – 0)] 2 using Cramer’s Rule. 1 i) x+y+z = 6, x–y+z = 2, x+2y–z = 2 ∴ ±9 = [3 + 3k] ∴ ±9 = 3k ∴ k = ±3 2 ii) x+y−2z = –10, Ex. 3) Find the area of triangle whose vertices are 2x+y–3z = –19, 4x+6y+z = 2 A(3, 7) B(4, −3) and C(5, –13). Interpret iii) x+z = 1, y+z = 1, x+y = 4 your answer. 74  −2 1 3 2 3 1 MISCELLANEOUS EXERCISE - 4 (A) iv) − − = 3 , − + = –13 x y z x y z 2 3 (I) Select the correct option from the given and − = –11 x z alternatives. Q.2) The sum of three numbers is 15. If the a b a+b second number is subtracted from the sum of first and third numbers then we get 5. Q.1 The determinant D = b c b+c When the third number is subtracted from a+b b+c 0 the sum of twice the first number and the = 0 lf second number, we get 4. Find the three A) a,b, c are in A.P. numbers. B) a , b, c are in G.P. Q.3) Examine the consistency of the following C) a, b, c are in H.P. equations. D) α is root of ax2 + 2bx + c = 0 i) 2x−y+3 = 0, 3x+y−2=0, 11x+2y−3 = 0 ii) 2x+3y−4=0, x+2y=3, 3x+4y+5 = 0 xk xk +2 x k +3 iii) x+2y−3 =0, 7x+4y−11=0, 2x+4y−6= 0 Q.2 If yk y k +2 y k +3 = (x–y) (y–z) (z–x) zk z k +2 z k +3 Q.4) Find k if the following equations are consistent. 1 1 1 ( + + ) then i) 2x+3y-2=0, 2x+4y−k=0, x−2y+3k =0 x y z ii) kx +3y+1=0, x+2y+1=0, x+y=0 A) k= –3 B) k = –1 C) k = 1 D) k = 3 Q.5) Find the area of triangle whose vertices are sinθ.cosφ sinθ.sinφ cosθ i) A(5,8), B(5,0) C(1,0) Q.3 Let D = cosθ.cosφ cosθ.sinφ − sinθ then 3 −1 − sinθ.sinφ sinθ.cosφ 0 ii) P( , 1), Q(4, 2), R(4, ) 2 2 iii) M(0, 5), N(−2, 3), T(1, −4) A) D is independent of θ B) D is independent of φ Q.6) Find the area of quadrilateral whose vertices are C) D is a constant A(−3, 1), B(−2, −2), C(3,−1), D(1,4) dD D) at θ =π/2 is equal to 0 Q.7) Find the value of k, if the area of triangle d whose vertices are P(k, 0), Q(2, 2), R(4, 3 Q.4 The value of a for which system of equation 3) is sq.unit a3x + (a + 1)3y + (a + 2)3z = 0 ax + (a +1) y 2 + (a + 2) z = 0 and x + y + z = 0 has non zero Q.8) Examine the collinearity of the following Soln. is set of points A) 0 B) –1 C) 1 D) 2 i) A(3, −1), B(0, −3), C(12, 5) ii) P(3, −5), Q(6, 1), R(4, 2) iii) L(0, 1 ), M (2, −1), N(−4, 7 ) 2 2 75  b+c c+a a+b C) Determinant is number associated to Q.5 q+r r+ p p+q = square matrix y+z z+x x+ y D) None of these c b a b a c a b c (II) Answer the following questions. A) 2 r q p B) 2 q p r C) 2 p q r 2 −5 7 1 −3 12 z y x y x z x y z Q.1) Evaluate i) 5 2 1 ii) 0 2 −4 9 0 2 9 7 2 a c b D) 2 p r q Q.2) Evaluate determinant along second column x z y 1 −1 2 3 2 −2 Q.6 The system 3x – y + 4z = 3, x + 2y –3z 0 1 −2 = –2 and 6x + 5y + λz = –3 has at least one Solution when 2 3 5 A) λ = –5 B) λ = 5 Q.3) Evaluate i) 400 600 1000 C) λ = 3 D) λ = –13 48 47 18 x 3 7 101 102 103 ii) 106 107 108 by using properties Q.7 If x = –9 is a root of 2 x 2 = 0 has 7 6 x 1 2 3 other two roots are Q.4) Find minor and cofactor of elements of the A) 2, –7 B) –2, 7 C) 2, 7 D) –2, –7 determinant. −1 0 4 1 −1 2 6i −3i 1 i) −2 1 3 ii) 3 0 −2 Q.8 If 4 3i −1 = x + iy then 1 0 3 0 −4 2 20 3 i Q.5) Find the value of x if A) x = 3 , y = 1 B) x = 1 , y = 3 C) x = 0 , y = 3 D) x = 0 , y = 0 1 4 20 1 2x 4x i) 1 −2 −5 = 0 ii) 1 4 16 = 0 Q.9 If A(0,0), B(1,3) and C(k,0) are vertices of 1 2 x 5x2 1 1 1 triangle ABC whose area is 3 sq.units then value of k is Q.6) By using properties of determinant prove A) 2 B) –3 C) 3 or −3 D) –2 or +2 x+ y y+z z+x Q.10 Which of the following is correct that z x y =0 1 1 1 A) Determinant is square matrix B) Determinant is number associated to matrix 76 Q.7) Without expanding determinant show that iii) (k−2)x +(k−1)y =17 , (k−1)x+ (k−2)y = 18 and x + y = 5 b + c bc b 2 c 2 xa yb zc i) c + a ca c a = 0 ii) a 2 2 2 b2 c2 Q.11) Find the area of triangle whose vertices are a + b ab a 2b 2 1 1 1 i) A(−1,2), B(2,4), C(0,0) ii) P(3,6), Q(−1,3), R (2,−1) x y z iii) L(1,1), M (−2,2), N (5,4) = a b c bc ca ab Q.12) Find the value of k i) If area of triangle is 4 square unit and l m n n f w vertices are P(k, 0), Q(4, 0), R(0, 2) iii) e d f = l e u ii) If area of triangle is 33/2 square unit and u v w m d v vertices are L (3,−5), M(−2,k), N (1,4) Q.13) Find the area of quadrilateral whose 0 a b vertices are A (0, −4), B(4, 0) , C(−4, 0), iv) −a 0 c = 0 D (0, 4) −b −c 0 Q.14) An amount of ` 5000 is put into three investments at the rate of interest of 6% , 7% a 1 1 and 8% per annum respectively. The total Q.8) If 1 b 1 =0 then show that annual income is ` 350. If the combined 1 1 c income from the first two investments is ` 70 more than the income from the third. 1 1 1 Find the amount of each investment. + + =1 1− a 1− b 1− c Q.15) Show that the lines x−y=6, 4x−3y=20 and 6x+5y+8=0 are concurrent.Also find the Q.9) Solve the following linear equations by point of concurrence Cramer’s Rule. Q.16) Show that the following points are collinear i) 2x−y+z = 1, x+2y+3z = 8, 3x+y−4z =1 by determinant 1 1 5 1 1 4 ii) 1 + 1 = 3 , + = , + = a) L (2,5), M(5,7), N(8,9) x y 2 y z 6 z x 3 b) P(5,1), Q(1,−1), R(11,4) iii) 2x+3y+3z=5 , x−2y+z = – 4 , Further Use of Determinants 3x– y– 2z=3 1) To find the volume of parallelepiped and iv) x–y+2z=7 , 3x+4y–5z=5 , 2x–y+3z=12 tetrahedron by vector method Q.10) Find the value of k if the following equation 2) To state the condition for the equation are consistent. ax2+2hxy+by2 +2gx+2fy+c =0 representing a i) (k+1)x+(k−1)y+(k−1) = 0 pair of straight lines. (k−1)x+(k+1)y+(k−1) = 0 3) To find the shortest distance between two (k−1)x+(k−1)y+(k+1) = 0 skew lines. 4) Test for intersection of two line in three ii) 3x+y−2=0 kx+2y−3=0 and 2x−y = 3 dimensional geometry. 77 5) To find cross product of two vectors and B is a matrix having 3 rows and 2 columns. scalar triple product of vectors The order of B is 3×2. There are 6 elements in matrix B. 6) Formation of differential equation by eliminating arbitrary constant. 1 + i 8  iii) C =  , C is a matrix of order 2×2.  i −3i  Let's Learn  −1 9 2  iv) D =  3 0 −3 , D is a matrix of order 2×3.  4.4 Introduction to Matrices : In general a matrix of order m x n is represented The theory of matrices was developed by a Mathematician Arthur Cayley. Matrices are by useful in expressing numerical information  a11 a12 … a1 j … a1n  in compact form. They are effectively used a … a 2 n   21 a 22 … a2j in expressing different operators. Hence in  a 31 a 32 … a3j … a 3n  Economics, Statistics and Computer science they A = [ aij]mxn =   are essential. … … … … … …  a i1 a i 2 … a ij … a in  Definition : A rectangular arrangement of mn   a m1 a m 2 … a mj … a mn  numbers in m rows and n columns, enclosed in [ ] or ( ) is called a matrix of order m by n. Here aij = An element in ith row and jth column. A matrix by itself does not have a value or any special meaning.  2 −3 9   a11 a12 a13   Ex. In matrix A = 1 0 −7  =  a  21 a 22 a 23  Order of the matrix is denoted by m × n, read as  4 −2 1   a 31 a 32 a 33  m by n. Each member of the matrix is called an element a11 = 2, a12 = –3, a13 = 9, a21 = 1, a22 = 0, a23 = of the matrix. –7, a31 = 4, a32 = –2, a33 = 1 Matrices are generally denoted by A, B, C ,… and their elements are denoted by aij, bij, cij, … etc. 4.4.1 Types of Matrices : e.g. aij is the element in ith row and jth column of 1) Row Matrix : A matrix having only one the matrix. row is called as a row matrix. It is of order  2 −3 9  1 x n, Where n ≥ 1. 1 0 −7  For example, i) A =   Here a32 = – 2 Ex. i) [–1 2]1×2 ii) [0 –3 5]1×3  4 −2 1  2) Column Matrix : A matrix having only one A is a matrix having 3 rows and 3 columns. The column is called as a column matrix. It is of order of A is 3×3, read as three by three. There order m x 1, Where m ≥ 1. are 9 elements in matrix A. 5  −1 −5 1      Ex. i)   ii)  −9  ii) B =  2 0  0   2 x1  −3 3 x1  6 9  78  Note : Single element matrix is row matrix 5) Diagonal Matrix : A square matrix in which as well as column matrix. e.g. 1x1 every non-diagonal element is zero, is called a diagonal matrix. 3) Zero or Null matrix : A matrix in which every element is zero is called as a zero or 5 0 0    null matrix. It is denoted by O. Ex. i) A = 0 0 0  = diag (5,0,9) 0 0 9  3 x 3 0 0 0    Ex. i) O = 0 0 0   −1 0  0 0 0  3 x 3 ii) B =    0 −5 2 x 2 0 0   −1 0 0  ii) O = 0 0   iii) C =  0 −2 0   0 0  3 x 2  0 0 −3 2 x 2 4) Square Matrix : A matrix with equal Note: If a11, a22, a33 are diagonal elements of a number of rows and coloumns is called a diagonal matrix A of order 3, then we write the square matrix. matrix A as A = Diag.  5 −3 i   1 0 −7  Examples, i) A =   6) Scalar Matrix : A diagonal matrix in which  2i −8 9  3 x 3 all the diagonal elements are same, is called as a scalar matrix.  −1 0  ii) C =  1 −5 5 0 0    2x 2   For Ex. i) A = 0 5 0  Note : A matrix of order n×n is also called as 0 0 5  3 x 3 square matrix of order n.  −2 0  Let A = [ aij]nxn be a square matrix of order n then ii) B =    0 −2  2 x 2 (i) The elements a11, a22, a33,… aii … ann are called the diagonal elements of matrix A. 7) Unit or Identity Matrix : A scalar matrix in which all the diagonal elements are 1(unity) , Note that the diagonal elements are defined is called a Unit Matrix or an Identity Matrix. only for a square matrix. An Identity matrix of order n is denoted (ii) Elements aij, where i ≠ j are called non by In. diagonal elements of matrix A. 1 0 0  1 0  (iii) Elements aij, where i < j represent elements Ex. i) I3 = 0 1 0  ii) I2 =     0 1  above the diagonal. 0 0 1  Note: (iv) Elements aij, where i > j represent elements below the diagonal. 1. Every Identity matrix is a Scalar matrix but every scalar matrix need not be Identity Statements iii) and iv) can be verified by observing matrix. However a scalar matrix is a scalar square matrices of different orders. multiple of the identity matrix. 79 2. Every scalar matrix is a diagonal matrix but  −3 1 every diagonal matrix need not be a scalar ii) B =    1 8 2 x 2 matrix. 8) Upper Triangular Matrix : A square matrix  2 4 −7  in which every element below the diagonal iii) C =  4 5 −1 is zero, is called an upper triangular matrix.  −7 −1 −3 3 x 3 Matrix A = [ aij]nxn is upper triangular if Note: aij = 0 for all i > j. The scalar matrices are symmetric. A null square  4 −1 2  matrix is symmetric.   For Ex. i) A =  0 0 3  12) Skew-Symmetric Matrix : A square matrix  0 0 9  3 x 3 A = [ aij]nxn in which aij = − aji, for all i and j, is called a skew symmetric matrix.  −3 1 ii) B =   Here for i = j, aii = − aii, 2aii = 0 aii = 0 for  0 8 2 x 2 all i = 1, 2, 3,……n. 9) Lower Triangular Matrix : A square matrix In a skew symmetric matrix each diagonal in which every element above the diagonal element is zero. is zero, is called a lower triangular matrix. Matrix A = [ aij]nxn is lower triangular if  0 5 e.g. i) A =   aij = 0 for all i < j.  −5 0  2 x 2  2 0 0  0 4 −7  For Ex. i) A =    −1 1 0  ii) B =  −4 0 5   −5 1 9  3 x 3  7 −5 0  3 x 3  7 0 ii) B =   Note : A null square matrix is also a skew  −1 3 2 x 2 symmetric. 10) Triangular Matrix : A square matrix is 13) Determinant of a Matrix : Determinant of called a triangular matrix if it is an upper a matrix is defined only for a square matrix. triangular or a lower triangular matrix. If A is a square matrix, then the same arrangement Note : The diagonal, scalar, unit and null matrices of the elements of A also gives us a determinant, are also triangular matrices. by replacing square brackets by vertical bars. It is denoted by |A| or det(A). 11) Symmetric Matrix : A square matrix A = If A = [ aij]nxn, then is of order n. [ aij]nxn in which aij = aji, for all i and j, is called a symmetric matrix.  1 3 Ex. i) If A =   a h g  −5 4  2 x 2   Ex. i) A =  h b f  1 3  g f c  3 x 3 then |A| = -5 4 80   2 −1 3 15) Transpose of a Matrix : The matrix   obtained by interchanging rows and columns ii) If B =  −4 1 5   7 −5 0  3 x 3 of matrix A is called Transpose of matrix A. It is denoted by A' or AT. If A is matrix of 2 -1 3 order m × n, then order of AT is n × m. then |B| = -4 1 5 If AT = A' = B then bij = aji 7 -5 0  −1 5  14) Singular Matrix : A square matrix A is   e.g. i) If A =  3 −2  said to be a singular matrix if |A| = det(A)  4 7  3×2 = 0, otherwise it is said to be a non-singular matrix.  −1 3 4  6 3 then AT =   Ex. i) If A =   5 −2 7  2×3  8 4  2 x 2 1 0 −2  6 3   then |A| = = 24–24 = 0. ii) If B = 8 −1 2  8 4  4 3 5  3×3 Therefore A is a singular matrix.  1 8 4 2 3 4   then B =  0 −1 3  T ii) If B =  3 4 5  then  4 5 6  3 x 3  −2 2 5  3×3 Remark: 2 3 4 |B| = 3 4 5 1) If A is symmetric then A = AT 4 5 6 2) If B is skew symmetric, then is B = −BT |B| = 2(24–25)–3(18–20) + 4(15–16) Activity : = – 2+6–4 = 0 Construct a matrix of order 2 × 2 where the aij th |B| = 0 Therefore B is a singular matrix. (i + j ) 2 element is given by aij =  2 −1 3 2+i   iii) A =  −7 4 5  then a a12   −2 1 6  3 X 3 Solution : Let A =  11  be the required a 21 a 22  2 X 2 2 -1 3 matrix. |A| = -7 4 5 (i + j ) 2 (……..) 2 4 Given that aij = , a11 = = , -2 1 6 2+i ……+ 1 3 |A| = 2(24-5)-(–1)(-42+10)+3(–7+8) (........)2 9 a12 = = =...... = 38–32+3 = 9........... 3 |A| = 9, As |A| ≠ 0 , A is a non-singular (2 + 1) 2........ matrix. a21 = = , 2+2 4 81  (……….) 2........  2 a 3 a22 = = =4 2+2........ Ex. 3) Find a, b, c if the matrix A =  −7 4 5   c b 6   4 