Chapter 4.1 Overview To every square matrix A = [aij] of order n, we can associate a number (real or complex) called determinant of the matrix A, written as det A, where aij is the (i, j)th element of A. ab If A , then determinant of A, denoted by |A| (or det A), is given by cd ab|A| = = ad – bc.cd Remarks (i) Only square matrices have determinants. (ii) For a matrix A, A is read as determinant of A and not, as modulus of A. 4.1.1 Determinant of a matrix of order one Let A = [a] be the matrix of order 1, then determinant of A is defined to be equal to a. 4.1.2 Determinant of a matrix of order two ⎡ab⎤ Let A = [aij] = ⎢⎥ be a matrix of order 2. Then the determinant of A is defined cd⎣⎦as: det (A) = |A| = ad – bc. 4.1.3 Determinant of a matrix of order three The determinant of a matrix of order three can be determined by expressing it in terms of second order determinants which is known as expansion of a determinant along a row (or a column). There are six ways of expanding a determinant of order 3 corresponding to each of three rows (R, R and R) and three columns (C, C and12312C3) and each way gives the same value. Consider the determinant of a square matrix A = [a], i.e., ij3×3aaa1112 13 A aaa2122 23 aaa31 32 33 Expanding |A| along C1, we get aa aa22 23 12 13 + a21 (–1)2+1|A| = a11 (–1)1+1 aaaa32 33 32 33 + a31 (–1)3+1 aa12 13 aa22 23 = – ) – a21 – ) + a31 – )a11(a22 a33a23 a32 (a12 a33a13 a32 (a12 a23a13 a22Remark In general, if A = kB, where A and B are square matrices of order n, then |A| = kn |B|, n = 1, 2, 3. 4.1.4 Properties of Determinants For any square matrix A, |A| satisfies the following properties. (i) |A′| = |A|, where A′ = transpose of matrix A. (ii) If we interchange any two rows (or columns), then sign of the determinant changes. (iii) If any two rows or any two columns in a determinant are identical (or proportional), then the value of the determinant is zero. (iv) Multiplying a determinant by k means multiplying the elements of only one row (or one column) by k. (v) If we multiply each element of a row (or a column) of a determinant by constant k, then value of the determinant is multiplied by k. (vi) If elements of a row (or a column) in a determinant can be expressed as the sum of two or more elements, then the given determinant can be expressed as the sum of two or more determinants. (vii) If to each element of a row (or a column) of a determinant the equimultiples of corresponding elements of other rows (columns) are added, then value of determinant remains same. Notes: (i) If all the elements of a row (or column) are zeros, then the value of the determinant is zero. (ii) If value of determinant ‘Δ’ becomes zero by substituting x = α, then x – α is a factor of ‘Δ’. (iii) If all the elements of a determinant above or below the main diagonal consists of zeros, then the value of the determinant is equal to the product of diagonal elements. 4.1.5 Area of a triangle Area of a triangle with vertices (x1, y1), (x2, y2) and (x3, y3) is given by x1 y11 1 x2 y21 .2 x3 y31 4.1.6 Minors and co-factors (i) Minor of an element aij of the determinant of matrix A is the determinant obtained by deleting ith row and jth column, and it is denoted by Mij. (ii) Co-factor of an element aij is given by Aij = (–1)i+j Mij. (iii) Value of determinant of a matrix A is obtained by the sum of products of elements of a row (or a column) with corresponding co-factors. For example |A| = a11 A + a A + a A11 12121313. (iv) If elements of a row (or column) are multiplied with co-factors of elements of any other row (or column), then their sum is zero. For example, A + a A + a A = 0.a11 21122213234.1.7 Adjoint and inverse of a matrix(i) The adjoint of a square matrix A = [a] is defined as the transpose of the matrix(ii) A (adj A) = (adj A) A = |A| I, where A is square matrix of order n. ijn×n[a], where A is the co-factor of the element a. It is denoted by adj A. n×nij ij ij If A 11 21 31 a a a 12 22 32 a a a 13 23 33 a a a, then adj 11 12 13 A AA A 21 22 23 A A A 31 32 33 A A , A where Aij is co-factor of aij. (iii) A square matrix A is said to be singular or non-singular according as |A| = 0 or |A| ≠ 0, respectively. (iv) If A is a square matrix of order n, then |adj A| = |A|n–1.(v) If A and B are non-singular matrices of the same order, then AB and BA are also nonsingular matrices of the same order. (vi) The determinant of the product of matrices is equal to product of their respective determinants, that is, |AB| = |A| |B|. (vii) If AB = BA = I, where A and B are square matrices, then B is called inverse of A and is written as B = A–1. Also B–1 = (A–1)–1 = A. (viii) A square matrix A is invertible if and only if A is non-singular matrix. 1 (ix) If A is an invertible matrix, then A–1 = |A| (adj A) 4.1.8 System of linear equations (i) Consider the equations: a1x + b1 y + c1 z = d1 a2x + b2 y + c2 z = d2 a3x + b3 y + c3 z = d3, In matrix form, these equations can be written as A X = B, where 1a 1b 1c x 1d A = 2a 2b 2c ,X y and B 2d 3a 3b 3c z 3d (ii) Unique solution of equation AX = B is given by X = A–1B, where |A| ≠ 0. (iii) A system of equations is consistent or inconsistent according as its solution exists or not. (iv) For a square matrix A in matrix equation AX = B (a) If |A| ≠ 0, then there exists unique solution. (b) If |A| = 0 and (adj A) B ≠ 0, then there exists no solution. (c) If |A| = 0 and (adj A) B = 0, then system may or may not be consistent. 4.2 Solved Examples Short Answer (S.A.) 2x 5 65 Example 1 If , then find x.8 x83 2x 5 65 Solution We have . This gives 8 x 83 2x2 – 40 = 18 – 40 ⇒ x2 = 9 ⇒ x = ± 3. 1 2xx 1 1 1 Example 2 If 1Δ= y 2 1,y Δ = yz zx xy , then prove that Δ + Δ1 = 0. 1 2zz x y z 1 1 1 Solution We have 1 yz zx xy x y z Interchanging rows and columns, we get x xyz x2 1 yz x 1 = y xyz y21 zx y1 xyz1 xyz z xyz z2 x1 x2 xyz y1 y2 Interchanging C1 and C2 = xyz z1 z2 1 xx2 (–1) 1 yy2 – = 1 zz2 ⇒Δ1+ Δ = 0 Example 3 Without expanding, show that cosec 2 cot 2 1 cot2 cosec 2 1 = 0. 42 40 2 Solution Applying C→ C – C – C, we have1 12322 2 0 cot 2θ 1cosec – cot –1 cot 1 22 2 0cosec 2θ−1 =0cot – cosec 1 cosec 1 = 040 20 40 2 xpq Example 4 Show that = (x– p) (x2 + px– 2q2)pxq qqxSolution Applying C1 → C1 – C2, we have xppq 1 pq pxxq (xp) 1 xq 0 qx 0 qx 0 px+ 2q ( ) 1xp=− − x q Applying R1 → R1 + R2 0 q x Expanding along C1, we have 22 22( )( xx 2q)= ( )( x2q)xppxpx p0 baca Example 5 If ,then show that is equal to zero.ab 0 cb ac bc 0 0 ab ac Solution Interchanging rows and columns, we get ba 0 bc ca cb 0 Taking ‘–1’ common from R1, R2 and R3, we get 0 baca (–1) 3 ab 0 cb – ac bc 0 ⇒ 2 = 0 or = 0 Example 6 Prove that (A–1)′ = (A′)–1, where A is an invertible matrix. Solution Since A is an invertible matrix, so it is non-singular. We know that |A| = |A′|. But |A| ≠ 0. So |A′| ≠ 0 i.e. A′ is invertible matrix. Now we know that AA–1 = A–1 A = I. Taking transpose on both sides, we get (A–1)′ A′ = A′ (A–1)′ = (I)′ = I Hence (A–1)′ is inverse of A′, i.e., (A′)–1 = (A–1)′ Long Answer (L.A.) x 23 1 x 1Example 7 If x= – 4 is a root of = 0, then find the other two roots. 32 x Solution Applying R1 → (R1 + R2 + R3), we get x 4 x 4 x 4 1 x 1 . 3 2 x Taking (x + 4) common from R1, we get 111 (x 4) 1 x 1 32 x Applying C2 → C2 – C1, C3 → C3 – C1, we get 1 0 0 (x 4) 1 x 1 0 . 3 1 x 3 Expanding along R1, Δ = (x + 4) [(x – 1) (x – 3) – 0]. Thus, Δ = 0 implies x = – 4, 1, 3 Example 8 In a triangle ABC, if 1 11 1sinA 1 sinB 1sinC 0 , 2 22sinA+sin A sinB+sin B sinC+sin Cthen prove that ΔABC is an isoceles triangle. 1 11 1sinA 1 sinB 1sinC Solution Let Δ = 22sinA+sin A sinB+sin B sinC+sin C 111 1sinA 1 sinB 1 sinC = R3 → R3 – R22 22cosA cosB cos C10 0 1 sinA sinB sinA sinC sinB = . (C3 → C3 – C2 and C2 → C2 – C1)222 22cos A cos A cos B cos B cos C Expanding along R1, we get Δ = (sinB – sinA) (sin2C – sin2B) – (sinC – sin B) (sin2B – sin2A) = (sinB – sinA) (sinC – sinB) (sinC – sin A) = 0 ⇒ either sinB – sinA = 0 or sinC – sinB or sinC – sinA = 0 ⇒ A = B or B = C or C = A i.e. triangle ABC is isoceles. 3 2 sin3 17 8 cos2 0 , then sinθ = 0 orExample 9 Show that if the determinant 2. 1114 2 Solution Applying R2 → R2 + 4R1 and R3 → R3 + 7R1, we get 3 2 sin3 5 0 cos2 4sin3 0 10 0 2+7sin3 or 2 [5 (2 + 7 sin3θ) – 10 (cos2θ + 4sin3θ)] = 0 or 2 + 7sin3θ – 2cos2θ – 8sin3θ = 0 or 2 – 2cos 2θ – sin 3θ = 0 sinθ (4sin2θ + 4sinθ – 3) = 0 or sinθ = 0 or (2sinθ – 1) = 0 or (2sinθ + 3) = 0 1 or sinθ = 0 or sinθ = (Why ?).2Objective Type Questions Choose the correct answer from the given four options in each of the Example 10 and 11. Example 10 Let (A) Δ1 = – Δ Axx21 ABC Byy2 1and 1 xyz , then zy zx xy Czz21 (C) Δ – Δ1 = 0 Solution (C) is the correct answer since 1 Axx2 xyz xyz1 Byy2 xyz = = xyzxyz 2Czz Example 11 If x, y ∈ R, then the determinant in the interval (A) 2,2 (C) 2,1 (B) (D) AB C A x yz x y z = B yzx zy zx xy C zxy xyz cos x sin x cos( xy) (B)[–1, 1] (D) 1, 2, Δ≠Δ1 None of these Axx21 Byy21 = Δ Cz z21 sin x 1 cos x 1 sin( xy)0 lies Solution The correct choice is A. Indeed applying R3→ R3 – cosyR1 + sinyR2, we get cos x sin x 1 sin x cos x 1 . 0 0 sin y cos y Expanding along R3, we have Δ = (siny – cosy) (cos2x + sin2x) = (siny – cosy) = 2 1 2 sin y 1 2 cos y = 2cos sin 4 y sin cos 4 y = 2 sin (y – 4 π ) Hence – 2 ≤ Δ ≤ 2 . Fill in the blanks in each of the Examples 12 to 14. Example 12 If A, B, C are the angles of a triangle, then 2sinA cotA 1 2sin B cotB 1 ................ 2sinC cotC 1 Solution Answer is 0. Apply R→ R – R, R→ R – R1.2 213 3Example 13 The determinant Δ= 15 + 46 5 10 is equal to ............... common from C and C and applyingSolution Answer is 0.Taking 523C1 → C3 – 3 C2, we get the desired result. Example 14 The value of the determinant 2 2sin 23 sin 67 cos180 2 22sin 67 sin 23 cos 180 .......... cos180 sin 223 sin 267 Solution Δ = 0. Apply C1 → C1 + C2 + C3. State whether the statements in the Examples 15 to 18 is True or False. Example 15 The determinant cos ( x )y sin ( x )y cos 2 y sin x cos x sin y cos x sin x cos y is independent of x only. Solution True. Apply R1 → R1 + sinyR2 + cosy R3, and expand Example 16 The value of 11 1 nn+2 n+4CCC111 is 8. nn+2 n+4CCC222 Solution True x 52 Example 17 If A 2 y 3, xyz = 80, 3x + 2y + 10z = 20, then 11 z 810 0 A adj. A 081 0 . 0 081Solution : False. 15422013 –11 3 Example 18 If A 12 x,A 232231 11 y22 then x= 1, y= – 1. Solution True 4.3 EXERCISE Short Answer (S.A.) Using the properties of determinants in Exercises 1 to 6, evaluate: x2 x 1 x 1 1. x 1 x 1 0 xy2 xz2 x2 y 0 yz2 3. 22xz zy 0 2. 4. axy z xay z x yaz 3x xy xz xy 3y zy xz yz 3z x 4 xx abc 2a 2a xx 4 x 2b bca 2b5. 6. x xx 4 2c 2c cab Using the proprties of determinants in Exercises 7 to 9, prove that: 22yz yz y z 7. 22zx zx z x 22xy xy x y 0 8. yz z y z z x x 4xyz y x x y 9. 2 2 2 1 3 a a a 2 1 2 3 a a 1 1 1 (a 31) 10. If A + B + C = 0, then prove that 1 cosC cos B cosC 1 cos A cos B cos A 1 0 11. If the co-ordinates of the vertices of an equilateral triangle with sides of length x1 y1 12 3a4 x2 y21 = ‘a’ are (x1, y1), (x2, y2), (x3, y3), then .4 x3 y31 1⎡ 1 sin 3 θ ⎤ 12. Find the value of θ satisfying 4⎢−⎢ 3 cos 2 0⎥θ = ⎥ . 7⎢⎣ 7− 2− ⎥⎦ 4 x 4 x 4 x 13. If 4 x 4 x 4 x 0 , then find values of x. 4 x 4 x 4 x 14. If a1, a2, , ..., a are in G.P., then prove that the determinant a3rra 1 ra 5 ra 9 ra 7 ra 11 ra 15 is independent of r. ra 11 ra 17 ra 21 15. Show that the points (a + 5, a – 4), (a – 2, a + 3) and (a, a) do not lie on a straight line for any value of a. 16. Show that the ΔABC is an isosceles triangle if the determinant ⎡ 1 11 ⎤ ⎢ ⎥Δ= 1+cosA 1+cosB 1+cosC =0.⎢ ⎥ ⎢ 2 22 ⎥cos A +cosA cos B +cosB cos C +cosC ⎣ ⎦ 011 2A 3I 17. Find A–1 if A 1 0 1 and show that A–1 .2110Long Answer (L.A.) ⎡ 120 ⎤ ⎢⎥18. If A =−2 −1 −2 , find A–1.⎢⎥ ⎢ 0 −11 ⎥⎣⎦ A–1Using , solve the system of linear equations x – 2y = 10 , 2x – y – z = 8 , –2y + z = 7. 19. Using matrix method, solve the system of equations 3x + 2y – 2z = 3, x + 2y + 3z = 6, 2x – y + z = 2. 224 110 20. Given A 4 2 4 ,B 2 3 4 , find BA and use this to solve the 2 15 012 system of equations y + 2z = 7, x – y = 3, 2x + 3y + 4z = 17. abc bca21. If a + b + c ≠ 0 and = 0 , then prove that a = b = c. cab 22 2bca ca b ab c 222cab ab c bca22. Prove that is divisible by a + b + c and find the 222abc bc a cab quotient. xayb zc a b c 23. If x + y + z = 0, prove that yc za xb xyz= c a b zb xc ya b c a Objective Type Questions (M.C.Q.) Choose the correct answer from given four options in each of the Exercises from 24 to 37. 24. If 2 8 x 5 x 6 7 2 3 , then value of x is (A) 3 (B) ± 3 (C) ± 6 (D) 6 ab b + ca− ba c + ab−25. The value of determinant ca a + bc− (A) a3 + b3 + c3 (B) 3 bc (C) a3 + b3 + c3 – 3abc (D) none of these 26. The area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq. units. The value of k will be (A) 9 (B) 3 (C) – 9 (D) 6 b2 ab bc bc ac ab a2 abb2 ab 27. The determinant equals bcac caab a2 (A) abc (b–c) (c – a) (a – b) (B) (b–c) (c – a) (a – b) (C) (a + b + c) (b – c) (c – a) (a – b) (D) None of these sin xcos xcos x cos xsin xcos x28. The number of distinct real roots of 0 in the interval cos xcos xsin x ππ x−≤≤ is44 (A)0 (B) 2 (C)1 (D) 3 29. If A, B and C are angles of a triangle, then the determinant 1 cosC cos B cosC 1 cos A is equal to cos B cos A 1 (A)0 (B) – 1 (C) 1 (D) None of these cos tt 1 f()t2sin 2tt t 30. Let f(t) = , then lim 2 is equal tot 0 tsin ttt (A)0 (B) – 1 (C)2 (D) 3 1 1 1 31. The maximum value of 1 1 sin 1 is (θ is real number) 1cos 1 1 13(A) (B)22 (C) 2 (D) 4 0 xaxb xa 0 x c32. If f (x) = , then xb xc0 (A) f (a) = 0 (B) f (b) = 0 (C) f (0) = 0 (D) f (1) = 0 23 33. If A = 02 5 , then A–1 exists if 11 3(A) λ = 2 (B) λ≠ 2 (C) λ≠ – 2 (D) None of these 34. If A and B are invertible matrices, then which of the following is not correct? (A) adj A = |A|. A–1 (B) det(A)–1 = [det (A)]–1 (C) (AB)–1 = B–1 A–1 (D) (A + B)–1 = B–1 + A–1 1 x 11 35. If x, y, z are all different from zero and 0 , then value of 1 11 z 11 y 1 x–1 + y–1 + z–1 is x–1(A) x y z (B) y–1 z–1 (C) – x – y – z (D) –1 x xyx 2y 36. The value of the determinant isx 2yx xy xy x 2y x(A) 9x2 (x + y) (B) 9y2 (x + y) (C) 3y2 (x + y) (D) 7x2 (x + y) 1–2 5 37. There are two values of awhich makes determinant, Δ = 2 a 1 = 86, then 0 4 2a sum of these number is (A) 4 (B) 5 (C)– 4 (D) 9 Fill in the blanks 38. If A is a matrix of order 3 × 3, then |3A| = _______ . 39. If A is invertible matrix of order 3 × 3, then |A–1 | _______ . 22x – xx – x22 221 22x – xx – x33 33 1 is40. If x, y, z∈ R, then the value of determinant 22x – xx – x44 441 equal to _______. 0 cos θ 2sin θ 41. If cos2θ = 0, then cosθ sin θ 0 _________. = sin θ 0 cos θ 42. If A is a matrix of order 3 × 3, then (A2)–1 = ________. 43. If A is a matrix of order 3 × 3, then number of minors in determinant of A are ________. 44. The sum of the products of elements of any row with the co-factors of corresponding elements is equal to _________. x37 45. If x= – 9 is a root of = 0, then other two roots are __________.2 x2 76 x0 x−yz xz yx− 0 yz−46. = __________. zx zy −0− 17 19 23(1+ x) (1 + x) (1 + x) 2329 34(1+ x) (1 + x) (1+ x)47. If f(x) = = A + Bx+ Cx2 + ..., then 4143 47(1+ x) (1 + x) (1+ x) A = ________. State True or False for the statements of the following Exercises: 48. A3–1 = A 1 3, where A is a square matrix and |A| ≠ 0. –149. (aA)–1 = 1A , where a is any real number and A is a square matrix. a 50. |A–1| ≠ |A|–1 , where A is non-singular matrix. 51. If A and B are matrices of order 3 and |A| = 5, |B| = 3, then |3AB| = 27 × 5 × 3 = 405. 52. If the value of a third order determinant is 12, then the value of the determinant formed by replacing each element by its co-factor will be 144. x+1 x+ 2 xa + x+ 2 x+3 xb +53. =0 , where a, b, c are in A.P. x+ 3 x+ 4 xc + 54. |adj.A| = |A|2 , where A is a square matrix of order two. sinA cosA sinA+cosB sinB cosA sinB+cosB 55. The determinant is equal to zero. sinC cosA sinC+cosB xapul f yb qv mg 56. If the determinant splits into exactly K determinants of +zcrw nhorder 3, each element of which contains only one term, then the value of K is 8. apx 57. Let bqy crz 16 , then Δ=1 11 pxaxap +++ qyby bq +++ =32 . rz cz cr +++ 1 1is .58. The maximum value of 1 (1 sin ) 1 21 1 1cos

RELOAD if chapter isn't visible.