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Chapter, , 4, , DETERMINANTS, 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., If A =, , a b, , then determinant of A, denoted by |A| (or det A), is given by, c d, , |A| =, , a b, = ad – bc., c d, , 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, a b , Let A = [aij] = ,  be a matrix of order 2. Then the determinant of A is defined, c d , 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 (R1, R2 and R3) and three columns (C1, C2 and, C3) and each way gives the same value., , 20/04/2018

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66, , MATHEMATICS, , Consider the determinant of a square matrix A = [aij]3×3, i.e.,, a11, A = a21, , a12, a22, , a13, a23, , a31, , a32, , a33, , Expanding |A| along C1, we get, |A| = a11 (–1)1+1, , a22, , a23, , a32, , a33, , + a21 (–1)2+1, , a12, , a13, , a32, , a33, , + a31 (–1)3+1, , a12, , a13, , a22, , a23, , = a11(a22 a33 – a23 a32) – a21 (a12 a33 – a13 a32) + a31 (a12 a23 – a13 a22), Remark 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., , 20/04/2018

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DETERMINANTS, , (vii), , 67, , 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 ‘∆’ becomes zero by substituting x = α, then x – α is a, factor of ‘∆’., , (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, 1, ∆ = x2, 2, x3, , y1 1, y2 1, ., y3 1, , 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 A11 + a12 A12 + a13 A13., , (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,, a11 A21 + a12 A22 + a13 A23 = 0., , 4.1.7 Adjoint and inverse of a matrix, (i), , The adjoint of a square matrix A = [aij]n×n is defined as the transpose of the matrix, , 20/04/2018

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68, , MATHEMATICS, , [aij]n×n, where Aij is the co-factor of the element aij. It is denoted by adj A., a11, If A = a21, , a12, a22, , a13, A11, a23 , then adj A = A12, , A 21, A 22, , A31, A 32 , where Aij is co-factor of aij., , a31, , a32, , a33, , A13, , A 23, , A33, , (ii), , A (adj A) = (adj A) A = |A| I, where A is square matrix of order n., , (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., (ix), , If A is an invertible matrix, then A–1 =, , 1, (adj A), |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, , A=, (ii), , a1, a2, , b1, b2, , c1, x, d1, c2 , X = y and B = d 2, , a3, , b3, , c3, , z, , d3, , Unique solution of equation AX = B is given by X = A–1B, where |A| ≠ 0., , 20/04/2018

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DETERMINANTS, , 69, , (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, , 4.2, , (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., , Solved Examples, , Short Answer (S.A.), Example 1 If, , 2x 5 6 5, =, , then find x., 8 x 8 3, , Solution We have, , 2x 5 6 5, =, . This gives, 8 x 8 3, , ⇒ x2 = 9 ⇒, , 2x2 – 40 = 18 – 40, , 1 x, Example 2 If ∆ = 1 y, , 1 z, , x2, , 1, y , ∆1 = yz, x, z2, , 1 1, zx xy , then prove that ∆ + ∆1 = 0., y z, , 2, , 1, , 1, , 1, , Solution We have ∆1 = yz, x, , zx, , xy, , y, , z, , x = ± 3., , Interchanging rows and columns, we get, , 1 yz, , x, , ∆1 = 1 zx, 1 xy, , y, z, , x, =, , 1, y, xyz, z, , xyz, , x2, , xyz, , y2, , xyz, , z2, , 20/04/2018

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DETERMINANTS, , 0, = ( x − p) −1, 0, , 71, , p + x 2q, x, , q, , q, , x, , Applying R1 → R1 + R2, , Expanding along C1, we have, ∆ = ( x − p) ( px + x 2 − 2q 2 ) = ( x − p) ( x 2 + px − 2q 2 ), , 0, Example 5 If ∆ = a − b, , a−c, , b−a c−a, c − b , then show that ∆ is equal to zero., 0, , 0, b−c, , 0, Solution Interchanging rows and columns, we get ∆ = b − a, , c−a, , a −b a−c, 0, , b−c, , c −b, , 0, , Taking ‘–1’ common from R1, R2 and R3, we get, , b−a c −a, , 0, , ∆ = (–1) a − b, 0, a −c b−c, , c −b = – ∆, 0, , ⇒, , or, , 3, , 2∆ =0, , ∆ =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 2 3, Example 7 If x = – 4 is a root of ∆ = 1 x 1 = 0, then find the other two roots., 3 2 x, , 20/04/2018

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72, , MATHEMATICS, , 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, , 1 1 1, ∆ = ( x + 4) 1 x 1, 3 2, , 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,, , ∆ = (x + 4) [(x – 1) (x – 3) – 0]. Thus, ∆ = 0 implies, x = – 4, 1, 3, Example 8 In a triangle ABC, if, 1, , 1, , 1, , 1 + sin A, , 1 + sin B, , 2, , 2, , 1 + sin C = 0 ,, , sinA +sin A sinB+sin B sinC+sin 2 C, then prove that ∆ABC is an isoceles triangle., 1, Solution Let ∆ =, , 1, , 1, , 1 + sin A, , 1 + sin B, , 2, , 2, , 1 + sin C, , sinA +sin A sinB+sin B sinC+sin 2C, , 20/04/2018

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74, , MATHEMATICS, , or, , sinθ = 0 or (2sinθ – 1) = 0 or (2sinθ + 3) = 0, , or, , sinθ = 0 or sinθ =, , 1, (Why ?)., 2, , Objective Type Questions, Choose the correct answer from the given four options in each of the Example 10 and 11., Ax, , Example 10 Let ∆ = By, Cz, , x2 1, y, , 2, , z2, , A, 1 and ∆1 = x, zy, 1, , B, y, zx, , C, z , then, xy, , (A), , ∆1 = – ∆, , (B), , ∆ ≠ ∆1, , (C), , ∆ – ∆1 = 0, , (D), , None of these, , A, , B, , C, , A, , x, , yz, , Solution (C) is the correct answer since ∆1 = x, zy, , y, , z, , =B, , y, , zx, , zx, , xy, , C, , z, , xy, , =, , Ax, , x2, , xyz, , 1, By, xyz, Cz, , y2, , xyz, , z, , 2, , xyz, , Ax, , x2 1, , xyz, By, = xyz, Cz, , y2 1, , cos x, Example 11 If x, y ∈ R, then the determinant ∆ =, , z2 1, , − sin x, , =∆, , 1, , sin x, cos x, 1 lies, cos( x + y ) − sin( x + y ) 0, , in the interval, (A) − 2, 2, , (B) [–1, 1], , (C) − 2,1, , (D) −1, − 2,, , Solution The correct choice is A. Indeed applying R3→ R3 – cosyR1 + sinyR2, we get, , 20/04/2018

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DETERMINANTS, , cos x − sin x, ∆ = sin x, 0, , cos x, 0, , 75, , 1, 1, ., sin y − cos y, , Expanding along R3, we have, , ∆ = (siny – cosy) (cos2x + sin2x), = (siny – cosy) =, , 1, 1, sin y −, cos y, 2, 2, , 2, , π, π, 2 cos sin y − sin cos y, 4, 4, , =, Hence – 2, , ≤∆≤, , =, , π, 2 sin (y – 4 ), , 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, sin 2 A cotA 1, ∆ = sin 2 B, 2, , sin C, , cotB 1 = ................, cotC 1, , Solution Answer is 0. Apply R2 → R2 – R1, R3 → R3 – R1., 23 + 3, Example 13 The determinant ∆ = 15 + 46, 3 + 115, Solution Answer is 0.Taking, C1 → C3 –, Example 14, , 5, , 5, , 5, , 10 is equal to ..............., , 15, , 5, , 5 common from C 2 and C 3 and applying, , 3 C2, we get the desired result., The value of the determinant, , 20/04/2018

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76, , MATHEMATICS, , sin 2 23°, , sin 2 67°, , cos180°, , ∆ = − sin 67° − sin 23° cos2 180° = .........., 2, , 2, , cos180°, , sin 2 23°, , sin 2 67°, , Solution ∆ = 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, , cos x, sin x, , sin y, cos y, , is independent of x only., Solution True. Apply R1 → R1 + sinyR2 + cosy R3, and expand, Example 16, , The value of, 1, n, , C1, , n, , C2, , 1, n+ 2, n+ 2, , 1, n+4, , C1, C2, , n+4, , C1, , is 8., , C2, , Solution True, , x 5 2, Example 17, , If A = 2 y 3 , xyz = 80, 3x + 2y + 10z = 20, then, 1 1 z, , 81, , 0, , 0, , A adj. A = 0, 0, , 81, , 0, , 0, , 81, , ., , Solution : False., , 20/04/2018

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DETERMINANTS, , 1, 2, 0 1 3, 1, –1, Example 18 If A = 1 2 x , A = −, 2, 2 3 1, 1, 2, , −4, 3, y, , 77, , 5, 2, 3, −, 2, 1, 2, , 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:, , a+x, , x 2 − x + 1 x −1, 1., , x +1, , xy 2, , 0, 2, , 3., , 0, , x y, 2, , x z, , zy, , x+4, 5., , 2., , x +1, , xz 2, yz, , 2, , x, , 2, , 4., , 0, , a+ y, z, y, a+z, , 3x, , −x+ y −x+ z, , x−y, , 3y, , z−y, , x−z, , y−z, , 3z, , 2a, , 2a, , b−c−a, 2b, 2c, c−a−b, , 2b, 2c, , 6., , z, , x, x, , a−b−c, , x, , x+ 4, x, x, x+4, , x, x, , y, , Using the proprties of determinants in Exercises 7 to 9, prove that:, , 7., , y2z2, , yz, , y+ z, , 2 2, , zx, , z+x = 0, , xy, , x+ y, , z x, 2, , x y, , 2, , 8., , y+z, , z, , y, , z, , z+x, , x, , y, , x, , x+ y, , = 4 xyz, , 20/04/2018

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78, , MATHEMATICS, , a 2 + 2a 2a + 1 1, 9., , 2a + 1, , a+2, , 3, , 3, , 1 = ( a − 1)3, 1, , 1, , cos C, , cos B, , 10., , 1, cos A = 0, If A + B + C = 0, then prove that cosC, cos B cos A, 1, , 11., , If the co-ordinates of the vertices of an equilateral triangle with sides of length, , x1, ‘a’ are (x1, y1), (x2, y2), (x3, y3), then x2, x3, , 12., , 2, , y1 1, 3a 4, y2 1 =, 4 ., y3 1, ,  1 1 sin 3θ , Find the value of θ satisfying  −4 3 cos 2θ  = 0 ., , , −2 ,  7 −7, , 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 a 1 , a 2 , a 3 , ..., a r are in G.P., then prove that the determinant, ar +1, ar + 7, , ar + 5, ar +11, , ar + 9, ar +15, , ar +11, , ar +17, , ar + 21, , is independent of r., , 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 ∆ABC is an isosceles triangle if the determinant, , 20/04/2018

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DETERMINANTS, , 79, , , , 1, 1, 1, , , ∆ =  1 + cos A, 1 + cos B, 1 + cos C  = 0 .,  cos 2 A + cos A cos 2 B + cos B cos 2 C + cos C , , , , 0 1 1, 17., , –1, Find A if A = 1 0 1 and show that A =, 1 1 0, –1, , A 2 − 3I, ., 2, , Long Answer (L.A.), , 18., , 1 2 0, If A =  −2 −1 −2  , find A–1., , ,  0 −1 1 , , Using, A –1 ,, solve, the, system, x – 2y = 10 , 2x – y – z = 8 , –2y + z = 7., 19., , linear, , Using matrix method, solve the system, 3x + 2y – 2z = 3, x + 2y + 3z = 6, 2x – y + z = 2 ., , 2, 20., , of, , 2, , −4, , of, , equations, equations, , 1 −1 0, , Given A = −4 2 −4 , B = 2, 2 −1 5, 0, , 3, 1, , 4 , find BA and use this to solve the, 2, , system of equations y + 2z = 7, x – y = 3, 2x + 3y + 4z = 17., , 21., , 22., , a b c, If a + b + c ≠ 0 and b c a = 0 , then prove that a = b = c., c a b, bc − a 2, , ca − b 2, , ab − c 2, , 2, Prove that ca − b, , ab − c 2, , bc − a 2 is divisible by a + b + c and find the, , ab − c 2, , bc − a 2, , ca − b 2, , quotient., , 20/04/2018

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80, , 23., , MATHEMATICS, , xa, , yb, , zc, , a b, , If x + y + z = 0, prove that yc, zb, , za, , xb = xyz c a b, , xc, , ya, , b, , c, , 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 x 5 6 −2, =, , then value of x is, 8 x, 7 3, , (A), , 3, , (B), , ±3, , (C), , ±6, , (D), , 6, , a−b b+c, 25., , 26., , The value of determinant, , b−a c+a b, c−a a+b c, , (A), , a3 + b3 + c3, , (B), , 3 bc, , (C), , a3 + b3 + c3 – 3abc, , (D), , none of these, , 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, , b 2 − ab b − c, , 27., , a, , The determinant ab − a, , 2, , bc − ac, , bc − ac, , a − b b 2 − ab equals, c − a ab − a 2, , (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, , 20/04/2018

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DETERMINANTS, , sin x, 28., , 81, , cos x cos x, , The number of distinct real roots of cos x sin x cos x = 0 in the interval, cos x cos x sin x, , π, π, − ≤ x ≤ is, 4, 4, , 29., , (A), , 0, , (B), , 2, , (C), , 1, , (D), , 3, , If A, B and C are angles of a triangle, then the determinant, , −1, , cos C, , cosC, , −1, , cos B, , cos A is equal to, cos B cos A, −1, (A), , 0, , (B), , –1, , (C), , 1, , (D), , None of these, , cos t, 30., , t, , Let f (t) = 2sin t t, sin t t, , 1, , f (t ), 2t , then lim 2 is equal to, t →0 t, t, , (A), , 0, , (B), , –1, , (C), , 2, , (D), , 3, , 1, 31., , The maximum value of ∆ =, , (A), , (C), , 1, 2, , 2, , 1, , 1, , 1, 1 + sin θ 1 is (θ is real number), 1 + cos θ, 1, 1, , (B), , 3, 2, , (D), , 2 3, 4, , 20/04/2018

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82, , MATHEMATICS, , 0, 32., , If f (x) = x + a, x+b, , x−a, , x−b, , 0, x+c, , x − c , then, 0, , (A), , f (a) = 0, , (B), , f (b) = 0, , (C), , f (0) = 0, , (D), , f (1) = 0, , 2 λ −3, 33., , 34., , 35., , If A = 0 2, 1 1, , 5, , , then A–1 exists if, , 3, , (A), , λ=2, , (B), , λ≠ 2, , (C), , λ≠–2, , (D), , None of these, , 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, , 1, , 1, , 1+ y, , 1, , 1, , If x, y, z are all different from zero and, , 1, 1 = 0 , then value of, 1+ z, , x–1 + y–1 + z–1 is, (A), , xyz, , (B), , x–1 y–1 z–1, , (C), , –x –y –z, , (D), , –1, , x, 36., , The value of the determinant x + 2 y, , x+ y, , x+ y, , x + 2y, , x, x + 2y, , x + y is, x, , (A), , 9x2 (x + y), , (B), , 9y2 (x + y), , (C), , 3y2 (x + y), , (D), , 7x2 (x + y), , 20/04/2018

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DETERMINANTS, , 1 –2, 37., , There are two values of a which makes determinant, ∆ = 2, , 5, −1 = 86, then, 2a, , a, 4, , 0, , 83, , 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 | _______ ., , 40., , If x, y, z ∈ R, then the value of determinant, , (2, (3, (4, , x, , + 2– x, , x, , + 3– x, , x, , + 4– x, , ) (2, ) (3, ) (4, 2, , x, , − 2– x, , 2, , x, , − 3– x, , 2, , x, , − 4– x, , ), ), ), , 2, , 2, , 2, , 1, 1 is, 1, , equal to _______., 2, , 41., 42., 43., 44., , 0, cos θ sin θ, 0 = _________., If cos2θ = 0, then cos θ sin θ, sin θ, 0, cos θ, If A is a matrix of order 3 × 3, then (A2)–1 = ________., If A is a matrix of order 3 × 3, then number of minors in determinant of A are, ________., The sum of the products of elements of any row with the co-factors of, corresponding elements is equal to _________., , x 3 7, 45., , If x = – 9 is a root of 2 x 2 = 0, then other two roots are __________., , 7 6 x, , 46., , 0, , xyz, , x− z, , y−x, , 0, , y−z, , z−x, , z− y, , 0, , = __________., , 20/04/2018

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84, , MATHEMATICS, , (1 + x )17, 47., , If f (x) =, , (1 + x )19, , (1 + x) 23, , (1 + x ) 23 (1 + x) 29, , (1 + x )34 = A + Bx + Cx 2 + ..., then, , (1 + x )41, , (1 + x ) 47, , (1 + x ) 43, , A = ________., State True or False for the statements of the following Exercises:, 48., , (A ), , 49., , 1 –1, (aA)–1 = a A , where a is any real number and A is a square matrix., , 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., , 53., , 3 –1, , ( ), , 3, , = A −1 , where A is a square matrix and |A| ≠ 0., , x +1, , x+2, , x+a, , x+2, , x+3, , x + b = 0 , where a, b, c are in A.P., x+c, , x+3 x+4, 54., , |adj. A| = |A|2 , where A is a square matrix of order two., , sin A cos A sin A + cos B, 55., , 56., , The determinant sin B cos A sin B+ cos B is equal to zero., sin C cos A sin C+ cos B, , x+ a, , p+u, , If the determinant y + b, z+c, , q+v, , l+ f, , m + g splits into exactly K determinants of, r +w n+h, , order 3, each element of which contains only one term, then the value of K is 8., , 20/04/2018

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DETERMINANTS, , 57., , a, , p, , Let ∆ = b, , q, r, , c, , p+ x a+ x a+ p, y = 16 , then ∆1 = q + y b + y b + q = 32 ., r + z c+ z c+r, z, , x, , 1, , 58., , 85, , 1, , 1, , 1, 1, is ., The maximum value of 1 (1+ sin θ), 2, 1, 1, 1 + cos θ, , 20/04/2018