Mathematics-4.Unit-3.02-Matrices and Determinants with Solution

DETERMINANTS SYLLABUS Determinants: Minors and Co-factors, Expansion of Determinants, Properties of Determinants, Application of Determinants, Matrices: Operations on Matrices, Transpose, Adjoint, Inverse, Solution of Linear Equations, Application of Matrices. BASIC CONCEPTS • Consider the equations a1x+b1y = 0 and a2x+b2y = 0. These give   a1b2 – a2b1 = 0 We express this eliminant as = 0. • • A determinant of order three consisting of 3 rows and 3 columns is written as and is equal to The numbers ai, bi, ci ( i =1,2,3 ) are called the elements of the determinant. The determinant obtained by deleting the ith row and jth column is called the minor of the element at the ith row and the jth column. The cofactor of this element is (-1)i+j (minor). Note that :  = = a1A1 + b1B1+c1C1 where A1, B1 and C1 are the cofactors of a1, b1 and c1 respectively. i.e., the sum of products of the elements of any row (column) of a determinant with the corresponding co-factors is equal to the value of the determinant. We can expand the determinant through any row or column. It means that we can write These results are true for determinants of any order. Example 1: Find the Co-factors of all the elements of Solution: (i) Minor of 1 Co-factor of 1 (ii) Minor of 2= Co-factor of 2 (iii) Minor of 3 Co-factor of 3 (iv) Minor of 4 = Co-factor of 4 (v) Minor of –5 = Co-factor of –5 Similarly the other Co-factors can be found. 1.1 PROPERTIES OF DETERMINANTS 1.2 PROPERTY -1 If rows be changed into columns and columns into the rows, the determinant remains unaltered. Property -2 If any two row (or columns) of a determinant are interchanged, the resulting determinant is the negative of the original determinant. Remark: If any line of a determinant D be passed over ‘m’ parallel lines, the resulting determinant D is equal to (-1)m D D =  D =  D = (-1) 2 D =D. Property -3 If two rows (or two columns) in a determinant have corresponding entries that are equal (or proportional), the value of determinant is equal to zero. Property –4 If each of the entries of one row (or columns) of a determinant is multiplied by a nonzero constant k, then the determinant gets multiplied by k. Property-5 If each entry in a row (or column) of a determinant is written as the sum of two or more terms then the determinant can be written as the sum of two or more determinants. Property-6 If to each element of a line (row or column) of a determinant be added the equimutiples of the corresponding elements of one or more parallel lines, the determinant remains unaltered = Property –7 If each entry in any row (or any column) of determinant is zero, then the value of determinant is equal to zero. Property-8 If a determinant D vanishes for x = a, then (x-a) is a factor of D, In other words, if two rows (or two columns) become identical for x = a. then (x- a) is a factor of D. For example, let D = , if we are putting a = b, we will get D = 0. i.e. a – b is a factor of D. Note: In general, if r rows (or r columns) become identical when a is substituted for x, then (x-a)r-1 is a factor of D. Property -9 If in a determinant (of order three or more) the elements in all the rows (columns) are in A.P. with same or different common difference, the value of the determinant is zero. Remarks: • It is important to know that all the properties applicable to rows are also equally applicable to columns but independently • Whenever rows are disturbed by applications of properties of determinants, at least one of the row shall remain in original shape. In other words all the rows shall not be disturbed at a time. • It is always desirable to try to bring in as many zeros as possible in any row ( or column) and then expand the determinant with respect to that row (column). Mere expansion from the outset should be avoided as far as possible. We can express a determinant as Where Ci ( i = 1,2, 3 ) are the columns and Rj ( j=1,2,3) are the rows of the determinant. Example - 2 Show that  = 0 if  = . Solution: Operating C2  C2 – ( C1 + C3) , we get  = = 0 Note: Using the A.P. property one can immediately write  = 0 directly Example - 3 Without expanding to any stage, prove that  = . Solution:  = = 1 -2 (say) 2 = = = 1 Hence  = 1 -1 = 0 Example - 4 Evaluate (where x, y, z being positive). Solution: Multiplying R1, R2, R3 by log x, log y and log z respectively   = 0 as all rows become identical. 1.3 PRODUCT OF TWO DETERMINANTS We can write Here we have multiplied rows by rows. We can also multiply rows by columns or columns by rows, or columns by columns. Note: If  = |aij| is a determinant of order n, then the value of the determinant |Aij|, where Aij is the cofactor of aij, is n1. This is known as power cofactor formula. Example - 5: Prove the following by multiplication of determinants and power co-factor formula. = Solution: (i) First determinant is equal to ( 2abc)2 (ii) Second determinant is direct multiplication of determinants in row to row. (iii) Third determinant is co-factors of the first determinant and therefore square of the first. Example- 6: Express  = as the product of two determinants and evaluate it. Solution:  = =  = 2 ( a – b) ( b – c) ( c – a)( x – y ) ( y – z)( z – x). 1.4 DIFFERENTIATION OF A DETERMINANT where a dash denotes derivative with respect to x. Example -7: Let  be a repeated root of the quadratic equation f(x) =0 and A(x), B(x), C(x) be polynomials of degree 3, 4, and 5 respectively. Then show that is divisible by f(x), where a dash denotes the derivative, with respect to x. Solution: Let g(x) = We find that g()=0, & g'()=0  g(x)=(x-)2 P(x), where P(x) is a polynomial of degree 3. Also f(x)=0 is a quadratic equation having repeated root   g(x) is divisible by f(x). Example -8: If fr (x), gr (x), hr (x) where r = 1, 2, 3 are polynomials in x such that fr(a) = gr(a) = hr(a), r = 1, 2, 3 and F (x) = , then find F (a). Solution: F(x) =  F (a) = = 0 + 0 + 0 = 0. Since gr (a) = hr (a), hence in first determinant R2 and R3 are identical; Since fr (a) = hr (a), hence in second determinant R1 and R3 are identical; Since fr (a) = gr (a), hence in third determinant R1 and R2 are identical. DETERMINANTS INVOLVING INTEGRATIONS Let (x) = where a, b, c, l, m and n are constants.  = Note: If the elements of more than one column or rows are functions of x then the integration can be done only after evaluation / expansion of the determinant. Example - 9: Let f (x) = , prove that at . Solution: Operate R1  R1 – sec x R3  f (x) = Example - 10: Prove that . Solution:  = + By applying C1  C1 – C2 and C3  C3 – C2 in both determinants , we get L.H.S. = = (a – b)( b – c) = (a – b)( b – c) { ( c – a)+ ( c – a) } = 2 (a – b) (b – c) ( c – a). Example -11: Given that a = cos  + i sin , b = cos 2 -i sin2, c = cos 3 + i sin 3 and if , show that  = 2n, n  Z. Solution:  =  a + b + c = 0 or a = b = c I f a + b + c = 0, we have cos  + cos 2 + cos 3 = 0, sin  - sin 2 + sin 3 = 0  cos 2 (2cos  + 1) = 0 and sin 2 (1 –2cos ) = 0 ....(1) which is not possible as cos 2 = 0 gives sin 2  0, cos   1/2 and cos = -1/2 gives sin 2  0, cos   1/2  Equation (1) does not hold simultaneously.  a + b + c  0  a = b = c or ei = e-2i = e3i which is satisfied only by ei = 1 i.e. cos  = 1, sin  = 0, so  = 2n, n  Z. LINEAR EQUATIONS The system of homogeneous simultaneous linear equations has a non-trivial solution (i.e. at least one of x,y,z, is different from zero ) if If   0, then the only solution of the above system of equations is x = 0, y = 0, z = 0 (called trivial solution). If  = 0, then the number of solutions of the system of equations is infinite (called non-trivial and includes trivial solution). Example -12: If the system of equations 3x + 10 y + 17z = 0 , –x + 6 y + 13 z = 0 and 20 x – 13y +  z = 0 has a non-trivial solution then find the solution. Solution: The observation for A.P. property reveals  = 46 to have  = 0 (non-trivial solution) Let z = k and by first two equations, we have 3x +10 y = - 17 k - 3x + 18y = - 39 k, therefore y = - 2k and x = k CRAMER'S RULE If  =  0, then the solution of the system of non-homogeneous simultaneous linear equations a1x + b1y + c1z = d1 a2x + b2y + c2z = d2 a3x + b3y + c3z = d3 is given by {where (d1, d2, d3)  (0, 0, 0)} If any of x, y, z  R and   0, system of equation will have unique solution and is said to be consistence independent If x= y= z = 0 and  is also zero then the system of equations will have infinitely many solutions and is said to be consistence dependent. If x, y, z is non zero and  is zero, then the system of equations will have no solution and is said to be inconsistant. Example -13: Prove that the system of the equations, 237x + 229y + 221z = 213, – 3x + 7y+17z = 27 and 7x + 19y + 31 z = 43, have infinitely many solutions. Solution: Coefficient of variable and the constant in the right hand side in each equation are in A.P. with different common difference 8, 10 , 12 respectively therefore 1 = x = y =z = 0 and hence system of equation have infinitely many solution. Example - 14: For what values of p and q, the system of equations 2x + py + 6z = 8 x + 2y + qz = 5 x + y + 3z = 4 has (i) no solution (ii) a unique solution (iii) infinitely many solutions Solution:  = = 2(6 – q) – p(3 – q) + 6(1 – 2) = 12 – 2q – 3p + pq – 6 = pq – 2q – 3p + 6 = (p –2)(q –3) 1 = = 8(6 –q) – p(15 – 49) + 6(5 – 8) = 48 – 8q – 15p + 4pq –18 = 4pq – 8q – 15p + 30 = 4q(p – 2) – 15(p –2) = (4q – 15)(p –2) 2 = = 2(15 – 4q) – 8(3 – q) + 6(4 – 5) = 0 3 = = 2(8 –5) – p(4 – 5) + 8(1 – 2) = p –2 Case –I: when q = 3, p  2,  = 0, 1  0.  given system of equations will have no solution. Case–II: When   0 , i.e. p  2, q  3,  given system of equation has unique solution Case–III: When  = 0, i.e. p = 2, or q = 3 When p = 2,  = 0, 1 = 0, 2 = 0, 3 = 0  given system of equation has infinitely many solutions. MATRICES A rectangular array of symbols (which could be real or complex numbers) along rows and columns is called a matrix. Thus a system of m  n symbols arranged in a rectangular formation along m rows and n columns and bounded by the brackets [.] is called an m by n matrix (which is written as m x n matrix). Thus, Equal Matrices: Two matrices are said to be equal if they have the same order and each element of one is equal to the corresponding element of the other. CLASSIFICATION OF MATRICES Row Matrix: A matrix having a single row is called a row matrix. e. g. [1 3 5 7] Column Matrix: A matrix having a single column is called a column matrix. e.g. . Square Matrix: An m x n matrix A is said to be a square matrix if m = n i.e. number of rows = number of columns. For example: is a square matrix of order 3  3. Note: • The diagonal from left hand side upper corner to right hand side lower side lower corner is known as leading diagonal or principal diagonal. In the above example square matrix containing the elements 1, 3, 5 is called the leading or principal diagonal. Trace of a Matrix: The sum of the elements of a square matrix A lying along the principal diagonal is called the trace of A i.e. tr(A) Thus if A = [aij]nn Then tr(A) = = a11 + a22 + ..... + ann Example -15: Find the trace of the matrix A = . Solution: tr (A) = 1 + (–1) + 4 = 4. Diagonal Matrix: A square matrix all of whose elements except those in the leading diagonal, are zero is called a diagonal matrix. For a square matrix A = [aij]nn to be a diagonal matrix, aij = 0, whenever i  j. For example: is a diagonal matrix of order 3  3. Scalar Matrix: A diagonal matrix whose all the leading diagonal elements are equal is called a scalar matrix. For a square matrix A = [aij]nn to be a scalar matrix aij = , where m  0. For example: is a scalar matrix. Unit Matrix or Identity Matrix: A diagonal matrix of order n which has unity for all its diagonal elements, is called a unit matrix of order n and is denoted by In. Thus a square matrix A = [aij]nn is a unit matrix if aij = For example: Triangular Matrix: A square matrix in which all the elements below the diagonal elements are zero is called Upper Triangular matrix and a square matrix in which all the elements above diagonal elements are zero is called Lower Triangular matrix. Given a square matrix A = [aij]nn, For upper triangular matrix, aij = 0, i > j and for lower triangular matrix, aij = 0, i < j Notes: • Diagonal matrix is both upper and lower triangular • A triangular matrix A = [aij]nn is called strictly triangular if aii = 0 for 1  i  n. For example: are respectively upper and lower triangular matrices. Null Matrix: If all the elements of a matrix (square or rectangular) are zero, it is called a null or zero matrix. For A = [aij] to be null matrix, aij = 0  i, j For example: is a zero matrix Transpose of a Matrix: The matrix obtained from any given matrix A, by interchanging rows and columns, is called the transpose of A and is denoted by A. If A = [aij]mn¬ and A = [b¬ij]nm then bij = aji,  i, j For example: If A = , then Properties of Transposes: (i). (A) = A. (ii). (A  B) = A  B, A and B being conformable matrices. (iii). (A) = A,  being scalar. (iv). (AB) = BA, A and B being conformable for multiplication. Conjugate of a Matrix: The matrix obtained from any given matrix A containing complex number as its elements, on replacing its elements by the corresponding conjugate complex numbers is called conjugate of A and is denoted by . For example: . Properties of Conjugates: (i). (ii). (iii). ,  being any number (iv). , A and B being conformable for multiplication Transpose Conjugate of a Matrix: The transpose of the conjugate of a matrix A is called transposed conjugate of A and is denoted by A. The conjugate of the transpose of A is the same as the transpose of the conjugate of A i.e. = A If A = [aij]mn, then A = [bji]nm where bji = i.e. the (j, i)th element of A = the conjugate of (i, j)th element of A. For example: If A = then A = Properties of Transpose conjugate: (i). (A) = A (ii). (A + B) = A + B (iii). (kA) = A, k being any number (iv). (AB) = BA ALGEBRA OF MATRICES Addition and Subtraction of Matrices: Any two matrices can be added if they are of the same order and the resulting matrix is of the same order. If two matrices A and B are of the same order, they are said to be conformable for addition. For example: Similarly, Notes: • Only matrices of the same order can be added or subtracted. Scalar Multiplication: The matrix obtained by multiplying every element of a matrix A by a scalar  is called the scalar multiple of A by  . For example: Thus, Properties: All the laws of ordinary algebra hold for the addition or subtraction of matrices and their multiplication by scalars. (i). If A and B be two matrices of the same order and if k be a scalar, then k(A + B) = kA + kB (ii). If k1 and k2 are two scalars and if A is a matrix, then (k1 + k2)A = k1A + k2A and k1(k2A) = k2(k1A) Multiplication of Matrices: Two matrices can be multiplied only when the number of columns in the first is equal to the number of rows in the second. Such matrices are said to be conformable for multiplication. where cij = ai1 b1j + ai2 b2j + .......+ ain bnj = Example -16: and , show that AB  BA. Solution: Thus A .B B . A. Notes: • Commutative law does not necessarily hold for matrices. • Matrix multiplication is associative. • Matrix multiplication is distributive with respect to addition. • If A is a square matrix of order n and if In is an identity matrix of order n, then AIn=InA=A. • If I be a unit matrix, then I = I2 = I3 = …… = In. SPECIAL MATRICES Symmetric and Skew Symmetric Matrices: A square matrix A = [aij] is said to be symmetric when aij = aji for all i and j, i.e. A = A. If aij = -aji for all i and j and all the leading diagonal elements are zero, then the matrix is called a skew symmetric matrix, i.e. A = – A. For example: is a symmetric matrix and is a skew-symmetric matrix. Singular and Non-singular Matrix: Any square matrix A is said to be non-singular if  0, and a square matrix A is said to be singular if = 0. Here |A| (or det(A) or simply det A) means corresponding determinant of square matrix A e.g. A = then |A| = = 10  12 = 2  A is a non-singular matrix Unitary Matrix: A square matrix is said to be unitary if A = I since   = A and  A =  A therefore if  A = I, we have | | |A| = 1. Thus the determinant of unitary matrix is of unit modulus. For a matrix to be unitary it must be non-singular. Hence A = I  A = I Hermitian and Skew-Hermitian Matrix: A square matrix A = [aij] is said to be Hermitian matrix if =  i, j i.e. A = A and a square matrix, A = [aij] is said to be a skew-Hermitian if = - ,  i, j i.e. A = A. For example: are skew-Hermitian matrices. are Hermitian matrices. Orthogonal Matrix: Any square matrix A of order n is said to be orthogonal if AA = A A = . Idempotent Matrix: A square matrix A is called idempotent provided it satisfies the relation A2 = A. For example: The matrix A = is idempotent as A2 = A.A = . = = A. Involutary Matrix: A square matrix A is said to be involutary if A2 = I. Nilpotent Matrix: A square matrix A is called a nilpotent matrix if there exists a positive integer m such that Am = O, where O is a null matrix. If m is the least positive integer such that Am = O, then m is called the index of the nilpotent matrix A. Adjoint of a Square Matrix: Let A = [aij] be a square matrix of order n and let Cij be cofactor of a¬ij in A. Then the transpose of the matrix of cofactors of elements of A is called the adjoint of A and is denoted by adj A. Thus, adjA = [Cij]T  (adj A)ij = Cji If A = , then, adjA = ; where Cij denotes the cofactor of aij in A. For Example: A = , C11 = s, C12 = r, C21 = q, C22 = p  adj A = . Theorem: Let A be square matrix of order n. Then A(adj A) = |A| In = (adj A)A Inverse of a Matrix: A nonsingular square matrix of order n is invertible if there exists a square matrix B of the same order such that AB = In = BA. In such a case, we say that the inverse of A is B and we write, A1 = B. The inverse of A is given by A1 = . adj A Properties of Inverse of a Matrix: (i). (Reversal Law) If A and B are invertible matrices of the same order, then AB is invertible and (AB)1 = B1A1. In general, if A, B, C, .... are invertible matrices then (ABC.....)1 = .....C1B1A1 (ii). The inverse of the inverse of the matrix is the original matrix itself, i.e. (A–1)–1 = A. SYSTEM OF SIMULTANEOUS LINEAR EQUATIONS Consider the following system of n linear equations in n unknowns: a11 x1 + a12 x2 + ………+ a1n xn = b1 a21 x1 + a22 x2 + ………+ a2n xn = b2 . . . . . . . . an1 x1 + an2 x2 + ………+ ann xn = bn This system of equation can be written in matrix form as or AX = B The n  n matrix A is called the coefficient matrix of the system of linear equations. Homogeneous and Non-Homogeneous System of Linear Equations: A system of equations AX = B is called a homogeneous system if B = O, where O is a null matrix. Otherwise, it is called a non-homogeneous system of equations. 1.5 1.6 SOLUTION OF A SYSTEM OF EQUATIONS: Consider the system of equation AX = B A set of values of the variables x1, x2,...,xn which simultaneously satisfy all the equations is called a solution of the system of equations. Consistent System: If the system of equations has one or more solutions, then it is said to be a consistent system of equations, otherwise it is an inconsistent system of equations. 1.7 Solution of a Non-Homogeneous System of Linear Equations: There are two methods of solving a non-homogeneous system of simultaneous linear equations. (i). Cramer’s Rule (ii). Matrix Method (i). Cramer’s Rule: is discussed in the Chapter Determinants. (ii). Matrix Method: Consider the equations … (i) If Then the system of equations given by AX = D has a unique solution given by X = A-1D. (i). If A is singular matrix, and (adjA)D = O, then the system of equations given by AX = D is consistent with infinitely many solutions. (ii). If A is singular matrix, and (adjA)D  O, then the system of equation given by AX = D is inconsistent and has no solution. Solution of Homogeneous System of Linear Equations: Let AX = O be a homogeneous system of n linear equation with n unknowns. Now if A is nonsingular then the system of equations will have a unique solution i.e. trivial solution and if A is a singular then the system of equations will have infinitely many solutions. Example -17: With the help of matrices, solve the equations; 3x + y + 2z = 3, 2x – 3y – z = –3, x + 2y + z = 4. Solution: We can write the given equations as AX = B …(1) Where, Since, = 3 (-3 + 2 ) –1 (2 + 1) + 2 (4 +3) = -3 –3 + 14 = 8 From (1), we have X = B …(2) Now, Let, adj Hence, from (2) MATRICES OF ROTATION OF AXES We know that if x and y axis are rotated through an angle  about the origin the new co-ordinates are given by x = X cos  –Y sin  y = X sin  + Y cos   , where is the matrix of rotation through an angle  1. If ,  and  are such that  +  +  = 0, then the value of is - (A) 0 (B) 1 (C)    (D) None of these Solution: Operating C2  C2 –cos  C1, C3  C3 –cos  C1,  =  = = Hence (A) is the correct answer. 2. If a  p, b  q, c  r and = 0 then the value of is : (A) 0 (B) 1 (C) 2 (D) None of these Solution: By operating R1 R1 – R2 and R2 R2 – R3 0 = = ( p – a){r ( q – b) –b ( c – r)} + a ( b – q ) (c – r) = (p – a) ( rq – rb) + a( b – q) (c – r) + b(pa) (rc) dividing throughout by (p – a)(q – b)(r – c), =  Therefore . Hence (C) is the correct answer. 3. The value of is : (A) 0 (B) 1 (C) 2 (D) can't find Solution: Operating R1 ® R1 + R2 + R3 and using trigonometric identities, the given determinant = = = 0 . Hence (A) is the correct answer. 4. Let a1, a2 and b1, b2 be the roots of ax2+bx+c=0 and px2+qx+r=0 respectively. If the system of equations a1y+a2z=0, b1y+b2z=0 has a non-trivial solution, then . (A) (B) (C) (D) None of these Solution: Since the system of equations possess non-trivial solution, = 0 Þ a1b2 – a2b1 = 0 Þ ……(1) Also a1 + a2 = – , a1a2 = and b1 + b2 = – , b1b2 = \ equation (1) Þ Þ . Hence (A) is the correct answer. 5. if = 0, then a, b, c are in (A) A.P (B) G.P (C) H.P (D) none of these Solution: Apply R1  R1 + R3 – 2R2 = 0  a + c –2b = 0  a, b, c are in A.P Hence (A) is the correct answer. 6. If = 0,then x is equal to (A) 0, –( +  + ) (B) 0, ( +  + ) (C) 1, ( +  + ) (D) 0, (2 + 2 + 2) Solution: applying C1  C1 + C2 + C3  (x +  +  + ) = 0 applying R2  R2 – R1, R3  R3 – R1  (x +  +  + ) = 0  (x +  +  + ).x2 = 0  x = 0, –( +  + ) Hence (A) is the correct answer. 7. Given that = (a –b) (b –c) (c –a) (a + b + c), the value of is (A) –9 (B) 42 (C) 18 (D) 24 Solution: In the problem a = 2, b = 3, c = 4 Hence (C) is the correct answer. 8. If = 0 then x is equal to (A) 9 (B) -9 (C) 0 (D) None of these Solution: By circulant determinant property a + b + c = 0  x + 3 + 6 = x + 2+ 7 = x + 4 + 5 = 0  x = -9. Hence (B) is the correct answer. 9. Let ax7 + bx6 + cx5 + dx4 + ex3 + fx2 + gx + h = (A) g = 3 and h = – 5 (B) g = –3 and h = – 5 (C) g = –3 and h = –9 (D) None of these Solution: By putting x = 0 an both sides of the equation we have h = = 9 Differentiating both sides and then putting x = 0, we get g = –3 Hence (D) is the correct answer. 10. If = k (xyz), then k is equal to (A) 4 (B) -4 (C) zero (D) None of these Solution: Putting x = 1 y = 1, and z = 1 on both sides, we get k = 4 Hence (A) is the correct answer 11: For what value of x, the matrix is singular. (A) x = 1, 2 (B) x = 0, 2 (C) x = 0, 1 (D) x = 0, 3. Solution: Since, the given matrix is singular R2 +  x {(3 – x) (1 + x – 4) – 0 + 2 (2 – 2)} = 0  x (3 – x) (x – 3) = 0  x = 0, 3 Hence (D) is the correct answer. 12: The adjoint of the matrix A = is (A) (B) (C) (D) Solution: The co-factors are given by, A11 = –5; A12 = –3; A21 = –3; A22 = 1 Let , Hence (B) is the correct answer. 13: If A = , then A1 is equal to (A) (B) (C) (D) none of these Solution: Since, =2(1 – 4) –5 (3 –2 )+( 6 –1 ) = –6 –5 + 15 =15 – 11 = 4  A is a non – singular matrix. Now, Let, Hence, Hence (A) is the correct answer. 14: The matrix X in the equation AX = B, such that A = and B = is given by (A) (B) (C) (D) Solution: Given A =  |A| = 1  0  A–1 exists A–1 = = AX = B  X = A–1B = = Hence (B) is the correct answer. 15: If the trace of the matrix: is 0 then x is equal to (A) {–2, 3} (B) (–2, 3) (C) {–3, 2} (D) (–3, 2) Solution: Trace of matrix is defined as = 2x2 + 2x –12 = 0  x = –3, 2 Hence (C) is the correct answer. 16: If A and B are square matrices of order 3, then (A) adj(AB) = adjA + adjB (B) (A + B)–1 = A–1 + B–1 (C) AB = O  |A| = 0 or |B| = 0 (D) AB = O  |A| = 0 and |B| = 0 Solution: If AB = O then either of A and B are necessarily singular Hence (C) is the correct answer. 17: If A and B are any two square matrices of the same order, then (A) (AB) = AB (B) adj(AB) = adj(A) adj(B) (C) (AB) = BA (D) AB = O  A = O or B = O Solution: Multiplication of square matrices is a square matrix of same order and by property of transpose. Hence (C) is the correct answer. 18: The system AX = B of n equations in n unknowns has infinitely many solutions if (A) det A  0 (B) det A = 0, (adjA)B  O (C) det A = 0, (adjA)B = O (D) det A  0, (adjA)B = O Solution: X = A–1B X = Clearly if the system has infinite solutions |A| = 0 and (adj A)B = O Hence (C) is the correct answer. 19: If and , then (A) A B = B A (B) AB  BA (C) A B = 1/2 B A (D) None of these Solution: AB = = = BA = = =  AB = BA Hence (A) is the correct answer. 20: If A = , then the value of |ATA1| is (A) cos4x (B) sec2x (C)  cos4x (D) 1 Solution: AT = A1 = ATA1 = | ATA1| = 1 Hence (D) is the correct answer. 21.  = is always (A) real (B) imaginary (C) zero (D) none of these Solution: Since  =   is real only Hence (A) is the correct answer 22. It a, b and c are the pth, qth and rth terms of an HP. then = (A) A term containing a, b, c, p, q, r (B) a constant (C) zero (D) none of these Solution: If A is the first term and D is the common difference of the corresponding A.P. then = A + (p – 1) D = A + (q – 1) D = A + (r –1)D Now  = abc = abc Operating R1  R1 – D(R2) – (A – D) R3  = abc = 0 Hence (C) is the correct answer . 23. If , then P is given by (A) xn (B) (n+ 1) (C) either A or B (D) both A and B Solution: C1 and C3 become equal for p = xn and R1 and R3 become equal for p = n + 1. Hence (D) is the correct answer. 24. The system of equations x + 2y + 3z = 4, 2x + 3y + 4z = 5, 3x + 4y + 5z = 6 has (A) Infinitely many solution (B) No solution (C) Unique solution (D) None of these Solution: Since , (row are in AP with common difference 1) Similarly x = y = z = 0. Hence (A) is the correct answer. 25. If the equations x = ay + z, y = az + x and z = ax + y are consistent having non-trivial solution, then (A) a3 = 1 (B) a3 + 1= 0 (C) a + 1 = 0 (D) None of these Solution:  =  a3 –3a = 0 Hence (D) is the correct answer. 26. The determinant  = is (A) 0 (B) independent of  (C) independent of  (D) independent of both  and  Solution: Apply R1  R1 + R2 sin -R3 cos Then taking 2cos common from R1 & then apply R1  R1 + R3 Then  = 2 cos Hence (B) is the correct answer. 27. If x, y, z are three integers in A.P, lying between 1 and 9 and x51, y41 and z31 are three digits numbers, then the value of is (A) x + y + z (B) x –y + z (C) 0 (D) None of these Solution:  = R2  R2 –100R3 –10R1 = x –2y + z = 0 ( x, y, z in A.P) Hence (C) is the correct answer. 28. Let m be a positive integer and . Then the value of is given by (A) 0 (B) m2 –1 (C) 2m (D) 2m sin2 (2m) Solution: Using concept of summation of determinant. We get R1, R2 are identical so is zero. Hence (A) is the correct answer. 29. (A) a2 + b2 + c2 –3abc (B) 3ab (C) 3a + 5b (D) 0 Solution: by applying R2  R2 – R1 R3  R3 – R1 Hence (D) is the correct answer. 30. Given x = cy + bz; y = az + cx; z = bx + ay where x, y, z are not all zero. The value of a2 +b2 +c2 +2abc is : (A) 1 (B) 0 (C) 2 (D) 3 Solution:  x + c y + b z = 0 c x  y + a z = 0 b x + a y  z = 0 since x, y, z are not all zero, using the condition for concurrency = 0   1 + 2 a b c + a2 + b2 + c2 = 0  a2 + b2 + c2 + 2 a b c = 1 Hence (A) is the correct answer.