Solved Examples. JEE Main/Boards. Similarly, Example 1: If x y 2x 3x + y 3z + 4w = 5 25, find x, y, z, w.

6.6 Matrices Solved Examples JEE Main/Boards Example : If x y x + z x + y z + 4w 5 5 5, find x, y, z, w. Sol. We know that in equal matrices the corresponding elements are equal. Therefore, by equating the elements of these two matrices which have the same number of rows and columns, we get the value of x, y, z and w. Given x y x + z x + y z + 4w 5 5 5 x y, x + z 5; x + y 5, z + 4w 5 By solving these equations, we get x, y, z, w 4 Example : Show that the matrix 5 6 is a nilpotent matrix of index Sol: Value of the index at which all elements of the matrix become 0, i.e. null matrix, is called the nilpotent matrix of that index. Here we calculate the n th power of the matrix, where n,,,. The value of n at which the matrix becomes null matrix is the index value. Given A 5 6 A A A 5 6 5 6 0 0 0 9 Similarly, 0 0 0 A A. A 9 5 6 0 0 0 0 0 0 0 0 0 A 0. i.e. A k 0 Here k Hence, A is nilpotent matrix of index Example : Solve the following system of homogeneous equations: x + y z 0, x y z 0 and x + y + z 0 Sol: In this problem we can write the given homogeneous equations in a matrix form, i.e. [A][X] [O] and then by calculating the determinant of matrix A we can find if that given system has a trivial solution or not. The given system can be written as x 0 y 0 or AX O z 0 Where, A Now, A ( + ) ( + 6) ( + ) 7 4 0 Thus A 0. x 0 X y and O 0 z 0 So the given system has only the trivial solution given by x y z 0

Mathematics 6.7 Example 4: Find x, y, z and a for which x + y + x z 4a 6 0 7 a Sol: We know that, in equal matrices the corresponding elements are equal. Therefore by equating the elements of these two matrices which have the same number of rows and columns we get the values of x, y, z and w. Given, x + y + x z 4a 6 0 7 a We know, that for equal matrices the corresponding elements are equal, therefore x + 0; y + x 7; z ; 4a 6 a; By solving these equations, we get x, z 4, y, a. Example 5: Compute the adjoint of the matrix 4 5 A 6 0 0 Sol: For this problem, we use the formula to get the cofactors of all the elements of matrix A. Then by taking the transpose of the co-factor matrix we can get the adjoint of matrix A. Consider C ij be a co-factor of a ij in matrix A. Then the co-factors of the elements of A are given by C 6 0 6 6, 0 C 6 0 0 0, C 0 0, C 4 5 0 5 5 0 ( ) C 5 0 0 0, C 4 0 ( 0), C 4 5 C 5 C 4 4 0 4, 6 ( ) 6 5 9, 6 ( ) 0, ( ) 6 0 6 5 4 adj A 5 0 0 0 9 4 9 0 0 Example 6: If A cos θ sinθ T sin θ cosθ, Then find lim n n An. Sol: For this problem, we first have to calculate the n th power of matrix A, i.e. A n, and multiply the matrix A n by n. Then, by with the given limit we can find the solution of this problem. Given A cos θ sinθ sin θ cosθ n A n cos θ sin θ cosn θ sinn θ sinθ cosθ sinnθ cosnθ cosnθ sinnθ n An n n sinnθ cosnθ n n But cos nθ, sin nθ ; cos nθ lim 0, n n lim n sinnθ n 0, lim n n An 0 0 0 0 Example 7: A trust fund has Rs. 50,000 that is to be invested into two types of bonds. The first bond pays 5% interest per year and the second bond pays 6% interest per year. Using matrix multiplication determine how to divide by Rs, 50,000 among the two types of bonds so as to obtain an annual total interest of Rs.,780. Sol: In this problem, investment amounts can be written in the form of a row matrix and interest amounts can

6.8 Matrices be written in the form of column matrix. By multiplying these two matrix we will get the equation for annual interest rates. By equating this to the given annual interest value we will get the required answer. Consider investment of first type of bond Rs. x And second type of bond Rs. 50,000 x These amounts can be written in the form of a row matrix A which is given by A x 50000 x The interest amounts per rupee, per year from the two 5 bonds are Rs. 00 and 6 which can be written in the 00 form of a column matrix B which is given by B 5 00 6 00 The total interest per year is given by 5 A.B [x 50,000 x] 00 6 00 [x. 5/00 + (50,000 x). 6/00] [000 x/00] Since the required total annual interest is Rs.,780. [000 x/00] [780] 000 x/00 780 x 00(000 780),000 Hence the required amounts to be invested in the two bonds are Rs.,000 and Rs. (50,000,000), i.e. Rs.,000 and Rs. 8,000 respectively. Example 8: If f(α) cos α sin α and if α, β, γ are sinα cosα the angles of a triangle, then prove that f(α). f(β), f(γ) I Sol: In this problem, by the methods of substitution and multiplication of matrices we can easily prove the given equation. Given that f(α) cos α sin α sinα cosα f(β) cos β sin β and sinβ cosβ cos γ sin γ f ( γ ) sin γ cos γ f(α)f(β) cos α sin α cosβ sinβ sinα cosα sinβ cosβ cos α cos β sin α sin β cos α sin β+ sin α cos β sinαcosβ cosαsinβ sinαsinβ+ cosαcosβ cos( α+β ) sin( α+β ) sin( α+β) cos( α+β) Similarly f(α) f(β) f(γ) cos( α+β+γ ) sin( α+β+γ ) sin( α+β+γ) cos( α+β+γ) cos π sin π and as α + β + γ p sinπ cosπ 0 0 0 0 I Example 9: If M(α) M(β) cosβ 0 sinβ 0 0 β β sin 0 cos cosα sinα 0 sinα cosα 0 0 0 then prove that [M(α) M (β)] M( β) M( α) Sol: In this problem, by finding the inverse of the matrix we can easily get the required answer. [M(α) M (β)] M(β) M(α) Given M(α) cosα sinα 0 sinα cosα 0 0 0 cosα sinα 0 M(α) sinα cosα 0 0 0 cosα sinα 0 sinα cosα 0 0 0 We can also write this in the form cos( α) sin( α) 0 sin( α) cos( α) 0 M( α) 0 0 T

Mathematics 6.9 Similarly, cosβ 0 sinβ M(β) 0 0 β β sin 0 cos cos( β) 0 sin( β) 0 0 M( β) β β sin( ) 0 cos( ) [M(α) M (β)] M( β) M( α) Example 0: Show that the homogeneous system of equations x y + z 0, x + y z 0, x + 6y 5z 0 has a non-trivial solution, Also, find the solution. Sol: In this problem we can write the given homogeneous equations in a matrix form, i.e. [A][X] [O] and then by calculating the determinant of matrix A we can find if that given system has a non- trivial solution or not. The given equations are x y + z 0, x + y z 0, x + 6y 5z 0, We can write these equations in the form of matrices as shown below 6 5 x 0 y 0 or AX O, where z 0 x 0 A, X y and O 0 6 5 z 0 Now, A 6 5 ( 5 + 6) + ( 5 + ) + (6 ) 0 Thus, A 0 Hence, the given system of equations has a non-trivial solution. To find the solution, we take z k in the first two equations and write them as follows: x y k and x + y k or x y k or AX B, k where A, X x y and B k Now, A k 0. So A exists; We have, adj A So, A A adj A A Now X A B x y x k/, y k/ k k/ k k / These values of x, y and z also satisfy the third equation. Hence x k/, y k/ and z k, where k is any real number and which satisfy the given system of equations. JEE Advanced/Boards Example : Let A and B be symmetric matrices of the same order. Then show that (i) A + B is symmetric (ii) AB BA is skew-symmetric (iii) AB + BA is symmetric Sol: In this problem, by using the conditions for symmetric and skew-symmetric matrices we can get the required result. As given, A and B are symmetric. A A and B B (i) (A + B) A + B A + B A + B is symmetric (ii) (AB BA) (A B) (BA) B A A B [by reversal law] BA AB [A A, B B] AB BA is skew-symmetric (iii) (AB + BA) (AB) + (BA) B A + A B BA + AB AB + BA AB + BA is symmetric.

6.0 Matrices Example : Solve the following equations: x y + z 9 x + y + z 6 x y + z Sol: In this problem, we can write the given homogeneous equations in a matrix form, i.e. [A][X] [O]. Then, by calculating the determinant of matrix A and adjoint of A, we get an inverse of matrix A, i.e. A. By multiplying this into [A][X] [O] we get the required values of x, y, and z. We can also find if that given system has a trivial solution or not. As given, x y + z 9 x + y + z 6 x y + z This system can be written as AX B, x Where, A X y z and B 9 6 A ()+(0)+(-) Adj A 4 0 5 A Now, Adj A A X A B 4 9 0 6 5 4 4 x, y, z-7 is the solution. T Example : Let A, prove that A 4A 5I 0, hence obtain A : Sol: In this problem, by using a simple multiplication method we can get the matrix A, then by substituting these in the given equation we will easily obtain the required result. A A. A + 4+ 4 + + 4 + 4+ 9 8 8 + + 4 4+ + 4 4+ + 8 9 8 + + + + + + 4 4 4 4 8 8 9 Now A 4A 5I 9 8 8 4 8 8 5 0 0 8 9 8 8 4 8 0 5 0 8 8 9 8 8 4 0 0 5 9 4 5 8 8 0 8 8 0 0 0 0 8 8 0 9 4 5 8 8 0 0 0 0 8 8 0 8 8 0 9 4 5 0 0 0 0 [Here 0 is the zero matrix] Thus A 4A 5I O A A 4A A 5A I A O O or (A A)A 4(A A) 5A I O; or IA 4I 5A O; 5A A 4I 4 0 0 0 4 0 0 0 4 A /5 /5 /5 5 /5 /5 /5 /5 /5 /5 Example 4: Find the product of two matrices 5 A and B where A 7 5 B and use it for solving the equations x + y + z, x + y + z 7 and x + y + z

Mathematics 6. Sol: As the given system of equations is in the form BX C, multiplying it by B, which is obtained by the multiplication of AB, we can get the required result. AB 5 7 5 5+ + 6 5+ + 0+ + 9 4 0 0 7 + 0 7 + 5 4 + 5 0 4 0 + + + 0 0 4 Also the given system of equations in matrix form is BX C (ii) x Where X y and C 7 z From (ii), X B C [Multiplying both sides of (ii) by B B B I From (), AB 4I A 4. B I B A 5/4 /4 /4 4 7/4 /4 5/4 /4 /4 /4 x y X B C z 5/4 /4 /4 7/4 /4 5/4 7 /4 /4 /4 5 7 6 + + 4 4 4 7 7 0 + 4 4 4 x, y, z 7 + 4 4 4 Example 5: Given A Find P such that BPA 0 0 0 4, B 4 Sol: Pre-multiplying both sides by B and Postmultiplying both sides by A in BPA 0 we can find P. 0 0 Given BPA 0 0 0 B BPA A B 0 0 0 A IPI B 0 0 0 A P B 0 0 0 A (i) To find B, B 4 B 4 8 9 0 Let C be the matrix of co-factors of elements in B ; C C C C C C 4 C C C C 4 B adjb B C' C 4 4 To Find A, Since A 4 A (4 ) ( ) + (6 8) 0 0 Let C be the matrix of co-factors of elements in A ; C C C C C C C C C C (ii)

6. Matrices 4 4 4 4 0 ; C 0 Adj A A AdjA A 0 0 Adj, A Substituting eq. (ii) and (iii) in eq. (i), we get P 4 0 0 0 0 P 4 4 0 P 4 + 0 8 8 + 4 + 8 0+ 6 6 + 9+ 6 P 4 7 7 5 5 cosα sinα 0 Example 6: If F(α) sinα cosα 0 then 0 0 show that F(x). F(y) F( x+ y). Hence, prove that [F(x)] F( x). (iii) Sol: By substituting x and y in place of α in given matrices we will get F(x) and F(y) respectively and then by multiplying them we will get the required result. F(x) F(y) cosx sinx 0 sinx cosx 0 0 0 cos y siny 0 siny cos y 0 0 0 cosx cos y sinx siny cosx siny sinx cos y 0 sinx cos y + cosx siny sinx siny + cosx cos y 0 0 0 cos(x + y) sin(x + y) 0 sin(x + y) cos(x + y) 0 F(x +y) 0 0 i.e. F(x).F(y) F(x +y) nd part. As we know that F(x) [F(x)] I Replacing y by x in (i), we get F(x). F( x) F(x x) F(0) cos0 sin0 0 0 0 sin0 cos0 0 0 0 0 0 0 0 i.e. F(x) F( x) I therefore from (ii) and (iii) [F(x)] F( x). (i) (ii) (iii) Example 7: Show that every square matrix A can be uniquely expressed as P + iq where P and Q are Hermitian matrices. Sol: By considering P (A + Aθ ) And Q i (A Aθ ) we get A P + iq Then, using the property of a Hermitian matrix we can prove the above problem. θ Now P θ θ (A + A ) (A + Aθ ) q {Aθ + (A θ ) q } (Aθ + A) (A + Aθ ) P P P θ, hence P is a Hermitian matrix. Similarly θ Q θ θ (A A ) (A A θ ) q i i i {Aθ (A θ ) q } i (Aθ A) i (A Aθ ) Q Q is also Hermitian matrix,

Mathematics 6. Therefore A can be expressed as P + iq,where P and Q are Hermitian matrices. Let A R + is where R and S are both Hermitian matrices We have A θ (R + is) θ R θ + (is) q R θ + is θ R θ is θ R is (since R and S are both Hermitian) A + A θ (R + is) + (R is) R R (A + Aθ ) P Also A A θ (R + is) (R is) is S (A A θ ) Q i Hence expression () for A is unique Example 8: If A is Hermitian such that A 0, show that A 0, Sol: As A is a Hermitian matrix therefore A θ A. By considering A [a ij ] n n to be a Hermitian matrix of order n and as given A 0, we can solve given problem as follows. A a a... an a a... an and............ a n a n... ann a a... an A θ a a... an............ a n a n... ann Since A 0; Let AA θ [b ij ] n n AA θ 0 Then each element of AA θ is zero and so all the principal diagonal elements of AA θ are zero b ii 0 for all i,,, n Now, b ii a i a i + a i a i + + a in a in a i + a i + a in b ii 0 a i + a i + + a in 0 a i a i a in 0 a i a i a in 0 each element of the i th row of A is zero, but b ii 0 i,. n Each element of each row of A is zero. Hence, A O Example 9: If the non-singular matrix A is symmetric, then prove that A is also symmetric. Sol: By using the conditions of non-singular and symmetric matrix we can easily find the required result. As given matrix A is a non-singular symmetric matrix. A 0 and A T A, So, A exists Now, AA I A A (AA ) T (I) T (A A) T (A ) T A T I A T (A ) T (A ) T A I A(A ) T [ A T A] A (A ) T A is symmetric. Example 0: Find the rank of the matrix A 4 6 0 7 Sol: given A 4 6 0 7 R R R,R R R and R 4 R 4 6R A 4 0 5 7 0 4 9 0 0 9 7 [Applying R 4 R 4 - R - R ] 4 0 5 7 0 4 9 0 0 0 0 0 [Applying R R R ] 4 0 6 0 4 9 0 0 0 0 0

6.4 Matrices [Applying R R 4R ] [Applying R / R ] 4 0 6 0 0 0 0 0 0 ~ 4 0 6 0 0 0 0 0 0 Since the equivalent matrix is in echelon form having three non-zero rows. Hence, r(a) JEE Main/Boards Exercise Q. Find x and y, if x 5 x+ y Q. A matrix has rows and columns. How many elements a matrix has? Find the number of elements of a matrix if it has rows and columns. Q. Order of matrix A is and order of matrix B is. Find the order of AB and BA, if defined. Q.4 Given a matrix A [a ij ], i and j, where a ij i + j. Write the element (i) a (ii) a (iii) a (iv) a 4 Q.5 A matrix has 8 elements. Write the possible orders of matrix. Q.6 Give an example of a diagonal matrix, which is not a scalar matrix. Also give an example of a scalar matrix. Q.7 For the matrix A, show that A + A T is a symmetric matrix. Q.8 For the matrix A, Show that A A T is a skewsymmetric matrix. Q.9 The total number of elements in a matrix represents a prime number. How many possible orders a matrix can have? Q.0 Find x and y, if x y + 4 0 Q. If f(x) x 9x + 7, then for a square matrix A, write f(a). Q. If A, B and AB are symmetric matrices, then what is the relation between AB and BA? Q. If A B, [ 4], find AB. Q.4 Are the matrix Give reasons. 4 and Q.5 Given a matrix A 4. Find matrix ka, where k Q.6 Simplify: tan θ sec θ tan θ tanθ secθ + sec θ tan θ sec θ. secθ tanθ 0 4 0 equal? 0 0 0 Q.7 If X m Y P 4 Z b, for three matrices X, Y, Z, find the values of m, p and b. 0 Q.8 Is matrix A 0 symmetric or skewsymmetric? Give 0 reasons. Q.9 If R (θ) cos θ sin θ, write (i) R π, (ii) R(x+y) sinθ cosθ

Mathematics 6.5 Q.0 For a skew-symmetric matrix A [a ij ], what is the nature of elements a ij, if i j. Q. If A a 0 0 0, find A6. Q. Find x, if [x ] 0 x 0 Q. Find the sum of matrix A and its additive inverse. 4 6 Q.4 Find X, if X + 4 5 0 Q.5 Evaluate, sin θ + cos θ 0 + 0 cot θ 0 cosec θ 0 Q.6 If A and B are symmetric matrices, show that AB is symmetric. Q.7 If a matrix has 8 elements, what the possible orders it can have? What if it has 5 elements? Q.8 Evaluate the following: [a, b] c + [a b c d] d Q.9 If A a b c d 0 0 0 0, find A. Hence find A 6 0 0 Q.0 Show that the element of the main diagonal of a skew-symmetric matrix are all zeros. Q.4 If A cos α sin α, then show that sinα cosα A cos α sin α sinα cosα Q.5 If A 0 0 0 and I 0 0, prove that (ai + ba) a I + a ba. Q.6 If A 0, find the values of p and q such that (pi + qa) A. 4 5 Q.7 If A 0 6 and B 6 4 5 5 4 find A B. Q.8 Construct a matrix [a ij ], whose elements are given by a ij i j. Q.9 If x x y x + z y w, find x, y, z, w. 4 7 Q.40 Find matrices X and Y, if X + Y 5 0 9 and X Y 6 0 Q.4 If A cos θ cosθsinθ ; θ θ θ cos sin sin cos φ cosφsinφ B, then show that AB is zero cosφsinφ sin φ matrix, provided (θ φ) is an odd multiple of π/. Q. Find AB, if A 0 4 0 and B 5 7 0 0 Q. If A 4, find values of x and y such that 5 A xa + yi O where I is a unit matrix and O is a zero matrix. Q. If A 5 5 7 find B. and A B 4 5 9, Q.4 If A and B 5 5 5 compute A B. Q.4 Find the matrix X such that, 8 0 0 X + 4 0 4 0 0 0 0 4, 4 4

6.6 Matrices Exercise Single Correct Choice Type Q. If number of elements in a matrix is 60 then how many dimensions of matrix are possible (A) (B) 6 (C) 4 (D) None of these Q. Matrix A has x rows and x + 5 columns. Matrix B has y rows and y columns. Both AB and BA exist, then (A) x, y 4 (B) x 4, y (C) x, y 8 (D) x 8, y Q. If A is square invertible matrix such that A A, then det.(a I) is (A) (B) (C) (D) None of these Q.4 Number of distinct matrices that can be formed using all the 4 distinct elements is (A) 4! (B) 4(4)! (C) (4)! (D) (4)! Q.5 If A A, then (I + A) 4 is equal to (A) I + A (C) I + 5A (B) I + 4A (D) None of these Q.6 If A α β is an orthogonal matrix, where α, β α β and the roots other than the common root of the equations x px + q 0 & x + px q 0, then (A) p±, q± (B) p 0, q ± (C) p ±, q 0 (D) None of these Q.7 A is a square matrix of order n and (det A). If det (λa) 8; where λ N, then possible value of n is (A) (B) 5 (C) (D) 7 Q.8 If A and f(x) + x, then f(a)is x (A) (C) (B) (D) None of these Q.9 If A is a skew symmetric matrix such that A T A I, then A 4n (n N) is equal to (A) A T (B) I (C) I (D) A T Q.0 A and B are matrices satisfying det A det B and tr(a) tr(b), further A A + 4I 0 and B λb + µi 0, then µ is equal to (A) (B) (C) (D) 4 Q. The false statement is - (A) The adjoint of a scalar matrix is scalar matrix. (B) The adjoint of upper triangular matrix is lower triangular matrix. (C) The adjoint of upper triangular matrix is upper triangular matrix. (D) adj(adj A) A, A is a square matrix of order. Q. If the matrices A, B, (A + B) are non-singular, then [A(A + B) B] is equal to (A) A + B (B) A + B (C) (A + B) (D) None of these Q. If A is an orthogonal matrix A, then A T is equal to (A) A (B) A (C) (adj A) (D) (adj A) Q.4 If A and B are square matrices of order, then (A) adj(ab) adj A+adj B (B) (A + B) A + B (C) AB 0 A 0 or B 0 (D) AB 0 A 0 and B 0 Q.5 If A a 0 0 0 a 0, then A adj A is equal to 0 0 a (A) a 5 (B) a 7 (C) a 8 (D) None of these Q.6 If A tanx, then the value of A T A is tanx (A) cos 4x (B) sec x (C) cos 4x (D)

Mathematics 6.7 Q.7 If A 5, then 9A is equal to (A) A T (B) A (C) A (D) A Q.8 If P is a two-rowed matrix satisfying P T P, then P is (A) cos θ sin θ (B) cos θ sin θ sinθ cosθ sinθ cosθ (C) cos θ sin θ sinθ cosθ (D) None of these Q.9 If A and B are two non-singular matrices of the same order such that B r I, for some positive integer r >, then A Br r A A B A is equal to. (A) 0 (C) A (B) I (D) None of these Q.0 If A and B are orthogonal matrices of same order, then: (A) A + B is also orthogonal. (B) A B is also orthogonal. (C) AB is also orthogonal. (D) AB + BA is also orthogonal. Q. If C is an orthogonal matrix and A is a square matrix of same order then, trace of C T AC is equal to (A) Trace of C (C) Trace of A (B) Trace of AC (D) None of these matrix Q. Let A and B are idempotent matrices such that A.B BA and A B is non singular then A + B is equal to (A) 0 (B) (C) (D) ± Q. If A and B are square matrices of same order and AA T I, then (A T BA) 0 is equal to (A) AB 0 A T (C) A 0 B 0 (A T ) 0 (B) A T B 0 A (D) 0A T BA 0 c b Q.4 If A c 0 a and b a 0 a ab ac B ab b bcthen AB is equal to ac bc c (A) A (B) B (C) O (D) I Q.5 If A, B, C are square matrices of same order & AB BA, C B, then (A CA) is equal to (A) B (B) A (C) C (D) C Q.6 A is a diagonal matrix of order, and tr(a). If all diagonal entries are positive then maximum value of det (A) is (A) 8 (B) 6 (C) (D) 64 Q.7 If A and B are two matrix such that AB B and BA A, then A + B is equal to (A) AB (B) BA (C) A + B (D) AB Q.8 A 0 and B is column matrix such 0 that (A 8 + A 6 + A 4 + A + I), B 0 where I is a unit matrix of order, then B is equal to (A) 0 (B) 0 (C) 0 (D) Q.9 If A and B are square matrices of same order such that AB BA and A I, then ABA is equal to (A) (AB) (B) I (C) B (D) B Previous Years Questions Q. The parameter, on which the value of the a a determinant cos(p d)x cospx cos(p + d)x sin(p d)x sinpx sin(p + d)x does not depend upon, is (997) (A) a (B) p (C) d (D) x

6.8 Matrices Q. If f(x) x x+ x x(x ) (x + )x x(x ) x(x )(x ) (x + )x(x ) then f(00) is equal to (999) (A) 0 (B) (C) 00 (D) 00 Q. The number of distinct real roots of sinx cosx cosx cosx sinx cosx cosx cosx sinx π 4 x π 4 0 in the interval is (00) (A) 0 (B) (C) (D) Q.4 The number of values of k for which the system of equations (004) (k + )x + 8y 4k, kx + (k + )y k (A) 0 (B) (C) (D) Q.5 Given, x y + z, x y + z 4, x +y + lz 4, then the value of λ such that the given system of equations has no solution is (004) (A) (B) (C) 0 (D) Q.6 If P / /, A / / 0 and Q PAP T, then P T Q 005 P is (005) (A) 005 0 (C) 0 005 (B) 005 005 (D) 0 0 0 0 Q.7 If A 0, 6A A + ca + di, then 0 4 (c, d) is (005) (A) ( 6, ) (B) (, 6) (C) (, 6) (D) (6, ) Q.8 Let α, α, β, β be the roots of ax + bx + c 0 and px + qx + r 0 respectively. If the system of equations α y + α z 0 and β y + β z 0 has a non-trivial b solution. Then prove that q ac (987) pr Q.9 Find the value of the determinant bc ca ab p q r where a, b, and c are respectively the p th, q th and r th terms of a harmonic progression (987) Q.0 Suppose, f(x) is a function satisfying the following conditions: (998) (a) f(0), f() (b) f has a minimum value at x 5, and (c) For all x, ax ax ax + b + f (x) b b+ (ax + b) ax + b + ax + b where a, b are some constants. Determine the constants a, b and the function f(x) Q. Prove that for all values of θ, (000) sinθ cosθ sinθ π π 4π sinθ+ cosθ+ sinθ+ 0 π π 4π sinθ cosθ sinθ Q. If A is an non singular matrix such that AA' A'Aand B A Athen BB equals: (04) (A) I+ B (B) I (C) B (D) (B ) Q. If A is a matrix satisfying the a b equation AA T I where I is identity matrix, then the ordered pair (a, b) is equal to: (05) (A) (, ) (B) (,), (C) (, ) (D) ( ) 5a b T Q.4 If A and AadjA AA, then 5a + b is equal to (06) (A) - (B) 5 (C) 4 (D)

Mathematics 6.9 JEE Advanced/Boards Exercise Q. (a) A is a matrix such that A a, B (adj A) such that B b. Find the value of (ab + a b + I) S Where S a b + a b + a 5 b + up to, and a (b) If A and B are square matrices of order, where A and B I, then find (A ) adj (B ) adj(a ) Q. Let A be the matrices given by A [a ij ] where a ij {0,,,, 4} such that a + a + a + a 4 (i) Find the number of matrices A Such that the trace of A is equal to 4. (ii) Find the number of matrices A such that A is invertible. (iii) Find the absolute value of the difference between maximum value and minimum value of det (A). (iv) Find the number of matrices A such that A is either symmetric or skew-symmetric or both and det (A) is divisible by. 4 4 5 Q. For the matrix A find A. 4 Q.4 (a) Given A 4,B 4, Find P such that BPA 0 0 0 (b) Find the matrix A satisfying the matrix equation. A. 5 4 Q.5 Let S be the set which contains all possible values of I, m, n, p, q, r for which A l p 0 0 m 8 q r 0 n 5 Be a non singular idempotent matrix. Find the absolute value of sum of the products of elements of the set S taken two at a time. cosx sinx 0 Q.6 If F(x) sinx cosx 0 0 0 F(y) F(x + y). Hence, Prove that [F(x)] F( x). then show that F(x). Q.7 Let A n and B n be square matrices of order, which are defind as A n [a ij ] and B n [b ij ] i + j i j where a ij and b n ij for all i and j, i.,j. n If l lim Tr (A + A + A +. n A n ) and n m lim Tr(B + B + B +. n B n ) n Then find the value of ( l + m). [Note: Tr (P) denotes the trace of matrix P] Q.8 Let A be a matrix such that a a and all the other a ij. Let A xa + ya + zi then find the value of (x + y + z) where I is a unit matrix of order. Q.9 Given that A, 0 C, D and that Cb D. 9 Solve the matrix equation Ax b. 0 7 x 4x 7x Q.0 Let A 0 0 and B 0 0 are x 4x x two matrices such that AB (AB) and AB (where I is an identity matrix of order ). Find the value of Tr (AB + (AB) + (AB) + + (AB) 00 ) where Tr.(A) denotes the trace of matrix A.

6.40 Matrices Q. Let M n [m ij ] denotes a square matrix of order n with entries as follows. For i n, m ii 0; For i n, m i+, i m i, i + ; And all other entries in M n are zero. Let D n be the determinant of matrix M n, then find the value of (D 9D ). Q. Find the product of two matrices A & B, 5 where A 7 5 & B and use it to solve the following system of linear equations x + y + z ; x + y + z 7; x + y + z Q. Determine the values of a and b for which the x b system 5 8 9 y a z (i) Has a unique solution; (ii) Has no solution and (iii) Has infinitely many solutions. Q.4 If A 4 ; B 0 ; C 4 and X x x 4 matrix equations. (a) AX B (c) CX A x then solve the following x (b) (B ) X IC Q.5 If A is an orthogonal matrix and B AP where P is a non singular matrix, then show that the matrix PB is also orthogonal. Q.6 Let M be a matrix such that M and M 0. If x and x (x > x ) are the two values x for which det (M xi) 0, where I is an identity matrix of order, then find the value of (5x + x ). Q.7 The set of natural numbers is divided into arrays of rows and columns in the form of matrices as A (), A 6 7 8 4 5, A 9 0 So on. 4 Find the value of T r (A 0 ). [Note: T r (A) denotes trace of A] I 0 x Q.8 Consider I n.m m x dx and J n.m I n x dx n > m and n, m N. m x + 0 (a) Consider a matrix A [a ij ], n I 6+ i, I i+,, i j where a ij 0, i j. Then find trace (A ). [Note: Trace of a square matrix is sum of the diagonal elements.] (b) Let A J 7 J J 6 J J 56 J 6,5,5 7,5,5 8,5,5 I 7 I and B I 6 I I 56 I then find the value of det (A) det (B) 6,5,5 7,5,5 8,5,5 Q.9 Consider the matrices A 4 and B a b and let P be any orthogonal matrix and 0 Q PAP T and R P T Q K P also S PBP T and T P T S K P Column I (A) If we vary K from to n then the first row first column elements of R will form (B) If we vary K from to n then the nd row nd column elements of R will form (C) If we vary K from to n then the first row first column elements of T will form (D) If we vary K from to n then the first row nd column elements of T will represent the sum of Column II (p) G.P. with common ratio a (q) A.P. with volume difference (r) G.P. with common ratio b (s) A.P. with volume difference

Mathematics 6.4 Q.0 Consider a square matrix A of order which has its elements as 0,, and 4. Let N denote the number of such matrices, all elements of which are distinct. Column I (A) Possible non-negative value of det (A) is (p) (B) Sum of values of determinants corresponding to N matrices is (C) If absolute value of (det(a)) is least, then possible value of adj(adj(adj A)) (D) If det (A) is algebraically least, then possible value of det (4A ) is Exercise Single Correct Choice Type Column II (q) 4 (r) (s) (t) 8 Q. Let A, B be two square matrices of the same dimension and let [A, B] AB BA, then for three matrices A, B, C, [[A, B],C] + [[B, C],A] + [[C, A],B] (A) (B) 0 (C) ABC CBA (D) None of these Q. A 4 5 7 & f(x) x 8x + bx + γ. If A α satisfies f(x) 0, then ordered pair (α, γ) is (A) (, 7) (B) (, 7) (C) (, 7) (D) (, 7) Q. If α β is a square root of the two rowed unit γ δ matrix, then δ is equal to (A) a (C) g (B) β (D) None of these Q.4 For A 4 i, (A I) (A I) is a i (A) Null-matrix (C) Unit matrix (B) Hermitian matrix (D) None of these Q.5 If α, β, γ are the real numbers and cos( α β) cos( α γ) A cos( β α) cos( β γ) then γ α γ β cos( ) cos( ) (A) A is skew symmetric (B) A is invertible (C) A is non singular (D) A 0 Q.6 The values of x for which the matrix x+ a b c a x+ b c is non-singular are + a b x c (A) R {0} (B) R { (a + b + c)} (C) R { 0, (a + b + c)} (D) None of these Q.7 Let A is a skew symmetric matrix such A A, and B is a square matrix such that B T B B; B 0. If X (A + B) (A B), then X T X is (A) A I (B) I A (C) A (D) None of these Q.8 For two uni-modular complex numbers z and z, z z z z z z (A) z z z z (C) / 0 0 / z z equal to (B) 0 0 Q.9 If / 5 0 x / 5 5 0 a 5 then the value of x is a (A) 5 (C) a 5 (D) None of these (B) a 5 (D) None of these,

6.4 Matrices Q.0 If A is square matrix such that A I, A and B (adj A) then incorrect statement is (A) AB BA (C) A B (B) AB I (D) B I Q. If A and B are square matrices of order and adj A B, then adj (AB) is equal to (A) B I (B) 9 B I (C) A I (D) 9 A I Q. Let A and B are square matrices of order n such that A T + B O, O is a null matrix, A adj B, tr (A) and A A then tr {adj(a T B)} is equal to (A) ( ) n (B) (C) ( ) n (D) None of these Q. If A is a non-singular matrix such that C A + B, C A I (A B) and AB BA, then (A) B is null matrix (B) A is null matrix (C) C A B (D) A B Previous Years Questions Q. Let ω be a cube root of unity and S be the set of all non-singular matrices of the form a b ω c, where each of a, b or c is ω ω either ω and w. Then, the number of distinct matrices in the set S is (0) (A) (B) 6 (C) 4 (D) 8 Q. Let M and N be two non-singular skewsymmetric matrices such that MN NM. If PT denotes the transpose of P, then M N (M T N) (MN ) T is equal to (0) (A) M (B) N (C) M (D) MN Q. Without expanding a determinant at any stage, show that x + x x+ x x + x x x x + x + x x Ax + B Where A and B are determinants of order not involving x. (98) Q.4 Show that the system of equations x y + 4z, x + y z, 6x + 5y + lz has at least one solution for any real number λ 5. Find the set of solutions, if λ 5. (98) Q.5 Consider the system of linear equations in x, y, z (sin θ) x y + z 0, (cos θ) x + 4y + z 0 and x + 7y + 7z 0. Find the values of θ for which this system has non-trivial solution. (986) Q.6 Let D a Show that n a a n 6 (a ) n 4n (a ) 8 n n n a c constant. (989) p b c Q.7 If a p, b q, c r and a q c 0 a b r p Then, find the value of p a + q q b + r r c Q.8 For a fixed positive integer n, if D D (n!) n! (n + )! (n + )! (n + )! (n + )! (n + )!, then show that (n + )! (n + )! (n + 4)! (99) 4 is divisible by n. (99) Q.9 Let λ and α be real. Find the set of all values of λ for which the system of linear equations lx + (sin α)y + (cos α)z 0 x + (cos α)y + (sin α)z 0 and x + (sin α) y (cos α)z 0 has a non-trivial solution for λ, find all values of α. (99) Q.0 Let a, b, c be real numbers with a + b + c. Show that the equation ax by c bx + ay cx + a bx + ay ax + by c cy + b 0 cx + a cy + b ax by + c represents a straight line. (00)

Mathematics 6.4 Q. Let ω be a cube root of unity and S be the set of all non singular matrices of the form a b ω c where each of a, b and c is either ω or ω ω ω. Then the number of distinct matrices in the set S is (0) (A) (B) 6 (C) 8 (D) 4 Q. Let M be a matrix satisfying 0 0 M,M and M 0 0 0 Then the sum of the diagonal entries of M is (0) (A) 5 (B) 6 (C) 9 (D) 8 Q. If P is matrix such that P T P + I where P T is he transpose of P and I is the identity matrix, then there exists a column matrix x 0 X y 0such that (0) z 0 (A) PX 0 0 0 (B) PX X (C) PX X (D) PX X Q.4 Let P a ij is matrix and let Q b ij, i j where b + a i, j.if the determinant of P is, ij ij then the determinant of the matrix Q is (0) (A) 0 (B) (C) (D) Q.5 Let ω be a complex cube root of unity with ω and P p i+ j be a n nmatrix with ω Then ij P 0, when n (0) (A) 57 (B) 55 (C) 58 (D) 56 Q.6 Let M and N be two matrices such that MN NM. Further, if M N and M N 4 then (04) (A) determinant of (M +MN ) is 0. (B) there is a non zero matrix U such that (M +MN ) P ij U is the zero matrix. (C) determinant of (M +MN ). (D) for a matrix U, if (M +MN ) U equals the zero matrix the U is the zero matrix. Q.7 The quadratic equation p(x)0 with real coefficients has purely imaginary roots. Then the equation p(p(x))0 has (A) Only purely imaginary roots. (B) All real roots. (C) Two real and two purely imaginary roots. (D) Neither real nor purely imaginary roots. + i Q.8 Let z, where,and r,s {,,}. Let ( ) r s z z P and I be the identity matrix of s r z z order. Then the total number of or dered pairs (r, s) for which P I is Q.9 Let P 0 α where 5 0 Suppose where R,k 0 k q 8 ij α R. Q q is a matrix such that PQ ki and det ( ) and I is the identity matrix of order. If k Q, then (A) α 0,k 8 (B) 4α k+ 8 0 9 (C) det ( Padj( Q) ) (D) det ( Padj( P) ) 0 0 Q.0 Let P 4 0 and I be the identity matrix 6 4 of order. If Q q 50 is a matrix such that p Q I then q + q q ij equals (06) (A) 5 (B) 0 (C) 0 (D) 05

6.44 Matrices MASTERJEE Essential Questions JEE Main/Boards Exercise Q.7 Q.8 Q.7 Q. Q. Q.5 Q.8 Q.4 Q.44 Exercise Q.4 Q. Q.4 Q.9 Q. Q.6 Previous Years Questions Q. Q. Q.6 Q.0 Q. JEE Advanced/Boards Exercise Q.7 Q.0 Q. Q.8 Q.9 Q.0 Q.7 Exercise Q. Q.5 Q.8 Q. Previous Years Questions Q. Q.4 Q. Q. Answer Key JEE Main/Boards Exercise Q. x, y Q. 6; 6 Q. Order of AB is ; order of BA is not defined Q.4 (i) (ii) 7 (iii) 8 (iv) Q.5 8, 9, 6, 6, 9, 8 Q.6 0 0 0 0 Q.9 Two Q.0 x, y Q. f(a) A 9S + 7I Q. AB BA Q. 4 4 6 8 6 9 Q.4 No Q.5 / Q.6 0 0

Mathematics 6.45 Q.7 m, p, b 4 Q.8 Skew-symmetric Q.9 (i) 0 0 (ii) cos(x + y) sin(x + y) sin(x + y) cos(x + y) Q. 0 0 0 0 Q.0 Each element is zero Q. ± 0 Q. 0 0 0 0 Q.4 0 5 Q.5 0 Q.7 8, 4, 4, 8 ;, 5 Q.8 [ac + bd + a + b + c + d ] Q.9 Q. 0 0 0 0 0 0 0 0; A 6 A 0 0 Q. x 9m, y 4 9 Q. B 5 8 Q.6 P ±, q ± 4 4 7 Q.7 6 0 Q.8 5 9 7 0 Q.9 x, y 7, z, w 4 Q.40 X 4 4 0 4, Y 0 5 Q.4A A, B I; A B AI A Q.4 X 5 4 0 Exercise Single Correct Choice Type Q. A Q. C Q. D Q.4 B Q.5 C Q.6 C Q.7 A Q.8 B Q.9 D Q.0 D Q. B Q. B Q. C Q.4 C Q.5 D Q.6 D Q.7 D Q.8 B Q.9 A Q.0 C Q. C Q. C Q. B Q.4 C Q.5 C Q.6 D Q.7 C Q.8 C Q.9 C Previous Years Questions Q. B Q. A Q. C Q.4 B Q.5 B 5 5 a,b f x x x+ 4 4 4 4 Q.6 A Q.7 A Q.9 0 Q.0 ( ) Q. B Q. D Q.4 B

6.46 Matrices JEE Advanced/Boards Exercise Q. (a) 5(b) 8 Q. (i) 5 (ii) 8 (iii) 8 (iv) 5 7 4 9 Q. 0 0 Q.4 (a) 4 7 7 5 5 5 (b) Q.5 9 Q.7 Q.8 Q.9 x, x, x Q.0 00 Q. Q. x, y, z Q. (i) a, b R (ii) a and b / (iii) a, b / Q.4 (a) X 5 (b) X (c) No solution Q.6 8` Q.7 55 Q.8 (a) 8 (b) 0 Q.9 A q; B s; C p; D p Q.0 A p, q, t; B s; C p, r; D r 48 5 9 70 4 Exercise Single Correct Choice Type Q. B Q. A Q. A Q.4 A Q.5 D Q.6 C Q.7 B Q.8 C Q.9 C Q.0 D Q. B Q. C Q. C Previous Years' Questions Q. A Q. C Q.4 0 Q.5 θ nπ, nπ + ( ) n π 6, n Z Q6. n a a c Q.7 Q.8 n(n + 4n + 5) Q.9 α nπ or nπ + π/4 Q.0 0 Q. A Q. 9 Q. D Q.4 D Q.5 B, C, D Q.6 A, B Q.7 D Q.8 A Q.9 B, C Q.0 B

Mathematics 6.47 Solutions JEE Main/Boards Exercise Sol : x 5 x+ y x x 4 x 4/ 5 x + y + y y 5 (x, y) (, ) Sol. row n Column m Then total elements mn if (n, m) (, ) nm. 6 if (n, m) (, ) mn. 6 Sol : A and B For AB order will be For BA row of B column of A So, BA does not exist Sol 4: A [a ij ], i, i j a ij i + j (i) a + (ii) a + () + 4 7 (iii) a + () 8 (iv) a 4 not a element i j but here 4 > Sol 5: Total element 8 Assume no of row n And no. of column m so n m 8 8 9 6 6 9 8 Sol 6: Diagonal matrix 0 0 scalar matrix 0 0 Sol 7: Matrix A [a ij ] assume A T [a ji ] So A + A T [a ij + a ji ] [b ij ] assume Here b ij a ij a ji b ji a ij + a ji Here b ij b ji Matrix is symmetric. Sol 8: A A T [a ij ] [a ji ] [b ij ] assume b ij a ij a ji b ji a ji a ij b ij b ij This matrix is known as symmetric matrix. Sol 9: A matrix row n, column m Total element mn mn is prime no. so mn could be, 5, 7, factor of or So for any prime no. of only order n and n (n prime no. ) Sol 0: x + 0 y 4 x 0 y + 4 x 0 x y + 4 y 4 Y Sol : f(x) x 9x + 7 f(a) if A is a matrix f(a) A 9A + 7I A is a square matrix so A is possible.

6.48 Matrices Sol : A, B and AB are symmetric matrices A a ij B b ij AB A ij B ji C ij BA B ij. A ji d ij but B and A ij ij B ji symmetric matrix s property A ji AB A ij B ij A ij. B ij BA AB BA Sol : A AB Sol 4: 4 B 4 4 and 0 4 0 0 0 0 4 4 4 8 6 6 Both have different orders. So they are not same. Sol 5: A 4, K KA K 4 4 ( ) (4) () Sol 7: X m Y p 4 Z b Column of x row of y p and b (m 4) So m ; b 4 Sol8: A 0 0 0 a a, a a a a, so A is skew symmetric. Sol 9: R(θ) cos θ sin θ sinθ cosθ π π cos sin π R 0 π π sin cos 0 R(x + y) cos(x + y) sin(x + y) sin(x + y) cos(x + y) Sol 0: Skew symmetric A [a ij ] For all skew symmetric Matrix dia. l element (a ij ) are zero so a ij 0 & when i j Sol : A a 0 0 0 a 0 A 0 0 A A a 0 0 0 a 0 0 0 A a 0 0 0 a 0 0 0 A4 4 a 0 0 0 Sol 6: tanθ sec θ tan θ tanθ secθ + secθ tan θ sec θ secθ tamθ A 6 6 a 0 0 0 tanθsecθ tanθsecθ tan θ sec θ tan θ sec θ tanθsecθ+ cosθsecθ Sol : [X ] 0 X 0 0 sin θ cos θ sin θ cos θ 0 0 0 X [ x 0 ] [x ] X 0 [(x )x ()] 0 x x 9 0

Mathematics 6.49 x ± 4( 9) Sol : A 4 6 ± 0 Additive inverse B which is A So, A + B A A 0 Sol 4: x + 4 5 0 Assume x x x x x 4 x + x x + x4 4 5 0 x + x 0 x 4 x + 4 5 x + 5 x 5 x 4 0 x 4 x x x Sol 5: 4 x 0 5 x sin θ + cos θ 0 + 0 cot θ 0 cosec θ 0 sin θ+ cos θ+ 0 + 0 cot θ cosec θ 0 + + 0 0 0 Sol 7: Matrix has 8 element m n 8 8 8 4 4 if m n 5 5 5 (only possible order) Sol 8: a b c d a b c [ac + bd] [a + b + c + d ] [a + b + c + d + ac + bd] a b c d 0 0 Sol 9: A 0 0 0 0 0 0 A 0 0 0 0 0 0 A 0 0 0 0 0 0 0 0 0 0 I A 6 [A ] [I] I A 6 I A Sol 0: Properties of skew symmetric matrix [a, j] All diagonal element are zero a ij a ji Sol : A 0 4 0. 5 7 B 0 0 AB 0 4 5 7 0 0 0 0.5 4.0 0( 7) 0 0 0(5) 0( ) 0 0 Sol : A 4 5 A XA +YI 0 A 4 4 5 5 4 +. 4. +.5.4 + 5.. + 5. A 7 8 A XA + YI 0 7 8 4 5 + 0 0 0 0 0 0 4x + y 7 x 0 0 8 x + x 5x + y 0 0 Compare elements 7 x 0

6.50 Matrices x 7 x 7 9 y 45 4 (x, y) (9, 4) Sol : A 5 5 7 A B 4 5 9 Assume B b b b b4 b5 b6 A B. b. b 5.b 4 b4.5 b5.7 b6 4 5 9 b 4 b 6 b b 6 5 b Same as b 9 4 b 4 5, b 8 5, b 6 9 So B 5 8 Sol 4: A cos α sin α sinα cosα 0 Sol 5: A 0 0, 0 F 0 For (ai + ba) ; ai a 0 0 a 0 0 a ba b 0 0 0 0 b 0 0 ai + ba a 0 0 a + 0 b 0 0 a b 0 a (ai + ba) a b a b a b 0 a 0 a 0 a a ab + ba a b 0 a 0 a a ab+ ab 0 a a a b 0 a and R. H. S. a I + a ba a 0 0 0 + a b 0 0 a 0 + 0 a b a a b 0 a 0 0 0 a L. H. S. R. H. S Sol 6: A 0 (pi + qa) A pi p 0 0 p, qa q 0 0 q q q A cos α sin α sinα cosα cosα sinα sinα cosα pi + qa p 0 0 p + 0 q q q p q q p+ q A cos α sin α cosαsinα+ sinαcosα sinαcosα sinαcosα sin α+ cos α we know cos α sin α cos α and cosα sinα sinα (pi + qa) p q p q q p+ q q p+ q p q pq + q(p + q) A (given) pq q(p + q) q + (q + q) so cosα sinα A sinα cosa A 0 So p q 0 p q p ± q

Mathematics 6.5 pq + qp + q q + qp q ± q ve q q q not possible +ve q + q q q / So p q ± 4 Sol 7: A 0 6 5 5 B 6 4 5 4 4 5 A B 0 6 6 4 5 5 4 4 6 0 9 7 Sol 8: A [a ij ] a ij i j \a () (), a () () 4 a () () 7, a () () a () (), a () () 5 a () (), a () () 0 a () () a a a So A a a a a a a 4 7 5 0 Sol 9: x x y x + z y w 7 Compare elements x x y () y 9 y y 9 7 x + z () + 7 6 + 7 4 7 4 6 y w (7) w 7 w 7 4 (x, y, z, w) (, 7,, 4) Sol 40: X + Y 5 0 9, X Y 6 0 sum of X + Y, X Y X + Y + X Y X 5 0 9 + 6 0 5 + + 6 0 9 X 8 8 0 8 4 4 0 4 X 4 4 0 4 Y 5 0 9 X 5 0 9 4 4 0 4 Y 5 4 4 0 9 4 0 5 Sol 4: A cos θ cosθsinθ cosθsinθ sin θ cos φ cosφsinφ B cosφsinφ sin θ AB (C ij ) C cos θ cos φ + cosθ sinθ cosφ sinφ C cosθ cosφ (cosθ cosφ + sinθ sinφ) C cosθ cosφ cos(θ φ) 0 Similarly C, C and C will also be zero So AB 0 0 0 0 0 4 Sol 4: A, B 5 5 4 4 A 5 5 5 5 + 5 + 5 + 5 9+ 5 + 9+ 5 9+ 5 + + + 5 5 5 5 5 5 5 5 5

6.5 Matrices B A 5 5 5 0 4 4 4 0 4 4 4 4 + + B + 4 + 9 + 9 + + 4 4 4 6 6 A B A,I A. I A 8 0 Sol 4: 0 X 4 0 4 0 0 0 X s number of row column of 0 4 order of X assume X X X X X4 X5 X6 X X X 0 4 X4 X5 X6 x x4 x x5 x x6 x4 x5 x6 + + + x 4x4 x 4x5 x 4x6 8 0 4 0 0 0 0 x 4, x 5 4, x 6 0 x + 4x 4 0 x + 4() 0 x 0 x x + 4x 5 x + 4(4) 0 x + 6 0 x 6 0 4 x 4 x + 4x 6 x + 4(0) 0 x 0 x 5 0 0 0 0 0 0 X X X X 5 X4 X5 X6 4 0 Exercise Single Correct Choice Type Sol : (A) Number of elements in a matrix 60 60 5 Number of order matrix can have ( + ) ( + ) ( + ) Sol : (C) A (x) (x + 5) B y ( y) AB and BA both exist for AB x + 5 y for BA y x y 8; x Sol : (D) A is a square invertible matrix A A Multiply A both sides A A A A I A I So A I A I 0 (zero matrix) Sol 4: (B) Total 4 elements all are different. 4 4 4 Total Number of order that exist 4 Number of way to arrange 4 elements 4! Total not of matrix 4 4 Sol 5: (C) A A (I + A) 4 (I + A + A) [I + A + A] [I + A] ( A A) I + 9A + 6A I + 9A + 6A I + 5A Sol 6: (C) A α β β α Since, A is orthogonal matrix (i) (ii)

Mathematics 6.5 So, AA A A I n AA α β α +β 0 a + b β α α β 0 0 α +β 0 x px + q 0 & x + px q 0 β α + β αβ + αβ α αβ + αβ α + β Sum of both equation for common roots x px + q + x + px q 0 x 0 x 0 (i) So if α is roots of x px + q 0 and β is roots of x + px q 0 α + 0 p and α(0) q 0 β + 0 p and β(0) q 0 α p and β p In (i) equation a + b p + ( p) p p ±, q 0 Sol 7: (A) (det A) (det λa) 8 if A s order n n then (det λa) λ n (det A) λ x 8 l n 8 7 l n 7 n λ N So, n Sol 8: (B) A, f(x) + x x I + A 0 0 + I A 0 0 0 0 0 0 (I A) det(i A) 0 0 0 det (I A) 0 ( ) ( ) 4 0 (I f(a) + A) (I A) 4 0 0 0+ + 0 0+ + 0 Sol 9: (D) A is skew symmetric matrix A T A I A T A (AT) (A) A T A A I Taking square of both sides A 4 I A 4n I ( A I are A A T ) \ A A 4n A I A 4n+ A ( A) I A 4(n + ) ( A)I A T I A I 4n - N A 4n A T Sol 0: (D) A B A, B Tr (A) Tr(B) A A + 4I 0 and B λb + µi 0 if A B, A B Tr A Tr(B) satisfied so A λa + µi 0 µ 4 ( A s order ) Sol : (B) The adjoint of upper triangular matrix is false. That is equal to upper triangular not lower triangular matrix. Sol : (B) A, B, (A + B) are non-singular [A(A + B) B] [A ((A + B) ) B ] (A (A + B) B ) [(A A + A B)B ) [(I + A B)B ) [B + A BB ] A + B Sol : (C) A is an orthogonal matrix A

6.54 Matrices AA T A T A I n A A T A T + A ( A is an orthogonal matrix) A T det(a) (adja) (adja) Sol 4: (C) A and B are square matrices of order (A) adj(ab) adj(a) + adj(b) is not necessary option (A) is wrong. + tan x 0 + tan x 0 0 0 A T A 0 0 Sol 7: (D) A 5 A ( ) (5) 5 4 9 (B) (A + B) A + B (C)AB 0 A, B are square matrix So if AB 0 A B 0 A 0 or B 0 a 0 0 Sol 5: (D) A 0 a 0 0 0 a a 0 0 adj(a) 0 a 0 0 0 a adj A (a ) a 6 \ A adj (A) a a 6 a 9 A a I a A A 5 Sol 8: (B) P T P T A 9 5 9 Assume P P P P P, PT P P 4 P P4 P P P 4 P P T P P 4 P P4 P (adj P) P P P P P P P P P P 4 P P P P P P4 P P P P P P 4 P P Sol 6: (D) A tanx tanx A + tan x A + tan x AT tanx tanx tanx tanx A ( + tan x) Only option (B) cos θ sin θ is correct. sinθ cosθ Sol 9: (A) B r I r > A B r A A B A A B r B A A B A A B A A B A 0 Sol 0: (C) A & B are orthogonal matrices tanx tanx A T A ( + tan x) tanx tanx ( + tan x) ( + tan x) + tan x tanx tanx tanx tanx tan x + + tan x 0 0 + tan x AA T A T A I n and BB T B T B I n AB (AB) (AB) T (AB) (B T A) T A(BB T )A t AIA T AA T I (AB) T (AB) B T A T AB B T IB B T B I n So (AB) T (AB)

Mathematics 6.55 ge AB(AB) T I n So, AB also satisfying property of orthogonal Sol : (C) C is an orthogonal matrix CC T C T C I n Tr (C T AC) Tr [(C T A)C] Tr [C(C T A)] Tr (CC T A) Tr (IA) Tr (A) Sol : (C) A and B are idempotent matrices so, A A and B B A, B or AB BA A B is non-singular A and B or A and B 0 Sol : (B) AA T I (A T BA) 0 (A T BA) (A T BA) (A T BA).. A T BIBIBI. BA A T B 0 A 0 c b Sol 4: (C) A c 0 a b a 0 a ab ca B ab b cb ac bc c abc abc b c b c c b bc AB a bc + ab c abc + abc ac + ac a b a b ab ab abc abc 0 0 0 AB 0 0 0 0 0 0 0 Sol 5: (C) AB BA, C B (A CA) (A CA) (A CA) A C ICA A - C A A BA A (AB) IB B C Tr(A) a + a + a and det A a a a for maximum of det (A) a a a a + a + a a a a a 4 det A 4 4 4 64 Sol 7: (C) AB B BA A (A + B) (AB + BA) A + B + AB + BA (AB) + (BA) + (AB) (BA) + (BA) (AB) A + B + AB + BA ABAB + BABA + AB + BA A + B AAB + BBA AB + BA A + B A + B Sol 8: (C) A 0, B is column matrix 0 0 (A 8 + A 6 + A 4 + A + I) B A 0 0 0 0 0 0 I A 4 A A I A 6 I, A 8 4 I A 8 + A 6 + A 4 + A + I I ( + + + + 4 ) I 0 IB 0 B Sol 9: (C) AB BA and A I ABA A(AB) A B IB B Previous Years Questions Sol : (B) + a a a cos(p d)x + cos(p + d)x cospx cos(p + d)x sin(p d)x + sin(p + d)x sinpx sin(p + d)x Sol 6: (D) Tr(A) a 0 0 Assume A 0 a 0 0 0 a ( A is diagonal matrix) + a a a cospxcosdx cospx cos(p + d)x sinpxcosdx sinpx sin(p + d)x

6.56 Matrices Applying C C cos dxc + a acosdx a a 0 cospx cos(p + d)x 0 sinpx sin(p + d)x ( + a a cos dx) [sin (p+ d) x cos px sin px cos (p + d) x] ( + a a cos dx) sin dx Which is independent of p Sol : (A) x x+ f(x) x x(x ) (x + )x x(x ) x(x )(x ) (x + )x(x ) cos x sin x or sin x cos x cot x / gives no solution in π 4 x π 4 and sin x cos x tan x x π/4 Sol 4: (B) For infinitely many solutions, we must have k + 8 k k+ 4k k k Sol 5: (B) Since, given system has no solution 0 and any one amongst D x, D y, D z is non-zero. Let λ 0 λ Applying C C (C + C ) x 0 x x(x ) 0 0 x(x ) x(x )(x ) 0 f(x) 0 (00) 0 sinx cosx cosx Sol : (C) Given, cosx sinx cosx 0 cosx cosx sinx Applying C C + C + C sinx + cosx cosx cosx sinx + cosx sinx cosx sinx + cosx cosx sinx cosx cosx (cos x+ sin x) sinx cosx 0 cosx sinx Applying R R R, R R R (cos x + sin x) cosx cosx 0 sinx cosx 0 0 0 0 sinx cosx (cos x + sin x) (sin x cos x) 0 cos x + sin x 0 or sin x cos x 0 Sol 6: (A) Now, P T P / / / / / / / / P T P 0 0 P T P I P T P Since, Q PAP T P T Q 005 P P T (PAP T )(PAP T ). 005 times]p T T T T T (P P)A(P P)A(P P)...(P P)A(P P) 005 times IA 005 A 005 [from eq.(i)] A 0 A 0 0 0 A 0 0 0..... A 005 005 0 P T Q 005 P 005 0 (i)

Mathematics 6.57 Sol 7: (A) Every square matrix satisfied its characteristic equation i.e. A λ I 0 λ 0 0 0 λ 0 4 λ 0 ( λ) {( λ) (4 λ) + } 0 6 + λ 6 0 A 6 A + A 6 I O A 6A + I 6A Sol 8: Since, a, a are the roots of ax + bx + c 0 a + a b a and a a c a Also, b, b are the roots of px + qx + r 0 b + b q p and b b r p Given system of equations a y + a z 0 And b y + b z 0, has non-trivial solution α β α β 0 α α β β Applying componendo-dividendo α +α α α β +β β β (a + a ) (b b ) (a a ) (b + b ) (a + a ) {(b b ) 4b b } (b + b ) {(a + a ) 4a a } From equation (i) and (ii), we get b q 4r a p p bq ap 4b r ap br a qc p b q q b 4c p a a bq ap ac pr 4q c ap (i) (i) (ii) Sol 9: A + (p )D a A + (q )D b A + (r )D c bc ca ab a b c Let p q r abc p q r, [From equation (i)] abc A + (p )D A + (q )D A + (r )D p q r Applying R R (A D) R DR abc 0 0 0 p q r bc ca ab p q r Sol 0: Given, f (x) 0 0 ax ax ax + b + b b+ (ax + b) ax + b + ax + b Applying R R R R, We get f (x) ax ax ax + b + b b+ 0 0 ax ax b b+ f (x) ax + b On integrating, we get f(x) ax + bx + c ax b (C C C ) Where c is an arbitrary constant Since, f has maximum at x 5/ (i)

6.58 Matrices f (5/) 0 5a + b 0 Also, f(0) c and f() a + b + c (i) (ii) 4π sinqcos sinqcos π+ π sinqcos π sin θ On solving equation (i) and (ii) for a, b, we get a 4, b 5 4 Thus, f(x) 4 x 5 4 x + Sol : sinθ cosθ sinθ π π 4π sinθ+ cosθ+ sinθ+ π π 4π + sin θ + cos θ + sin θ π π 4π sin θ cos θ sin θ π Now, sinθ+ + sin θ π π π π π θ+ +θ θ+ θ+ sin cos π sinθ cos sinθ cos π π sinθ cos π sinθ and cos θ+ π + cos θ π π π π π θ+ +θ θ+ θ+ cos cos cosθ cos π cosθ cosθ and sin 4 θ+ π + sin 4 θ π 4π 4π 4π 4π θ+ + θ θ+ θ+ sin cos sinθ cosθ sinθ sinθ cosθ sinθ π π 4π sinθ cosθ sinθ 0 (since, R and R are proportional) Sol : (B) T T T ( )( ) T T BBT A A A A ( A A ) A. ( A ) T ( ) ( ) ( )( ) T T T T A. A A A. A A AA A T ( ) T ( A A A. A ) A. ( A ) ( A A) T I Sol : (D) T AA 9T a A 9I a b b 9 0 a+ 4+ b 9 0 0 0 9 a+ b 0 9 0 a + 4 + b a + b a + 4 + b 0 0 9 Equation a + 4 + b 0 a + b 4... (i) a + b 0 a b... (ii) & a + 4 + b 0 a + b 5... (iii) Solving a,b 5a b Sol 4: (B) A 5a b T A AadjA AA 5a b b 5a b 5a 5a b 0a + b 0 5a + b 5 b 0 0a + b 5 b Equate, 0a + b 5a + b and 0a + b and 5a b 0

Mathematics 6.59 a b k (let) 5 Solving a,b 5 So, 5a + b 5 + 5 5 JEE Advanced/Boards Exercise Sol : (a) A a, (B) (adja), B b a a a S + + +... b 5 b b a a a + + +... b b b a a b ab b a b b a b a b ab S b a b > a, a B A n A () 9 b ()9 (ab + a b + ) 9 (. (9) +. 9 + ) ( + 8 + 4) 6.9 78 6.9 78 5 (b) A, B (A ) (adj B ) adj(a + ) A adj B adj A 4 4 ( ) 6 8 Sol : A [a ij ], a ij {0,,,, 4} a + a + a + a 4 (i) T r (A) a + a 4 and a, a {0,,,, 4} ()9 9 total possibilities 0 + 4 4 + 0 + + + 5 (ii) A is invertible so A 0 a a a a 0 a a a a (4 + ) total way. 7 8 C (iii) A max A min A max a a a a (a a ) max 4 ( a + a + a + a 4) and + ( a. a ) max a a 0 with a a 4 () () So A max () 0 4 A min 0 () 4 A max A min 4 ( 4) 8 (iv) A is symmetric or skew symmetric or both det A is divisible by So det A can be 0,, 4 4 0 A 0 0, 0 0 0 4, 0 0, 0, 0 0 Total no. 5 4 4 5 Sol : A 4 A 4 A.. A 9 + 8 A 5 + 6 A 6 5, A + 0 A 5, A 0 + A 8 4 adja + 0 4 A 4[A ] 4(A) + 5[A ] 4() 4( ) + 5( ) + 4 5 A adja A + 0 4 + + 7 4 9 A 0 0 0 4 0 4 5

6.60 Matrices Sol 4: (a) A 4, B 4 BPA Assume P 0 0 0 BPA 0 0 0 P P P P P P 4 5 6 0 B BPA PA B 0 0 B 8 9 adjb 4 4 PA 4 0 0 0 4 4 A [4 ] [ ] + [ 6 8] 4 adja 0 0, + 6 8 + A 0 PAA 4 4 A P P 4 4 0 P 4 7 7 5 5 (b) A 4 5 assume B C 5 B 4 C 9 0 9 adjb, B adjc 5, C 9 BAC 4 B BAC AC AC 9 0 4 ACC A 9 0 4 5 5 9 + 45 7 A 9 70 4 Sol 5: A 4 4 8 + 6 + 6 48 5 9 70 4 P 0 0 m 8 q r 0 n 5 A A[ A is idempotent matrix] A ( ) + 0 p( ) + p[m 8] pq qr (m 8) q(m 8) + q(n 5) r( ) + r(n + 5) rp (n 5) p 0 0 m 8 q r 0 n 5 compare elements ( ) 0 or ± or ± 4 ± p[l + m 8] p p 0 or l + m rp 0 r 0 or p 0 (n 5) n 5 n 5 or 0 q[(m 8) + n 5] q q 0 or m + n 0 + (m 8) m 8 m 8 or m ± 8 or ± 9 ± if, l m n, q, q, r z S {0, ±, ±, ±4}

Mathematics 6.6 Sum of products of elements + +4 9 cosx sinx 0 Sol 6: F(x) sinx cosx 0 0 0 F(y) cos y siny 0 siny cos y 0 0 0 cosx sinx 0 cos y siny 0 F(x). F(y) sinx cosx 0siny cos y 0 0 0 0 0 cosxcos y sinxsiny cosxsiny sinxcos y 0 sinxcos y + cosxsiny sinxsiny + cos y cosx 0 0 0 cos(x + y) sin(x + y) 0 sin(x + y) cos(x + y) 0 0 0 [F(x)] F( x) L. H. S. F(x) cos (x) + sin x cosx sinx 0 adj[f(x)] sinx cosx 0 0 0 [F(x)] adjf(x) L. H. S. R. H. S. Hence proved Sol 7: A n [a ij ], B n [b ij ] a ij i + j n, b ij i i n cos( x) sin( x) 0 sin( x) cos( x) 0 F( x) 0 0 l Lim Tr[A + A + A + + n An+. ] n For A n Tr(A) a + a + a (() + ) + () + + () + n n 8 8 Tr( na n) n n 9+ 6+ 8 n n Tr n An Tr( A ) 8 + +... i n n 8 8 8 8 Tr (B n ) () + () + () n Tr( n B n ) n a n m [Tr( B n )] + +... n l + m + 9 Sol 8: A is matrix A a, all other a ij R R R A R R R 0 0 A R R R R 0 0 0 0 A 0 I 0 0 A I xa + ya + zi I (x + y + z)i (x + y + z) Sol 9: A 0 C, D, Cb D 9 + 4 + 6 n n C [ ] + [ ] + [ ]

6.6 Matrices 0 adjc 0 C C C 0 C Cb C D b b 0 0 0 0 9 0 + + 8 5 0 9 A [6 + ] + [ 6] + [ ] 9 6 8 5 6+ 8 9 8 adja 4 4 4 A AX b A adja 5 adja X A b 5 9 8 4 5 X 9 4 + 0 + + 5 5 5 5 + 5 + 4 9 0 5 0 7 x 4x 7x Sol 0: A 0 0, B 0 0 x 4x x x + 7x 8x 8x 4x 4x AB 0 0 + x x 4x 4x 7x x 5x 0 0 0 0 0 0x 5x AB 5x [5x] 5x (AB) 5x 5x 0 0 0 5x 0 0 50x + 0x 5x 0 0 5x 5x 0 0 (AB) 0 0 AB 0 0 0x + 0 0 0x 5x 5x 5x x /5 5 0 0 5 0 0 AB 0 0 0 0 5 0 4 0 4 5 (AB) (AB) (AB) (AB) (AB) I Tr[AB+(AB) +(AB) + +(AB) 00 ] Tr[AB+I+AB+I+ +I] Tr[50AB+50I]50 Tr(AB)+50Tr(I) 50[ + ]+50 [++] 50+(50)00 Sol : M n [m ij ]order n i n, m ij 0; i n, m i +, I m i, i + All other entries in Mn are zero 0 0 M 0, M 0[00 9] + [ 0] 0 0 000 90 90 80 M 0 0 M 00 9 9 D 9D 80 9(9) 80 89 5 Sol : A 7 5 & B x + y + z 4 0 0 x + y + z 7 AB 0 4 0 4I x + y + z 0 0 4 5 4 4 4 A 7 5 B 4 4 4 4 4 4 4

Mathematics 6.6 B [6 ] + [ 9] + [ 4] 5 7 4 x X B, X 7 + [ 4] + 5 7 6 8 x 8, Y 7 4 [6 ] + 7[ ] [ ] + [ 9] + [6 4] 9 7 6 4 x b 5 8 9 y z R R + R 5+ 4 8+ 9 6 9 6 x y z x b y z 6 b y 4, Z 7 4 [4 7] + [4 6] + [ 4] + 8 4 or Bx C x B C z 4 4 Compare row nd and st x y + z b 9x 6y + z x y + z From equation (i) and (ii) b (i) (ii) (x, y, z) (,, ) For no. solution a and b x Sol : 5 4 4 4 7 5 4 4 4 4 4 4 5 8 9 a 7 x 6 y z + D 5 8 9 [ 8a 9] [8 5a] + [5 + 6] \ a 4a 7 6 + 0a + 4a 4 (i) System has a unique solution 0 0 40 4 0 a 4 4 a and b R (ii) At a has no solution a (iii) Has infinitely solution so a and b so D 0 and D x 0 Sol 4: A 4, B 0, C 4 X X X X X4 (a) AX B I X X 4 X X4 A 4 6 adja 4 4 A so A AX X A (B I) X 0 0 0 A adja