Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 50 -
CHAPTER 3: MATRICES QUICK REVISION (Important Concepts & Formulae) MARKS WEIGHTAGE 03 marks Matrix A matrix is an ordered rectangular array of numbers or functions. The numbers or functions are called the elements or the entries of the matrix. We denote matrices by capital letters. Order of a matrix A matrix having m rows and n columns is called a matrix of order m n or simply m n matrix (read as an m by n matrix). In general, an m n matrix has the following rectangular array: a a... a n a a... a n........ am am... a mn or A = [a ij ] m n, i m, j n i, j N Thus the ith row consists of the elements a i, a i, a i3,..., a in, while the j th column consists of the elements a j, a j, a 3j,..., a mj, In general a ij, is an element lying in the i th row and j th column. We can also call it as the (i, j) th element of A. The number of elements in an m n matrix will be equal to mn. x or x y y We can also represent any point (x, y) in a plane by a matrix (column or row) as, Types of Matrices (i) Column matrix A matrix is said to be a column matrix if it has only one column. In general, A = [a ij ] m is a column matrix of order m. (ii) Row matrix A matrix is said to be a row matrix if it has only one row. In general, B = [b ij ] n is a row matrix of order n. (iii) Square matrix A matrix in which the number of rows are equal to the number of columns, is said to be a square matrix. Thus an m n matrix is said to be a square matrix if m = n and is known as a square matrix of order n. In general, A = [a ij ] m m is a square matrix of order m. If A = [a ij ] is a square matrix of order n, then elements (entries) a, a,..., a nn are said to constitute the diagonal, of the matrix A. (iv) Diagonal matrix A square matrix B = [b ij ] m m is said to be a diagonal matrix if all its non diagonal elements are zero, that is a matrix B = [b ij ] m m is said to be a diagonal matrix if b ij = 0, when i j. (v) Scalar matrix Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 5 -
A diagonal matrix is said to be a scalar matrix if its diagonal elements are equal, that is, a square matrix B = [b ij ] n n is said to be a scalar matrix if b ij = 0, when i j b ij = k, when i = j, for some constant k. (vi) Identity matrix A square matrix in which elements in the diagonal are all and rest are all zero is called an identity if i j matrix. In other words, the square matrix A = [aij] n n is an identity matrix, if aij 0if i j We denote the identity matrix of order n by I n. When order is clear from the context, we simply write it as I. Observe that a scalar matrix is an identity matrix when k =. But every identity matrix is clearly a scalar matrix. (vii) Zero matrix A matrix is said to be zero matrix or null matrix if all its elements are zero. We denote zero matrix by O. Equality of matrices Two matrices A = [a ij ] and B = [b ij ] are said to be equal if (i) they are of the same order (ii) each element of A is equal to the corresponding element of B, that is a ij = b ij for all i and j. Operations on Matrices Addition of matrices The sum of two matrices is a matrix obtained by adding the corresponding elements of the given matrices. Furthermore, the two matrices have to be of the same order. a a a3 b b b3 Thus, if A = is a 3 matrix and B = is another 3 matrix. Then, a a a3 b b b3 we define a b a b a3 b3 A + B =. a b a b a3 b3 In general, if A = [a ij ] and B = [b ij ] are two matrices of the same order, say m n. Then, the sum of the two matrices A and B is defined as a matrix C = [c ij ] m n, where c ij = a ij + b ij, for all possible values of i and j. If A and B are not of the same order, then A + B is not defined. Multiplication of a matrix by a scalar If A = [a ij ] m n is a matrix and k is a scalar, then ka is another matrix which is obtained by multiplying each element of A by the scalar k. In other words, ka = k [a ij ] m n = [k (a ij )] m n, that is, (i, j)th element of ka is ka ij for all possible values of i and j. Negative of a matrix The negative of a matrix is denoted by A. We define A = ( ) A. Difference of matrices If A = [a ij ], B = [b ij ] are two matrices of the same order, say m n, then difference A B is defined as a matrix D = [d ij ], where d ij = a ij b ij, for all value of i and j. In other words, D = A B = A + ( ) B, that is sum of the matrix A and the matrix B. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 5 -
Properties of matrix addition (i) Commutative Law If A = [a ij ], B = [b ij ] are matrices of the same order, say m n, then A + B = B + A. (ii) Associative Law For any three matrices A = [a ij ], B = [b ij ], C = [c ij ] of the same order, say m n, (A + B) + C = A + (B + C). (iii) Existence of additive identity Let A = [a ij ] be an m n matrix and O be an m n zero matrix, then A + O = O + A = A. In other words, O is the additive identity for matrix addition. (iv) The existence of additive inverse Let A = [a ij ] m n be any matrix, then we have another matrix as A = [ a ij ] m n such that A + ( A) = ( A) + A= O. So A is the additive inverse of A or negative of A. Properties of scalar multiplication of a matrix If A = [a ij ] and B = [b ij ] be two matrices of the same order, say m n, and k and l are scalars, then (i) k(a +B) = k A + kb, (ii) (k + l)a = k A + l A Multiplication of matrices The product of two matrices A and B is defined if the number of columns of A is equal to the number of rows of B. Let A = [a ij ] be an m n matrix and B = [b jk ] be an n p matrix. Then the product of the matrices A and B is the matrix C of order m p. To get the (i, k) th element c ik of the matrix C, we take the ith row of A and k th column of B, multiply them elementwise and take the sum of all these products. In other words, if A = [a ij ] m n, B = [b jk ] n b k b k p, then the ith row of A is [a i a i... a in ] and the kth column of B is. then. b nk c ik = a i b k + a i b k + a i3 b 3k +... + a in b nk = The matrix C = [c ik ] m p is the product of A and B. If AB is defined, then BA need not be defined. In the above example, AB is defined but BA is not defined because B has 3 column while A has only (and not 3) rows. If A, B are, respectively m n, k l matrices, then both AB and BA are defined if and only if n = k and l = m. In particular, if both A and B are square matrices of the same order, then both AB and BA are defined. Non-commutativity of multiplication of matrices Now, we shall see by an example that even if AB and BA are both defined, it is not necessary that AB = BA. Zero matrix as the product of two non zero matrices We know that, for real numbers a, b if ab = 0, then either a = 0 or b = 0. If the product of two matrices is a zero matrix, it is not necessary that one of the matrices is a zero matrix. Properties of multiplication of matrices The multiplication of matrices possesses the following properties: The associative law For any three matrices A, B and C. We have (AB) C = A (BC), whenever both sides of the equality are defined. The distributive law For three matrices A, B and C. A (B+C) = AB + AC (A+B) C = AC + BC, whenever both sides of equality are defined. The existence of multiplicative identity For every square matrix A, there exist an identity matrix of same order such that IA = AI = A. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 53 -
Transpose of a Matrix If A = [a ij ] be an m n matrix, then the matrix obtained by interchanging the rows and columns of A is called the transpose of A. Transpose of the matrix A is denoted by A or (A T ). In other words, if A = [a ij ] m n, then A = [a ji ] n m. Properties of transpose of the matrices For any matrices A and B of suitable orders, we have (i) (A ) = A, (ii) (ka) = ka (where k is any constant) (iii) (A + B) = A + B (iv) (A B) = B A Symmetric and Skew Symmetric Matrices A square matrix A = [a ij ] is said to be symmetric if A = A, that is, [a ij ] = [a ji ] for all possible values of i and j. A square matrix A = [a ij ] is said to be skew symmetric matrix if A = A, that is aji = aij for all possible values of i and j. Now, if we put i = j, we have a ii = a ii. Therefore a ii = 0 or a ii = 0 for all i s. This means that all the diagonal elements of a skew symmetric matrix are zero. Theorem For any square matrix A with real number entries, A + A is a symmetric matrix and A A is a skew symmetric matrix. Theorem Any square matrix can be expressed as the sum of a symmetric and a skew symmetric matrix. Elementary Operation (Transformation) of a Matrix There are six operations (transformations) on a matrix, three of which are due to rows and three due to columns, which are known as elementary operations or transformations. (i) The interchange of any two rows or two columns. Symbolically the interchange of ith and jth rows is denoted by R i R j and interchange of ith and jth column is denoted by C i C j. (ii) The multiplication of the elements of any row or column by a non zero number. Symbolically, the multiplication of each element of the ith row by k, where k 0 is denoted by R i k R i. The corresponding column operation is denoted by C i kc i (iii) The addition to the elements of any row or column, the corresponding elements of any other row or column multiplied by any non zero number. Symbolically, the addition to the elements of i th row, the corresponding elements of j th row multiplied by k is denoted by R i R i + kr j. The corresponding column operation is denoted by C i C i + kc j. Invertible Matrices If A is a square matrix of order m, and if there exists another square matrix B of the same order m, such that AB = BA = I, then B is called the inverse matrix of A and it is denoted by A. In that case A is said to be invertible. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 54 -
A rectangular matrix does not possess inverse matrix, since for products BA and AB to be defined and to be equal, it is necessary that matrices A and B should be square matrices of the same order. If B is the inverse of A, then A is also the inverse of B. Theorem 3 (Uniqueness of inverse) Inverse of a square matrix, if it exists, is unique. Theorem 4 If A and B are invertible matrices of the same order, then (AB) = B A. Inverse of a matrix by elementary operations If A is a matrix such that A exists, then to find A using elementary row operations, write A = IA and apply a sequence of row operation on A = IA till we get, I = BA. The matrix B will be the inverse of A. Similarly, if we wish to find A using column operations, then, write A = AI and apply a sequence of column operations on A = AI till we get, I = AB. In case, after applying one or more elementary row (column) operations on A = IA (A = AI), if we obtain all zeros in one or more rows of the matrix A on L.H.S., then A does not exist. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 55 -
CHAPTER 3: MATRICES MARKS WEIGHTAGE 03 marks NCERT Important Questions & Answers. If a matrix has 8 elements, what are the possible orders it can have? What, if it has 5 elements? Since, a matrix containing 8 elements can have any one of the following orders : 8, 8, 9, 9, 3 6,6 3 Similarly, a matrix containing 5 elements can have order 5 or 5.. Construct a 3 4 matrix, whose elements are given by: (i) a ij = 3i + j (ii) a ij = i j (i) The order of given matrix is 3 4, so the required matrix is a a a3 a4 A a a a3 a 4, where a ij = 3i + j a3 a3 a33 a 34 34 Putting the values in place of i and j, we will find all the elements of matrix A. a 3, a 3, a3 3 3 0 5 a4 3 4, a 6, a 6 3 a3 6 3, a4 6 4, a3 9 4 7 5 a3 9, a33 9 3 3, a34 9 4 0 5 3 Hence, the required matrix is A 7 5 4 3 34 a a a3 a4 (ii) Here, A a a a3 a 4,where a ij = i j a3 a3 a33 a 34 34 a = =, a = = 0, a 3 = 3 =, a 4 = 4 =, a = 4 = 3, a = 4 =, a 3 = 4 3 =, a 4 = 4 4 = 0 a 3 = 6 = 5, a 3 = 6 = 4, a 33 = 6 3 = 3 and a 34 = 6 4 = Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 56 -
Hence, the required matrix is 0 A 3 0 5 4 3 34 a b a c 5 3. Find the value of a, b, c and d from the equation: a b 3c d 0 3 a b a c 5 Given that a b 3c d 0 3 By definition of equality of matrix as the given matrices are equal, their corresponding elements are equal. Comparing the corresponding elements, we get a b = (i) a b = 0 (ii) a + c = 5 (iii) and 3c + d = 3 (iv) Subtracting Eq.(i) from Eq.(ii), we get a = Putting a = in Eq. (i) and Eq. (iii), we get b = and + c = 5 b = and c = 3 Substituting c = 3 in Eq. (iv), we obtain 3 3 + d = 3 d = 3 9 = 4 Hence, a =,b =, c = 3 and d = 4. 5 3 6 4. Find X and Y, if X + Y = 0 9 and X Y = 0. 5 3 6 ( X Y ) ( X Y ) 0 9 0 8 8 8 8 X X 0 8 0 8 4 4 X 0 4 5 3 6 Now,( X Y) ( X Y) 0 9 0 4 4 Y X 0 0 0 0 Y 0 5 5. Find the values of x and y from the following equation: x 5 3 4 7 6 7 y 3 5 4 x 5 3 4 7 6 7 y 3 5 4 x 0 3 4 7 6 4 y 6 5 4 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 57 -
x 3 6 7 6 5 y 4 5 4 or x + 3 = 7 and y 4 = 4 or x = 7 3 and y = 8 or x = 4 and y = 8 i.e. x = and y = 9. 0 3 5 6. Find AB, if A 0 and B 0 0. 0 3 5 0 0 We have AB 0 0 0 0 0 Thus, if the product of two matrices is a zero matrix, it is not necessary that one of the matrices is a zero matrix. 3 7. If A 3, then show that A 3 3A 40 I = O 4 3 3 9 4 8 A A. A 3 3 8 4 4 4 6 5 3 9 4 8 63 46 69 3 So, A A. A 3 8 69 6 3 4 4 6 5 9 46 63 63 46 69 3 0 0 3 Now, A 3A 40I 69 6 3 3 3 40 0 0 9 46 63 4 0 0 63 46 69 3 46 69 40 0 0 69 6 3 69 46 3 0 40 0 9 46 63 9 46 3 0 0 40 63 3 40 46 46 0 69 69 0 69 69 0 6 46 40 3 3 0 9 9 0 46 46 0 63 3 40 0 0 0 0 0 0 0 0 0 0 0 8. If x y 3 5, find the values of x and y. 0 x y x y 0 3 5 3x y 5 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 58 -
By definition of equality of matrix as the given matrices are equal, their corresponding elements are equal. Comparing the corresponding elements, we get x y = 0 (i) and 3x + y = 5 (ii) Adding Eqs. (i) and (ii), we get 5x = 5 x = 3 Substituting x = 3 in Eq. (i), we get 3 y = 0 y = 6 0 = 4 x y x 6 4 x y 9. Given 3 z w w z w 3, find the values of x, y, z and w. By definition of equality of matrix as the given matrices are equal, their corresponding elements are equal. Comparing the corresponding elements, we get 3x = x + 4 x = 4 x = and 3y = 6 + x + y y = 6 + x y= 6 x Putting the value of x, we get 6 8 y 4 Now, 3z = + z + w, z = + w w z (i) Now, 3w = w + 3 w = 3 Putting the value of w in Eq. (i), we get 3 z Hence, the values of x, y, z and w are, 4, and 3. cos x sin x 0 0. If F( x) sin x cos x 0, show that F(x) F(y) = F(x + y). 0 0 cos x sin x 0 cos y sin y 0 LHS F( x) F( y) sin x cosx 0 sin y cos y 0 0 0 0 0 cos x cos y sin xsin y sin y cos x sin x cos y 0 sin x cos y cos xsin y sin xsin y cos x cos y 0 0 0 cos( x y) sin( x y) 0 sin( x y) cos( x y) 0 F( x y) RHS 0 0 0. Find A 5A + 6I, if A 3 0 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 59 -
0 0 5 A A. A 3 3 9 5 0 0 0 5 0 0 0 A 5A 6I 9 5 5 3 6 0 0 0 0 0 0 5 0 0 5 6 0 0 9 5 0 5 5 0 6 0 0 5 5 0 0 0 6 5 0 6 0 0 5 0 3 9 0 0 5 6 5 5 0 0 0 5 0 5 0 0 6 5 4 4 0. If A 0, prove that A 3 6A + 7A + I = 0 0 3 0 0 5 0 8 A A. A 0 0 4 5 0 3 0 3 8 0 3 5 0 8 0 0 34 3 A A. A 4 5 0 8 3 8 0 3 0 3 34 0 55 0 34 5 0 8 0 0 0 3 A 6A 7A I 8 3 6 4 5 7 0 0 0 34 0 55 8 0 3 0 3 0 0 0 34 30 0 48 7 0 4 0 0 8 3 4 30 0 4 7 0 0 34 0 55 48 0 78 4 0 0 0 30 7 0 0 0 0 34 48 4 0 0 0 0 0 0 8 4 4 3 30 7 0 0 0 0 O 34 48 4 0 0 0 0 0 55 78 0 0 0 3 0 3. If A 4 and I 0, find k so that A = ka I Given than A = ka I 3 3 3 0 k 4 4 4 0 9 8 6 4 3k k 0 8 8 4 4k k 0 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 60 -
3k k 4 4 4k k By definition of equality of matrix as the given matrices are equal, their corresponding elements are equal. Comparing the corresponding elements, we get 3k = k = k = k = 4k = 4 k = 4 = k k = Hence, k = 0 tan 4. If A and I is the identity matrix of order, show that tan 0 I + A = (I A) cos sin sin cos 0 x Let A where x tan x 0 tan tan Now, x x cos and cos tan x tan x cos sin RHS ( I A) sin cos x x 0 0 x x x 0 x 0 x x x x x x x x x x( x ) x x x x x x x x x( x ) x x x x x x x 3 3 x x x x x x x( x ) x x x x x 3 3 x x x x x x x( x ) x x x x x 0 0 x x LHS RHS 0 x 0 x 5. Express the matrix matrix. 4 B 3 4 3 as the sum of a symmetric and a skew symmetric Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 6 -
4 B 3 4 B ' 3 3 4 4 3 3 3 4 3 3 Let P ( B B ') 3 6 3 3 3 6 3 3 3 3 Now P ' 3 3 P 3 3 Thus P ( B B ') is a symmetric matrix. 0 5 0 5 Also, let Q ( B B ') 0 6 0 3 5 6 0 5 3 0 0 5 Now Q ' 0 3 Q 5 3 0 Thus Q ( B B ') is a skew symmetric matrix. 3 3 0 5 4 Now, P Q 3 3 0 3 3 4 B 3 5 3 3 3 0 Thus, B is represented as the sum of a symmetric and a skew symmetric matrix. 6. Express the following matrices as the sum of a symmetric and a skew symmetric matrix: 6 3 3 3 5 5 ( i) ( ii) 3 ( iii) ( ) iv 3 4 5 (i) 3 5 Let A, then A P Q where, P ( A A') and Q ( A A') 3 5 3 6 6 3 3 Now, P ( A A') 5 6 3 3 3 3 3 P ' 3 P 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 6 -
Thus P ( A A ') is a symmetric matrix. 3 5 3 0 4 0 Now, Q ( A A') 5 4 0 0 0 Q ' 0 Q Thus Q ( A A ') is a skew symmetric matrix. Representing A as the sum of P and Q, 3 3 0 3 5 P Q 3 A 0 (ii) 6 Let A 3, then A P Q 3 where, P ( A A') and Q ( A A') 6 6 4 4 6 Now, P ( A A') 3 3 4 6 3 3 3 4 6 3 6 6 P ' 3 3 P 3 3 Thus P ( A A ') is a symmetric matrix. 6 6 0 0 0 0 0 0 Now, Q ( A A') 3 3 0 0 0 0 0 0 3 3 0 0 0 0 0 0 0 0 0 Q ' 0 0 0 Q 0 0 0 Thus Q ( A A ') is a skew symmetric matrix. Representing A as the sum of P and Q, 6 0 0 0 6 P Q 3 0 0 0 3 A 3 0 0 0 3 (iii) 3 3 Let A, then A P Q 4 5 where, P ( A A') and Q ( A A') Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 63 -
3 5 3 3 3 4 6 5 Now, P ( A A') 3 5 4 4 4 5 5 4 4 5 3 5 3 5 P ' P 5 5 Thus P ( A A ') is a symmetric matrix. 0 5 3 3 3 3 4 0 5 3 Now, Q ( A A') 3 5 5 0 6 5 0 4 5 3 6 0 3 0 0 5 3 Q ' 5 0 Q 3 0 Thus Q ( A A ') is a skew symmetric matrix. Representing A as the sum of P and Q, 3 5 0 5 3 3 3 P Q 5 0 A 5 3 4 5 0 (iv) 5 Let A, then A P Q where, P ( A A') and Q ( A A') 5 4 Now, P ( A A') 5 4 4 P ' P Thus P ( A A ') is a symmetric matrix. 5 0 6 0 3 Now, Q ( A A') 5 6 0 3 0 0 3 Q ' 3 0 Q Thus Q ( A A ') is a skew symmetric matrix. Representing A as the sum of P and Q, Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 64 -
0 3 5 P Q A 3 0 cos sin 7. If A = sin cos and A + A = I, find the value of α. cos sin cos sin A A sin cos sin cos Now, A A' I cos sin cos sin 0 sin cos sin cos 0 cos 0 0 0 cos 0 Comparing the corresponding elements of the above matrices, we have cos cos cos 3 3 8. Obtain the inverse of the following matrix using elementary operations: 0 0 0 Write A = I A, i.e., 3 0 0 A 3 0 0 3 0 0 0 0 0 A (applying R R ) 3 0 0 3 0 0 0 0 0 A (applying R 3 R 3 3R ) 0 5 8 0 3 0 0 0 0 0 A (applying R R R ) 0 5 8 0 3 0 0 0 0 0 A (applying R 3 R 3 + 5R ) 0 0 5 3 0 0 0 0 0 A (applying R 3 0 0 5 3 R 3) 0 A 3 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 65 -
0 0 0 0 0 A 0 0 5 3 (applying R R + R 3 ) 0 0 0 0 4 3 A 0 0 5 3 (applying R R R 3 ) A 4 3 5 3 9. Using elementary transformations, find the inverse of 6 3, if it exists. 6 3 Let A We know that A = IA 6 3 0 0 A 0 6 A Using R R 6 0 0 6 A Using R R R 0 0 3 Now, in the above equation, we can see all the elements are zero in the second row of the matrix on the LHS. Therefore, A does not exist. Note Suppose A = IA, after applying the elementary transformation, if any row or column of a matrix on LHS is zero, then A does not exist. 0. Show that the matrix B AB is symmetric or skew-symmetric according as A is symmetric or skew-symmetric. We suppose that A is a symmetric matrix, then A = A Consider (B AB) ={B (AB)} = (AB) (B ) [ (AB) = B A ] = B A (B) [ (B ) = B] = B (A B) = B (AB) [ A = A] (B AB) = B AB which shows that B AB is a symmetric matrix. Now, we suppose that A is a skew-symmetric matrix. Then, A = A Consider (B AB) = [B (AB)] = (AB) (B ) [(AB) = B A and (A ) = A] Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 66 -
= (B A )B = B ( A)B= B AB [ A = A] (B AB) = B AB which shows that B AB is a skew-symmetric matrix.. If A and B are symmetric matrices, prove that AB BA is a skew-symmetric matrix. Here, A and Bare symmetric matrices, then A = A and B = B Now, (AB BA) = (AB) (BA) ((A B) = A B and (AB) = B A ) = B A A B = BA AB ( B = Band A = A) = (AB BA) (AB BA) = (AB BA) Thus, (AB BA) is a skew-symmetric matrix. 3 3. Using elementary transformations, find the inverse of 3, if it exists. 3 3 3 Let A 3. We know that A = IA 3 3 3 0 0 3 0 0 A 3 0 0 4 3 0 0 A 3 0 0 (Using R R + R R 3 ) 4 0 0 5 A 0 5 0 3 3 4 (Using R R R and R 3 R 3 3R ) 4 0 5 0 3 3 4 A 0 0 5 (Using R R 3 ) 4 3 3 4 0 A (Using R R and R3 R 3 ) 5 5 5 5 5 0 0 5 5 5 3 3 0 5 5 5 0 0 0 A 5 5 0 0 5 5 5 (Using R R R 3 and R R 4R 3 ) Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 67 -
3 0 0 0 5 5 0 0 0 A (Using R 5 5 R R ) 0 0 5 5 5 3 0 5 5 0 3 A 0 0 5 5 5 5 5 5 3 4 n n 4n 3. If A, then prove that A n n, where n is any positive integer. We are required to prove that for all n N n 4n P( n) n n () 4() 3 4 Let n =, then P()...( i) () which is true for n =. Let the result be true for n = k. k k 4k P( k) A...( ii) k k Let n = k + k ( k ) 4( k ) k 3 4k 4 P( k ) A k ( k ) k k k k k 4k 3 4 Now, LHS A A A k k ( k).3 ( 4 k). ( k).( 4) ( 4 k)( ) k.3 ( k). k.( 4) ( k)( ) 3 k 4 4k ( k ) 4( k ) k k k ( k ) Therefore, the result is true for n = k + whenever it is true for n = k. So, by principle of mathematical induction, it is true for all n N. 0 0 0 O? 0 x 0 0 0 O 0 x 4. For what values of x : Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 68 -
0 4 0 Since Matrix multiplication is associative, therefore 0 0 x O 0 0 x 4 x O x 4 x x O 4 4x O 4 4x 0 4x 4 x 3 5. If, show that A 5A + 7I = 0. 3 Given that A 3 3 8 5 A. 5 3 8 5 3 0 A 5A 7I 5 7 5 3 0 8 5 5 5 7 0 5 3 5 0 0 7 8 5 7 5 5 0 0 0 O 5 5 0 30 7 0 0 6. Find x, if x 5 0 x 0 4 O? 0 3 Given that x 5 0 x 0 4 O 0 3 Since Matrix multiplication is associative, therefore x 5 x 0 0 8 O x 0 3 x 5 x 9 O x 3 x( x ) ( 5).9 ( )( x 3) O x 48 O x 48 0 x 48 x 48 4 3 3 7 8 3 7. Find the matrix X so that X 4 5 6 4 6 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 69 -
3 7 8 3 Given that X 4 5 6 4 6 The matrix given on the RHS of the equation is a 3matrix and the one given on the LHS of the equation is as a 3 matrix. Therefore, X has to be a matrix. a c Now, let X b d a 4c a 5c 3a 6c 7 8 3 b 4d b 5d 3b 6d 4 6 Equating the corresponding elements of the two matrices, we have a + 4c = 7, a + 5c = 8, 3a + 6c = 9 b + 4d =, b + 5d = 4, 3b + 6d = 6 Now, a + 4c = 7 a = 7 4c a + 5c = 8 4 8c + 5c = 8 3c = 6 c = a = 7 4( ) = 7 + 8 = Now, b + 4d = b = 4d and b + 5d = 4 4 8d + 5d = 4 3d = 0 d = 0 b = 4(0) = Thus, a =, b =, c =, d = 0 Hence, the required matrix X is 0 cos sin n cos n sin n 8. If A sin cos, then prove that A, n N sin n cos n We shall prove the result by using principle of mathematical induction. cos sin n cos n sin n We have P(n) : If A sin cos, then A, n N sin n cos n cos sin Let n =, then P() A sin cos Therefore, the result is true for n =. Let the result be true for n = k. So k cos k sin k P( k) A sin k cos k Now, we prove that the result holds for n = k + k cos( k ) sin( k ) i.e. P( k ) A sin( k ) cos( k ) k k cos k sin k cos sin Now, P( k ) A A. A sin k cos k sin cos cos cos k sin sin k cos sin k sin cos k sin cos k cos sin k sin sin k cos cosk cos( k ) sin( k ) cos( k ) sin( k ) sin( k ) cos( k ) sin( k ) cos( k ) Therefore, the result is true for n = k +. Thus by principle of mathematical induction, we have n cos n sin n A, n N sin n cos n Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 70 -
5 5 9. Let A, B, C 3 4 7 4 3 8. Find a matrix D such that CD AB = O. Since A, B, C are all square matrices of order, and CD AB is well defined, D must be a square matrix of order. a b 5 a b 5 Let D, c d then CD AB 0 O 3 8 c d 3 4 7 4 a 5c b 5d 3 0 0 0 3a 8c 3b 8d 43 0 0 a 5c 3 b 5d 0 0 3a 8c 43 3b 8d 0 0 By equality of matrices, we get a + 5c 3 = 0... () 3a + 8c 43 = 0... () b + 5d = 0... (3) and 3b + 8d = 0... (4) Solving () and (), we get a = 9, c = 77. Solving (3) and (4), we get b = 0, d = 44. a b 9 0 Therefore D c d 77 44 30. By using elementary operations, find the inverse of the matrix A In order to use elementary row operations we may write A = IA. 0 0 A, then A 0 0 5 (applying R R R ) 0 A 0 (applying R R ) 5 5 5 0 5 5 A 0 (applying R R R ) 5 5 Thus A 5 5 = 5 5 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 7 -
CHAPTER 3: MATRICES Previous Years Board Exam (Important Questions & Answers). Use elementary column operation C C C in the matrix equation 4 0 3 3 0 3 4 0 Given that 3 3 0 3 Applying C C C, we get 4 6 4 3 3 0 3 a 4 3b a b. If 8 6 8 a 8b write the value of a b. a 4 3b a b Give that 8 6 8 a 8b On equating, we get a + 4 = a +, 3b = b +, a 8b = 6 a =, b = Now the value of a b = ( ) = - = 0 MARKS WEIGHTAGE 03 marks 3. If A is a square matrix such that A = A, then write the value of 7A (I + A) 3, where I is an identity matrix. 7A (I + A) 3 = 7A - {I 3 + 3I A + 3I.A + A 3 } = 7A {I + 3A + 3A + A A} [ I 3 = I = I, A = A] = 7A {I + 6A + A } = 7A {I + 6A + A} = 7A {I + 7A} = 7A I 7A = I x y z 4 4. If x y w 0 5, find the value of x + y. x y z 4 Given that x y w 0 5 Equating, we get x - y = - (i) x - y = 0 (ii) z = 4, w = 5 (ii) (i) x y x + y = 0 + x = and y = x + y = + = 3. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 7 -
5. Solve the following matrix equation for x : x 0 Given that x O 0 x 0 0 0 x = 0 x = 0 O 0 3 4 7 0 6. If y 5 x 0 0 5, find (x y). 3 4 7 0 Given that y 5 x 0 0 5 6 8 y 7 0 0 x 0 0 5 7 8 y 7 0 0 x 0 5 Equating we get 8 + y = 0 and x + = 5 y = 8 and x = x y = + 8 = 0 0 7. For what value of x, is the matrix A 0 3 a skew-symmetric matrix? x 3 0 A will be skew symmetric matrix if A = A' 0 0 x 0 x 0 3 0 3 0 3 x 3 0 3 0 3 0 Equating, we get x = 8. If matrix A and A = ka, then write the value of k. Given A = ka k k k k = Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 73 -
a b a c 5 9. Find the value of a if a b 3c d 0 3 a b a c 5 Given that a b 3c d 0 3 Equating the corresponding elements we get. a b = (i) a + c = 5 (ii) a b = 0 (iii) 3c + d =3 (iv) From (iii) a = b b a b Putting in (i) we get b b b a = (ii) c =5 - =5 - = 3 (iv) d =3 3 (3) =3 9 = 4 i.e. a =, b =, c = 3, d = 4 9 4 0. If A 3 0 4 9, then find the matrix A. 9 4 Given that A 3 0 4 9 9 4 A 3 0 4 9 8 3 5 A 3 6. If A is a square matrix such that A = A, then write the value of (I + A) 3A. (I + A) 3A = I + A + A 3A = I + A - A = I + A - A [ A = A] = I = I. I = I 0. If x y 3 5, write the value of x. 0 Given that x y 3 5 x y 0 3x y 5 x y 0 3x y 5 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 74 -
Equating the corresponding elements we get. x y = 0...(i) 3x + y = 5...(ii) Adding (i) and (ii), we get x y + 3x + y = 0 + 5 5x = 5 x = 3. 5 3 4 7 6 3. Find the value of x + y from the following equation: x 7 y 3 5 4 5 3 4 7 6 Given that x 7 y 3 5 4 x 0 3 4 7 6 4 y 6 5 4 x 3 6 7 6 5 y 4 5 4 Equating the corresponding element we get x + 3 = 7 and y 4 = 4 7 3 4 4 x and y x = and y = 9 x + y = + 9 = 3 4 T 4. If A and B 3, then find A T B T. 0 T Given that B 3 B 3 3 4 4 3 T T Now, A B 3 0 0 3 3 3 4 6 5. If 5 7 4 9 x, write the value of x 3 3 4 6 Given that 5 7 4 9 x 6 6 4 6 5 4 5 8 9 x 4 6 4 6 9 3 9 x Equating the corresponding elements, we get x = 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 75 -
cos sin sin cos 6. Simplify: cos sin sin cos cos sin cos sin sin cos cos sin sin cos cos sin cos sin cos sin sin cos sin cos cos sin cos sin cos sin 0 0 0 cos sin 0 7. Write the values of x y + z from the following equation: By definition of equality of matrices, we have x + y + z = 9... (i) x + z =5... (ii) y + z =7... (iii) (i) (ii) x + y + z x z = 9 5 y = 4... (iv) (ii) (iv) x y + z = 5 4 x y + z = x y z 9 x z 5 y z 7 y x 5 7 5 8. If x 3 3, then find the value of y. y x 5 7 5 Given that x 3 3 By definition of equality of matrices, we have y + x = 7 - x = - x = y + () = 7 y = 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 76 -