35113923_BCA114

146 Mathematics where, x1 = Sum of minor's of order one along main diagonal of A x1 = a11 + a22 + a33 x2 = Sum of minors of order two of the diagonal element of A x2 = Minor of a11 + minor of a22 + minor of a33 x2 = a 22 a 23 + a11 a13 + a11 a12 a32 a33 a31 a33 a12 a 22 A = a11(a22a33 – a33a32)– a12(a21a33 – a23a31)+ a13(a21a32 – a22a31) We have came across following three type of problem's. Type I: When all eigen values are distinct and matrix A may be symmetric or non-symmetric. Type II: When eigen values are repeated and A is non-symmetric. Type III: When eigen values are repeated and A is symmetric. Note: Crammer's Rule will be used quite often. It is given as follows: Two homogeneous equation in three variable x1, x2, x3 can be solved as follows: Let, a1x1 + a2x2 + a3x3 = 0 then, or, b1x1 + b2x2 + b3x3 = 0 x1 =  x2 = x3 a 2 b3  a3b2 a1b3  a3b1 a1b2  a 2 b1 x1 =  x2 = x3 a 2 b3  a3b2 a3b1  a1b3 a1b2  a 2 b1 Type I: All Roots are Non-repeated Matrix is Symmetric or Non-symmetric Exercise 6: Find Eigen values and Eigen vectors for CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 4 147 2  2 3  A = 1 1 1  1 3  1 Solution: Step I: Characteristic equation of A in  is A  I = 0 2    2 3  i.e.,  1 1  1  = 0    1 3  1   Instead of evaluating the determinant directly, we use the formula for its expansion which is as follows: 3 – (sum of diagonal element of A)2 + (sum of minors of diagonal element A);  – A = 0; Sum of diagonal element of A = 2 + 1 + (–1) = 2 Now, diagonal elements are 2, 1, –1  Minor of 2 = (–1 – 3) = –4  Minor of 1 = (–2 – 3) = –5  Minor of 1 = (2 + 2) = 4  Sum of minor diagonal elements = (–4) + (–5) + (4) = –5 And, det (A) = A = 2(–4) + 2(–2) + 3(2) = –6  Characteristic equation is 3 – 22 – 5 + 6 = 0 Since, sum of co-efficient of all terms = 0  ( – I) is a factor, of L.H.S. To find other factors, use synthetic division CU IDOL SELF LEARNING MATERIAL (SLM)

148 Mathematics 1 1 2 5 6 1 1  6 1 1  6 0  ( – 1)(2 –  – 6) = 0  ( – 1)( – 3)( + 2) = 0   = 1, –2 and 3 are the roots (Roots are not-repeated) Step II: Now, we find Eigen vectors for this we use matrix equation of A is . Matrix equation is (A – I)x = 0 2   2 3  x1  0 1 1  i.e.,  1  x  = 0   2  1 3  1  x3  0  Case I: When,  = 1 matrix equation is obtained by putting  = 1 in equation (1) 11 2 3  x1  0 0  x  0  1   = 2 1 3  2 x3  0  Expanding from R1 and R2, we have x1 – 2x2 + 3x3 = 0 x1 + 0x2 + x3 = 0 Solving these two equations by Cramer's rule, we get x1 =  x2 = x3 2 2 2 i.e., x1 = x2 = x3 1 1 1 CU IDOL SELF LEARNING MATERIAL (SLM)

    149 Matrices 4   11or 1    x  x =  1  1 1  1   1  Case II: When 2 = –2 Matrix equation is obtained by putting  = –2 in Equation (1) 41 2 3 x1  0  3 1 x  0  = 2 1 3 1 x3  0  To apply Cramer's rule, we need not write the two equation by expanding by R1 and R2. We apply Cramer's rule directly to the entries of R1 and R2. Applying Cramer's rule to R1 and R2, we get  x1 =  x 2 = x3  2  9 4  3 12  2 i.e. x1 = x 2 = x3 11 1 14  x  111 or x = 111 2   2    4   4   Case III: When, 3 = 3 Matrix equation is obtained by putting  = 3 in equation (1), 11 2 3  x1  0 2  x  0  1   = 2  1 3  4 x3  0  Apply Cramer's rule to R1 and R2, we get  x1 =  x2 = x3 4 4 4  x1 = x 2 = x3 111 CU IDOL SELF LEARNING MATERIAL (SLM)

150 Mathematics  1  11or x3 1  x3  =  1 1  2  2 3  Exercise 7: Find the Eigen values and Eigen vectors of the following matrix: A = 1 1 1  1 3  1  Solution: The characteristic equation is A  I =0 2    2 3    1 1  1  = 0    1 3  1   which gives 3 – 22 – 5 + 6 = 0. Solving, we get 1 = 1, 2 = –2, 3 = 3 All roots are distinct. The eigen vector is A  Ix = 0  For 1 = 1 11 2 3  x1  0 0  x  0  1   = 2 1 3  2 x3  0  x1 – 2x2 + 3x3 = 0, x1 + x3 = 0, x1 + 3x2 – 2x3 = 0 Using first two equations x1 =  x 2 = x3  2 3 1 3 1 2 0 1 11 1 0   x 1 =  x 2 = x3  2 2 2 CU IDOL SELF LEARNING MATERIAL (SLM)

 151  Matrices 4  x1 = x2 = x3 = K 1 1 1  x1 = –K, x2 = x3 = K For K = 1, x1 = [–1, 1, 1]T For K = 2, x2 = [–2, 2, 2]T For 2 = –2, 14 2 3 x1  0  3 1 x  0  = 2 1 3 1 x3  0   4x1 – 2x2 + 3x3 = 0, x1 + 3x2 + x3 = 0, x1 + 3x2 + x3 = 0 x1 =  x 2 = x3 2 3 4 3 4 2 31 11 1 3  x1 =  x2 = x3 = K 11 1 14  x1 = –11K, x2 = –K, x3 = 14K For K = 1, x1 = [–11, –1, 14]T For K = 2, x2 = [–22, –2, 28]T Similarly, for  = 3, we get For K = 1, x1 = [1, 1, 1]T For K = 3, x2 = [3, 3, 3]T For different values of K, we get different vectors. CU IDOL SELF LEARNING MATERIAL (SLM)

152 Mathematics Exercise 8: Find the Eigen values and Eigen vectors of the following matrix: (I) 14 10 A = 5 1  Solution: The characteristic equation of A is A  I 14   10 = 0, i.e., =0 5 1  (14 – )(–1 – ) + 50 = 0 or 2 – 13 + 36 = 0 or ( – 9)( – 4) = 0 or 2 – s  + A = 0, where, s = 14 – 1 = 13 and A = –14 + 50 = 36 11 2 – 13 + 36 = 0 Hence,  = 9,  = 4 are the required Eigen values of A. To find Eigen vectors, consider: (A – I) x = 0  14  λ 10 x = 0 5 1  λ y 0  For  = 9 in Equation (I):  5 10 x 0 5 10 y = 0   R–R 5 10 x 0 2 10 0 y = 0   5x – 10y = 0  5x = 10y  x = 2y  2 x1 = 1  Putting = 4 in Equation (I): CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 4 153  10 10 x = 0 5 5 y 0  R – 2R 0 1 25 0 x 0  5 y = 0   5x – 5 = 0  x = y 1  x2 = 1  Type II: Repeated Eigen Value and A is Non-symmetric Exercise 9: Find Eigen values and given vector for 2 1 1 A = 2 3 2 3 3 4  Solution: Characteristic equation of A in  is Step I: A  I = 0 2   1 1  i.e.,  2 3  2  = 0    3 3 4    i.e., 3 – (9)2 + (6 + 5 + 4) – 7 = 0 2 1 1 A = 2 3 2 = 0 3 3 4  i.e., 3 – 92 + 15 – 7 = 0 Since, the sum of co-efficient of all terms = (1) + (–9) + (15) + (–7) = 0  ( – 1) is a factor of L.H.S. CU IDOL SELF LEARNING MATERIAL (SLM)

154 Mathematics We use Synthetic division 1 1 9 15  7 1 8 7 18 7 0  ( – 1)(2 – 8 + 7) = 0 i.e., ( – 1)( – 7) ( – 1) = 0  1 = 7, 2 = 1, 3 = 1 (Hence, 2 root's are repeated first, we find Eigen vector for non-repeated root) Step II: Matrix equation of A in  is (A – I) x = 0 2  1 1  x1  0 2 3    x  0 i.e.,  2   = 2  3 3 4   x3  0  Case I: For 1 = 7 matrix equation is 25 1 1  x1  0 4  x  0  2   = 2  3 3  3 x3  0 Applying Cramer's rule to R1 and R2,we get  x1 =  x 2 = x3 6 12 18 i.e., x1 = x 2 = x3 1 2 18  x = 12 1   3 CU IDOL SELF LEARNING MATERIAL (SLM)

 155  Matrices 4 Case II:Let, 2 = 1, matrix equations is obtained by putting  = 1 in Equation (1), 21 1 1 x1  0  2 2 x  0  = 2 3 3 3 x3  0 When the Eigen values are repeated, the system gives essentially a single homogeneous equation and Cramer's rule is not applicable. In such cases, we choose convenient values to two variables and find the value of the third variable form the single equation. Expanding by R1, we have x1 + x2 + x3 = 0 We assume any one element to be zero say x1 and give any convenient value say 1 to x2 and find Let, x1 = 0; x2 = 1;  x3 = – 1  0  x2   1   1 Case III: Let 3 = 3, (same matrix equation) Again, consider x1 + x2 + x3 = 0 Now, let x2 = 0; x1 = 1 and x3 = – 1  x   1  0 3    1  6  2 2  Exercise 10: A =  2 3 1  2 1 3  CU IDOL SELF LEARNING MATERIAL (SLM)

 Mathematics  156 Solution: The characteristic matrix is [A – I] = 1    2  5 4      The characteristic equation is A  I = 0  1 λ  2 4  λ = 0  5    (1 – )(4 – ) – (–2) (–5) = 0  2 – 5 – 6 = 0  ( – 6)( + 1) = 0  The characteristic values of matrix A are 1 = 6 and 2 = – 1. The characteristic vector x has satisfy the equation [A – I] x = 0, which 1  λ 4  2 xx1  = 0  5   2 0    For, 1 = 6   5  22 xx1  = 00 5     2     which gives one equation –5x1 – 2x2 = 0  – 5x1 = 2x2 = K  Proportional values are x1 = x2 = K 2 5  x1 = 2K, x2 = –5K For K = 1, x1 = 2, x2 = –5 x1 = [2, 5]T For K = 2, x1 = 4, x2 = –10 x2 = [4, –10]T CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 4 157 For different values of K, we will get different vectors. For  = –1,  2  2 x1  = 0 5 5 x  0 2    2x1 – 2x2 = 0  x1 = x2  –5x1 + 5x2 = 0  x1 = x2 This gives only I equation  x1 = x 2 = K  x = x = K 11 12 For K = 1, x = x = 1  x = [1, 1]T 12 1 For K = 3, x = x = 3  x = [3, 3]T 12 2 Exercise 11: Find the Eigen values and the corresponding Eigen vectors for the following matrix 1  6  4 A = 0 4 2  0  6   3  Solution: The characteristic equation A  λI = 0 1   6  4 i.e., 0 4   2 = 0 0  6  3    3s 2 + s  – A = 0, where s = 1 + 4 – 3 = 2 12 1 4 2 1 4 1 6 s2 =  6 + + 3 0 3 0 4 =0 – 3 + 4 =1 CU IDOL SELF LEARNING MATERIAL (SLM)

158 Mathematics A = 1(0)=0 Hence, the characteristic equation of A is 3 – 22 +  = 0 …. (I) For the factors of Equation (I), we have 3 – 22 +  = 0 (2 – 2 + 1) = 0 ( – 1)2 = 0 Hence,  = 0,  = 1, 1 are the required Eigen value of A Note that, 1 + 2 + 3 = a11 + a22 + a33 i.e. 0 + 1 + 1 = 1 + 4 – 3 Again, denoting the non-repeated Eigen value by 1 = 0 and 2 = 3 = 1 To find the Eigen vectors for the corresponding Eigen values, we will consider the matrix equation (A – I) x = 0 xy 1   6  4  x 0 0 4   where, x = i.e. 2  y = 0 … (II)    z  0  6  3   z 0  For  =  = 0, let the corresponding Eigen vector be x  yx Put,  =  = 0, 1  1  z    The matrix equation (II) 10 6  4 x 0  4 2  y = 0  0  6  3 z 0  AX = Z, which is homogeneous system of equation consider the first two equations. CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 4 159 x – 6y – 4z = 0 0x + 4y + 2z = 0 Solving by Cramer's rule x = y = z 6 4 1 4 1 6 42 02 04 x = y = z, x = y = z =t 4  2 4 2 1 2 x = 2t, y = – t, z = 2t For  = 0 = x = 21 1 1    2  For  = 2 = 1, let the corresponding Eigen vector be x = yx Put  =  = 1, 2  2  z    The matrix equation II will become 00 6  4 x 0  3 2 y = 0 0  6  4 z 0   2t 3 By R2 0x + 3y + 2z = 0, let z = t, y = i.e., z = 3t, y = –2t and x can take any value, say x = t  For  = 1, x =  21 2   2  3 CU IDOL SELF LEARNING MATERIAL (SLM)

 Mathematics  160 For  = 3 = 1 = 2, we arrive at same question 0x + 3y + 2z = 0 Let z = t y =  2 t 3 i.e., z = 3t, y = – 2t Again, the matrix is non-symmetric.  The corresponding Eigen vector x2 and x3 must be linearly independent. This can be done by choosing x = 2t, y = – 2t, z = 3t  x = 22 3    For  = 1, 3  3   Type III: A is Symmetric and Eigen Values are Repeated Before we proceed, we define: Definition: Let x and y be two column matrix, then x and y are said to be orthogonal it x'y = 0. Let x = ab1  and y = ba 2   1   2  c1  c2   We known that x and y are orthogonal x'y = 0 i.e., a1 b1 c1  ab22  = 0   c2   i.e., [a1a2 + b1b2 + c1c2] = 0 i.e., a1a2 + b1b2 + c1c2 = 0 CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 4 161 Exercise 12: Find the Eigen values and Eigen vectors of the following matrix  2 5 4  A=  5 7 5     4 5  2  Solution: Here, note that ‘A' is a symmetric matrix; Eigen vectors x1, x2, x3 are orthogonal to each other. The characteristic equation A is A  I = 0.  2   5 4  5 7   5 = 0   4 5  2    3s 2 + s  – A = 0, where s = – 2 + 7 – 2 = 3 12 1 s2 = 7 5 2 4 2 5 = 39 – 12 – 39 = – 90 5 + 4 + 5 7 2 2 A = – 2(– 39) – 5(– 30) + 4(– 3) = 78 + 150 – 12 = 216 Hence, the characteristic equation of A is 3 – 32 – 90 – 216 = 0 For the factor of 3 – 32 – 90 – 216 = 0, try  =  1,  2,  3.  = – 3 satisfies Equation (I)  We divide by  = – 3. Hence, using synthetic division,  3 1 3  90  216  3 18 216 1 6  72 0 ( + 3)(2 – 6 – 72) = 0 ( + 3)( – 72)( + 6) = 0 CU IDOL SELF LEARNING MATERIAL (SLM)

162 Mathematics  = – 3, – 6, 12 are Eigen values of A. Let 1 = – 3, 2 = – 6, 3 = 12 are Eigen values. To find the Eigen vector, we will consider the matrix equation (A – I) x = 0 xy  2   5 4 x 0 5 7   5 y = 0 where, x = i.e.   5  2   z 0 z  4 – 3, x  =  y For = let the corresponding Eigen vector be x = 2 1  z  Put  = 1 = – 3  The matrix Equation (II) will become 51 5 4 x 0  10 5 y = 0 1 z 0 4 5 AX = Z, which is homogeneous system to equation. Consider the first two equations. x + 5y + 4z = 0 5x + 10y + 5z = 0 By using Cramer's rule, x =  y = z 54 14 1 5 10 5 5 5 5 10 x = y = z , x = y = z = t i.e. x = t, y = – t, z = t 15 15 15 1 1 1 CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 4 163 Hence, for  = – 3, the Eigen vector is x = 11 1 1    1    x For  =  =– 6, let the corresponding Eigen vector be x y = 1 2 1  z  Put,  = 2 = – 6  The matrix equation (2) will be become  54 5 4 x 0 AX = Z  13 5 y = 0 4 5 4 z 0  which is homogeneous system to equation. Consider the first two equations. 4x + 5y + 4z = 0 5x + 13y + 5z = 0 By Cramer's rule x =  y = z 54 44 4 5 13 5 5 5 5 13 x =y= z, x =y=z  27 0 27 1 0 1 – 6, 1   Hence, for  = the Eigen vector is x =  0  2 2  1    x For  =  = 12, let the corresponding Eigen vector be x y = 3 3  z  CU IDOL SELF LEARNING MATERIAL (SLM)

 Mathematics  164 Put,  = 3 = 12  The matrix equation (II) will be become 154 5 4  yx = 00AX = Z 5 5        4 5  14 z 0  Which is homogeneous system to equation. Consider the first two equation: – 14x + 5y + 4z = 0 5x – 5y + 5z = 0 By Cramer's rule, x = y = z 54 14 4 14 5 1 1 1 1 5 1 x= y =z 9 18 9 x= y = z=t 121 i.e., x = t, y = 2t, z = t  1 2 For  = 12, x = 3 3 1  Check whether x1, x2 and x3 are orthogonal to each other. x1 x = 0 i.e., (1–11) 01 = 0   22  1  CU IDOL SELF LEARNING MATERIAL (SLM)

 165  Matrices 4 x1 x = 0 i.e., (–101) 21 = 0   23 1   ( x1 × 1) = 0 i.e., (121) 11 = 0 3    1   Hence, x1, x2 and x3 are orthogonal. Exercise 13: Find the characteristic roots and characteristic vectors of A where 2 2 1 A = 1 3 1 1 2 2  Solution: Step I: Characteristic equation of A in  is A  I = 0 2   2 1 i.e.,  1 3   1 = 0   1 2 2    i.e., 3 – (Sum of diagonal element of A)2 + (Sum of minor of diagonal element A) –A =0  3 – (7)2 + (4 + 3 + 4) – (5) = 0  3 – 72 + 11 – 5 = 0  ( – 1)(2 – 6 + 5) = 0  ( – 1)( – 5)( – 1) = 0 [ A = 2(4) – 2(1) + 1(–1) = 5]  The roots are 5, 1 and 1. CU IDOL SELF LEARNING MATERIAL (SLM)

166 Mathematics Here, two roots are required and matrix A is non symmetric. This problem is of type (II). Step II: The matrix equation of A in  is A  I x = 0 … (1) 2   2 1 x1  0 1 3   i.e.,  1 x  = 0  2  1 2 2   x3  0  Case I: Let 1 = 5  Matrix equation is  3 2 1 x1  0 1 2 1 x 2  0   =  1 2  3 x3  0  From R1 and R2 using Cramer's rule, we get, x1 =  x2 = x3 4 4 4 1  x1 = 1 1  Case II: Let, 2 = 1  Matrix equation is 11 2 1 x1  0 2 1 x  0   = 2 1 2 1 x3  0  Expanding by R1, x1 + 2x2 + x3 = 0 Let, x = 0 and x = 1  x = –2  0     x =  1  12 3 2  2 CU IDOL SELF LEARNING MATERIAL (SLM)

 167  Matrices 4 Case III: Let, 3 = 1 Here, we have the same matrix equation Again, x1 + 2x2 + x3 = 0 Let x = 0 and x = 1  x = –1  0     x =  1  21 3 3   1  Exercise 14: Find the Eigen values and Eigen vector of   2 5 4  A=  5 7 5     4 5  2  Solution: Note: A is a symmetric matrix. Hence, eigen vectors are orthogonal to each other. Step I: The characteristic equation of A in  is A  I = 0  2   5 4  i.e.,  5 7 5  = 0    4 5  2     3 – (– 2 + 7 – 2)2 + (– 39 – 12 – 39) – (216) = 0  3 – 3 – 90 – 216 = 0 is the characteristic equation of A in . Note that  = – 3 is the root of this equation. To find other roots, we use synthetic division  3 1  3  90  216  3 18 216 1  6  72 0  ( + 3)(2 – 6 – 72) = 0  ( + 3)( – 12)( + 6) = 0 CU IDOL SELF LEARNING MATERIAL (SLM)

168 Mathematics  1 = – 3, 2 = – 6, 3 = 12 are the Eigen values of A. (1) (Note that all Eigen values are distinct) Step II: The matrix equation of A in  is A  I x = 0  2   5 4 x1  0 5 7    5 x  = 0 2    4 5  2   x3  0 Case I: Let, 1 = – 3, the matrix equation is 15 5 4 x1  0 10 5 x  0   = 2 4 5 1 x3  0  By Cramer's rule, x1 =  x 2 = x3 15 15 15 x1 = x 2 = x3 1 1 1  x = 11 1    1  Case II: For, 2 = – 6, the matrix equation is 45 5 4 x1  0 13 5 x  0   = 2 4 5 4 x3  0  By Cramer's rule, x1 =  x 2 = x3  27 0 27 CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 4 169 x1 = x 2 = x3 1 0 1  x =  1  0 2    1 Case III: Let, 3 = 12, the matrix equation is, 145 5 4 x1  0 5  5 x  = 0 2   4 5  14 x3  0  By Cramer's rule, x1 =  x 2 = x3 45  90 45 x1 = x 2 = x3 121  x = 12 3   1 Thus, x1, x2 and x3 are the Eigen vectors corresponding to 1 = – 3, 2 = – 6 and 3 = 12 respectively. 1  6  4 Exercise 15: A = 0 4 2   0  6 3  Solution: The characteristic matrix is [A – I] = 1   2  5 4      Then characteristic equation is CU IDOL SELF LEARNING MATERIAL (SLM)

170 Mathematics [A – I] = 0  1  2 =0   5 4      (1 – )(4 – ) – (–2)(–5) = 0  2 – 5 – 6 = 0  ( – 6)( + 1) = 0  The characteristic values of matrix A are 1 = 6 and 2 = –1. The characteristic vector x has to satisfy this equation [A – I] x = 0 which is 15 4  2 xx1  = 00      2        2 For  = 6,  2 1  5 x1  = 0  5  x  0   2  which gives one equation –5x1 – 2x2 = 0 – 5x1 = 2x2 = K Proportional values are x1 = x2 = K 2 5 x1 = 2K, x2 = – 5K For K = 1, x1 = 2 and x2 = –5  x 1= [2, –5]T For K = 2, x1 = 4and x2 = –10  x2= [4, –10]T For different values of K, we will get different vectors For 2 = –1, 25  2 x1  = 0   x  0 5  2  2x1 – 2x2 = 0  x1 = x2  –5x1 + 5x2 = 0  x1 = x2. This gives only one equation. CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 4 171  x1 = x 2 = K  x =x =K 11 12 For K = 1, x = x = 1  x = [1 1]T 12 1 For K = 3, x = x = 3  x = [3 3]T 12 2 Note that: Eigen vectors are not unique. 11.7.7 Cayley-Hamilton Theorem This theorem is useful for calculation of different power of a square matrix and also to find inverse of non-singular matrix. If  is an eigen value and x an eigen vector of a square matrix, then AX = X Now, A(AX) = A(X) (AA)X = (AX)  A2X= 2X Similarly, A3X= 3X .. .. AnX = nX Thus, the effect of , 2, 3, …….., n on X is same as that of A, A2, A3, ……. An on X. This is precisely Cayley-Hamilton theorem. Exercise 16: Verify Cayley-Hamilton theorem for 1 2 2 A = 2 1 2 2 2 1 CU IDOL SELF LEARNING MATERIAL (SLM)

 Mathematics  172 Solution: The characteristic equation is A  I X= 0 1   2 2  2 1  2 = 0 2 2 1    3 – 32 – 9 – 5 = 0 Now, consider A3 – 3A2 – 9A – 5I A2 = 12 2 22 21 2 22 = 89 8 88 1 1 9       2 2 1 2 2 1 8 8 9  A3 = 98 8 88 12 2 22 = 4412 42 4422 9 1 41       8 8 9 2 2 1 42 42 41   A3 – 3A2 – 9A – 5I = 4412 42 4422 – 3 89 8 88 – 9 21 2 22 – 5 01 0 00 = 00 0 00 = 0 41 9 1 1 0           42 42 41 8 8 9 2 2 1 0 0 1 0 0 0  Hence, Cayley-Hamilton theorem is verified. Exercise 17: Verify Calyely-Hamilton and use it to find A4 & A5 when 1 0 1 A = 0 1 0 0 0 1 CU IDOL SELF LEARNING MATERIAL (SLM)

 173  (1) Matrices 4 Solution: Now, [A – I] x = 0 characteristic equation. 1  0 1   0 1  0 = 0   0 0 1    (1 – )3 = 0  1 – 3 + 32 – 3 = 0 Replacing  by A, we get A3 – 3A2 + 3A– I = 0 Now, 1 0 1 1 0 1 1 0 2 A2 = 0 1 0 0 1 0 = 0 1 0 0 0 1 0 0 1 0 0 1  1 0 1 1 0 2 1 0 3 A3 = A.A2 = 0 0 0  0 0 1 1 0  = 1 0 0 1 0 0 1 0 0 1  Now, A3 – 3A2 + 3A– I 1 0 3 1 0 2 1 0 1 1 0 0 0 1 0 – 3 0 1 0 + 3 0 1 0 – 0 1 0 0 0 1 0 0 1 0 0 1 0 0 1   1  3  3 1 0  0  0  0 3  6  3  0 0 0 0 = 0  0  0  0 1  3  3 1 0  0  0  0 = 0 0 0 = 0 0  0  0  0 0  0  0  0 1 3  3  1 0 0 0  Hence, Cayley-Hamilton theorem verified. Now, to find A4, multiply by A on both sides of equation (1), we get A(A3 – 3A2 + 3A – I) = A.0 CU IDOL SELF LEARNING MATERIAL (SLM)

174 Mathematics A4 – 3A3 – 3A2 – A = 0 A4 = 3A3 – 3A2 + A 1 0 3 1 0 2 1 0 1 = 3 0 1 0 – 3 0 1  0 0 1 0 0 0  + 1 1 0 0 1 0 0  3 0 9 3 0 6 1 0 1 = 0 3 0 – 0 3 0 + 0 1 0 0 0 3 0 0 3 0 0 1   3  3  1 0  0  0 9  6  1 = 0  0  0 3  3  1 0  0  0 0  0  0 0  0  0 3 31  1 0 4 A4 = 0 1 0 0 0 1  Now, A5 equation (1) multiply by A2 both sides, we get A2 (A3 – 3A2 + 3A– I) = A2.0 A5 – 3A4 + 3A3 – A2 = 0 A5 = 3A4 – 3A3 + A2 1 0 4 1 0 3 1 0 2 A5 = 3 0 1 0 – 3 0 1 0 + 0 1 0 1 0 0 1 0 0 1 0 0  331 0 0 0 12  9  2 1 0 5 A5 = 0  0  0 331 0  0  0  = 0 1 0 0 0 0 3 31  0 0 0  0  0 1 CU IDOL SELF LEARNING MATERIAL (SLM)

  Matrices 4  175 Exercise 18: Verify Cayley-Hamilton theorem and find A–1 and A–2 when A = 1 12. 1 Solution: Consider, A  I = 0 1  2 = 0  2 – 2 – I = 0  1 1     Now, A2 – 2A – I = 0 … (I) A2 = A.A = 1 2 1 2 3 4 1    =  1 2 1 1 3  A2 – 2A– I= 3  4  1 2  1 0       – 2 – 2 3 1 1 0 1  0 =0 =  3  2 1 4  4  0 = 0 2  2  0 3  2  1 0 0  Cayley-Hamilton theorem verified. Now, for A–1 multiply equation (I) by A–1,  A – 2I – A–1 = 0  A–1 = A – 2I … (II) … (III) Now, A–2 = A–1(A – 2I) = I – 2A–1 = I – 2(A – 2I) = 5I – 2A A–3 = A–1(5I– 2A)= 5A–1 – 2I = 5(A – 2I) – 2I = 5A – 12I A–3 = 5A – 12I Now, A–1 =A – 2I = 1 2 – 2 1 0 1 1 0 1 CU IDOL SELF LEARNING MATERIAL (SLM)

 Mathematics  … (1) 176 A–1 = 1 2 1  1   1 0 A–2 = I – 2A–1 = 1 – 211 21 0    A–2 = 3  4  2 3  A–3 = 5A– 12I =5 1 2 – 12 1 0 1 1 0 1  A–3 =  7 10  7  5   Exercise 19: Verify Cayley-Hamilton theorem and hence find A–1.  1 2  2 A = 1 3 0    0  2 1   Solution: Now, [A – I] x = 0 1  2  2    1 3 0 = 0   0  2 1    3 – 52 + 9 – I = 0 Now, put  = A A3 – 5A2 + 9A– I = 0 CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 4 177 Now,  1 2  2  1 2  2  1 12  4 A2 = 1 3 0 1 3 0 =  4 7 2  0  2 1  0  2 1  2  8 1   1 12  4  1 2  2 13 42  2 A3 = A2  A =  4 7 2 1 3 0 = 11 9 10  2  8 1  0  2 1  10  22 3   A3 – 5A2 + 9A– I = 1131 42 102 – 5 41 12 24 + 9 11 2 02 – 10 0 00 9 7 3 1          10  22 3  2  8 1   0  2 1  0 0 1  13  5  9 1 42  60  18  0  2  20 18  =  11  20  9 9  35  27 1 10 10    10 10  22  40 18  3  5  9  1  0 0 0 = 0 0 0 = 0 0 0 0  Cayley-Hamilton theorem verified. Now, from equation (I),  A3 – 5A2 + 9A– I = 0  A–1[A3 – 5A2 + 9A – I] = A–1.0  A2 – 5A+ 9I – A–1 = 0  A–1 = A2 – 5A+ 9I CU IDOL SELF LEARNING MATERIAL (SLM)

178 Mathematics 3 2 6  A–1 = 1 1 2 2 2 5  Exercise 20: Verify Cayley-Hamilton theorem and find A6 – 7A5 + 4A4 + 12A3 + A2 + A + I. Also, find A–1 where, A = 5 33. 1  Solution: The characteristic equation A  I = 0  5   3  2 – 8 + 12 = 0 =0  1  3    Now, A2 – 8A + 12I = 0 A2 = 5 3 5 3 28 24     =   1 3 1 3  8 12  28 =8 A2 – 8A + 12I 2124 – 8 15 3 1 0   3 + 12 0 1  0 =0 28  40  12 24  24  0 0 = 8  8  0 12  24  12 = 0 0   Now, divide the expression by A2 – 8A+ 12I, we get A6 – 7A5 + 4A4 + 12A3 + A2 + A + I = (A2 – 8A + 12I) (A4 + A3) + A2 + A + I = 0(A4 + A3)+ A2 + A + I = A2 + A + I  A2 + A + I = 28 24 5 3 1 0 34 27   +   +   =    8 12 1 3 0 1  9 16  Now, find A–1 Consider, A2 – 8A + 12I CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 4 179 A–1(A2 – 8A + 12I) = A–1.0  A2 – 8I + 12A–1 = 0 12A–1 = (8I – A) A–1 = 1 (8I – A) 2 A–1 = 1  3  3 2 1 5  11.7.8 Transpose of a Matrix The transpose of one matrix is another matrix that is obtained by using by using rows from the first matrix as columns in the second matrix. Interchange of rows with columns is called transpose of a matrix. 4 5 and A–1 = 4 6 A = 6 7 5 7   11.7.9 Symmetric Matrix A symmetric matrix is a square matrix that is equal to its transpose. Matrix A is symmetric because equal matrices have equal dimensions, only square matrices can be symmetric. The Following 3 × 3 matrix is an example of a Symmetric matrix- Matrix A = 71 7 35 4   3 5 6   Since A = AT, matrix A is symmetric. If in a case of square matrix A = A, then it is called symmetric matrix. [aij] = [aji]  i, j  [A] CU IDOL SELF LEARNING MATERIAL (SLM)

180 Mathematics i.e., A = 45 5 36 , A = 54 5 36 2 2     6 3 7 6 3 7  11.7.10 Skew Symmetric Matrix If in case of a square matrix, all diagonal elements are zero and [aij = – aji], then it called Skew Symmetric Matrix. 0 1 2 A = 3 0 4 diagonal elements 0. 5 6 0 11.7.11 Hermitian Matrix A square matrix A = [aij] is said to be Hermitian, if (i, j)th element of A is equal to the complex conjugate of (i, j)th element of A. i.e., aij = a ji 11.7.12 Skew Hermitian Matrix A square matrix A = [aij] is said to be Skew Hermitian, if (i, j)th element of A is equal to the negative complex conjugate of (i, j)th element and all diagonal elements must be zero. i.e.,  0 a  ib  a  ib 0  Exercise 21: Express A = P + iQ such that P and Q both Hermitian.  2 3  i 2  i A =  i 0 1  i 1 2i 1 3i CU IDOL SELF LEARNING MATERIAL (SLM)

 181  Matrices 4 Solution:  2 3  i 2  i Given, A =  i 0 1  i  1 2i 1 3i   2 i 1 2i  A = Transpose of A = 3  i 0 1 2  i 1  i 3i   2  i 1 2i A = A = complex conjugate of A = 3  i 0 1 2  i 1  i  3i   Let, A = P + iQ = 1 (A + A)+ 1 (A – A) 2 2i 1  4 3  2i 3  i 2  P= 3  2i 0 2  i   3 i 2i 0    P is a Hermitian matrix. Q = 1 (A – A) = 1  0 3 1  3i 3 0 i  2i    1 3i  i 6i 2i  1  0  3i 3  i 2 1 =  3i 0 6  1 3 i   Q is Hermitian matrix. CU IDOL SELF LEARNING MATERIAL (SLM)

182 Mathematics 11.7.13 Unitary Matrix A square matrix A with complex elements is said to be unitary if AA= I where, A = complex conjugate transpose of A i.e., A = A–1  A–1A= I 11.7.14 Orthogonal Matrix A square matrix is said to be orthogonal, if AA = I and when, A = transpose of A. i.e., A = A–1 Exercise 22: Show that the matrix A is unitary and also find A–1.  2 i 2 0 1 2  A= 2 i 2 0 0  0 2  Solution:  Given, A = 1  2 i 2 0 2 i 2 2   0 0 0 2    A = A = complex conjugate of transpose of A.  2 i 2 0 1 2   A= i 2 0 2  0 2  0 CU IDOL SELF LEARNING MATERIAL (SLM)

  Matrices 4  183   0 i 0 1 2 i 2 0  2 i 2 2 4  2  2   2 A.A = 0 0 0 i 2 0 2   0 2  2 0 0 4 0 0 1 1 0 = 4  0 22 0 = 4 4 0  0 0 2 2 0 0 4 1 0 0 = 0 1 0 = I 0 0 1  A is unitary matrix.  A–1A = I and now AA= I  A–1 = A = 1  2 i 2 0 2 i 2 2   0 0 0 2  11.7.15 Diagonalizable Matrix If A is a square matrix of order “n” and D is in a diagonal of order matrix n and D = M–1AM, then A is called a diagonalizable matrix, i.e., D is similar to A. If A and B are similar matrix, then (1) A = B (2) A2 is similar to B2 (3) A and B have same eigen value.  8  6 2  Exercise 23: Show that A =  6 7  4 is diagonalizable.  2  4 3  CU IDOL SELF LEARNING MATERIAL (SLM)

184 Mathematics Solution: Consider, A  I = 0 8    6 2   6 7    4 = 0   2  4 3     3 – 182 + 45 = 0  ( – 3)( – 15) = 0   = 0, 3, 15 Now, for  = 0, (A – I) x = 0  8 6  24 yx = 00 6 7        2  4 3 z 0  By Cramer's rule, x=y=z 6 2 82 8 6 7 4 6 4 6 7  x = y = z, 1 122 x 2 = 1 2  For  = 3   5 6  24 yx = 00 6 4        2  4 0 z 0  R +R  5 6  24 6 4  2 3  0  2  4 CU IDOL SELF LEARNING MATERIAL (SLM)

 185  Matrices 4 R  1,R  1 24 32 51 6 2 x 0  0 1 y = 0 1  2 0 z 0   2 Now, x + 2, x – 2y = 0 x2 =  1   2    7  6 2  6  8  4 For  = 15,  2  4  12    7  6 2 1  2  6 R2 – R1  2  4  12  R – 2R , R – 3R 101 0 206 2 3 21 2    0 0 0  R  1 , R +R 10 0  24 1 10 2 1 2     0 0 0 1 0 2 R2  1  0 1 2 2   0 0 0  10 0 2 x 0  1 2 y = 0  0 0 0 z 0 CU IDOL SELF LEARNING MATERIAL (SLM)

 Mathematics  186 –  + 2z= 0, y + 2z = 0 x = 22 3    1   y = – 2z = M–1AM = D  D = 00 0 0  , and M = 12 2 22 3 0 1     0 0 15 2  2 1   11.7.16 Minimal Polynomial The harmonic polynomial of lowest degree that annihilate a matrix A is called minimal polynomial, if f(x) is minimal polynomial of A. Then, f(x) = 0 is called minimal equation of A. Derogatory and Non-derogatory Matrices An “n=rowed” square matrix is said to be derogatory, if the degree of its minimal polynomial is less than “n”. The matrix is said to be non-derogatory, if the degree of its polynomial is equal to n. Exercise 24: Diagonalize the matrices and find modal matrix P. 0 10 where, A = 0  Solution: Now, A  I = 0 0   1 = 0  2 = 0,  = 0.0  0 0     This matrix is not diagonalizable since eigen values are not distinct. CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 4 187 8  8  2 Exercise 25: If A = 4  3  2 is diagonalized, find P. 3  4 1 Solution: Now, A  I = 0 8    8  2  4  3    2 = 0   3  4 1    3 – (Sum of diagonal element of A)2 + (Sum of minors of A) – A = 0 A = 6, Sum of diagonal = 8 – 3 + 1 = 6 Sum of Minor = (– 3 – 8) + (8 + 6) + (– 24 + 32) = 11 So, 3 – 62 + 11 – 6 = 0 So,  = 2, 3, 1 For  = 2, 46 8  22 yx = 00 5        3  4  1 z 0    3 x1 x2 x13 , x 1=  2 3 = 2 =   1  For  = 3, 45 8  22 yx = 00 6        3  4  2 z 0 CU IDOL SELF LEARNING MATERIAL (SLM)

 Mathematics  188 x = 21 2   1  For  = 1, 74 8  22 yx = 00 4        3  4  0 z 0  3 x2 = 4 x1  4x – 4 34x – 2x = 0 1   1 3  1  x3 = 2 x1  x = 1, x = 3 , x = 1 1 2432 x = 31/ 4 3   1/ 2   3 2 1 2 0 0  P =  2 1 3/ 4 and D = 0 3 0  1 1 1/ 2 0 0 1  P–1 we can find by any method.  1 0 2 P–1 =  1 2 1   4 4 4 CU IDOL SELF LEARNING MATERIAL (SLM)

  Matrices 4  189   1 0 2 8  8  2  1 0 2 2 0 0 Consider, P–1AP =  1 2 1 4 3  2   1 2 1 = 0 3 0      4 4 4 3  4 1  4 4 4 0 0 1   P–1AP = D   1 1 3 Exercise 26: If P =  5 2 6, matrix P is derogatory of non-derogatory.   2  1  3  Solution:  1 1 3 Given, P=  5 2 6   2  1  3 1  1 3  P  I =  5 2   6   2 1  3    = 3 + 2 = (2 + 2)   = 0,  =   2 =  i 2   = 0, i 2 , – i 2 Now, for  = 0  Minimal polynomial is  ,  2,  3 if n(  ) = P  0 matrix n(  ) =  is not a minimal polynomial Let, n() = 2 i.e., n(P) = P2  1 1 3   1 1 3   0 0 0     6  = 3  P2 = P.P =  5 2 6   5 2 3 9   0  2  1  3  2 1  3  1 1  3   n()= 2 is not minimal polynomial. CU IDOL SELF LEARNING MATERIAL (SLM)

190 Mathematics  Consider, n()= 3,  m(P) = A3  0 0 0  1 1 3 0 0 0  9  6 = 0 0 0 = 0 A3 = A2.A =  3 3  5 2   1  1  3  2 1  3 0 0 0  n()= 3 is minimal polynomial matrix P degree of minimal polynomial = 3 = order of P. 11.8 Summary Apart from all types of matrix here we have the Eigen Values and Eigen Vectors, are often introduced to students in the context of linear algebra courses focused on matrices. The row or column matrix elements will form a vector. The set of all n-vector of a field to be denoted by Vn(F) is called n-vector space over F. Turn to matrices in particular. The eigen value idea not restricted to matrices, but is of wider application. The number  is called the eigen value and y is the eigen vector or eigen function. The theorem of Cayley-Hamilton is useful for calculation of different power of a square matrix and also to find inverse of nonsingular matrix . The Cayley–Hamilton theorem states that if one defines an analogous matrix equation, p(A), consisting of the replacement of the scalar variable λ with the matrix A, then this polynomial in the matrix A results in the zero matrix, 11.9 Key Words/Abbreviations  For an orthogonal matrix, A·AT = I.  If A is a square matrix of order “n” and D is a diagonal of order matrix n and D = M–1 AM, then A is called a diagonalizable matrix, i.e., D is similar to A.  If A and B are similar matrix, then (i) |A| = |B| (ii) A2 is similar to B2 (iii) A and B have same eigen values. CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 4 191 11.10 Learning Activity 8 8  22is diagonalizable, then find P. 1. If A  4 3  3  4 1  2  2 3  2. Find eigen value and eigen vectors for A  1 1 1 .  1 1 3        11.11 Unit End Questions (MCQ and Descriptive) A. Descriptive Type Questions Exercise 4 7 1 8 then find: (i) 3A + 2B and (ii) 2A – 4B. 1. If A = 1 9 and B = 7 2,   6 8 4 3 then find: (i) A + 4B, (ii) 2A + B and (iii) A – 2B. 2. If A = 7 1 and B = 3 4,   1  3 and B =  1 0 1 , then verify (AB) = BA. If A = 3.   2 1 2 1 3 4. Express the matrix A as the sum of symmetric and skew symmetric, when 1 7 1 A = 2 3 4  5 0 5 CU IDOL SELF LEARNING MATERIAL (SLM)

 Mathematics  192 1 1 that A2 – 4A + 5I = 0. 5. Given that A = 2  show 3 6. Show that given matrix is orthogonal, when  1 1  1   3  62 2 1 0  A =  3  6   1 1 1   3 6 2   7. Find co-factor of given matrix  1 2  2  1 2 3  1  3 2 I. A =  1 3 0 II. A = 2 4 5 III. A =  3 1 1 0 2 1  2 1 0 5 6 7 3 1 1  1 3  2   1  2  2 Ans: I. 1 2  3 II.  3 3 1 III.  2  4  5  2  1 3  2 1 0 3  5  6  8. Find inverse of given matrices by using Adjoint Method. 1 1 1 I. A = 1 2  3 2 1 3  3  4  5 Ans: A–1 = 1  9 1 4 11  5 3 1 0 1 2 II. Find A–1 A = 1 2 3 3 1 1  1 1 1 Ans: A–1 =  1  8  6 2 2 3  1  5 CU IDOL SELF LEARNING MATERIAL (SLM)

  Matrices 4  193  9. Find A–1 cos   sin  0 A = sin  cos  0  0 0 1  cos  sin  0 A–1 =  sin  cos  0  0 0 1  10. Find Eigen values and Eigen vectors of following: 0 1 0  4 2  2 (a) 0 0 1 (b)  5 3 2 0 0 0  2 4 1   2  3 1 cos   sin  (d)  3 1 3 (c) sin  cos   5 2  4  1 1 2 11. Find the Eigen values and Eigen vectors of  0 2 1.   0 0  3  Ans:  = –1, –2, –3; 01 , 11 ,  1  2       0  0  2   4 6 6 12. Determine the Eigen value and Eigen vector of A =  1 3 2.   1  4  3   1 2 2 13. Find the Eigen value and Eigen vector of matrix A =  0 2 1.   1 2 2  14. Find the Eigen value and the corresponding Eigen vectors of the following matrix CU IDOL SELF LEARNING MATERIAL (SLM)

194 Mathematics  4 1 1 A =  1 4 1  1 1 4   4 0 1 15. Find the spectrum of the matrix, A =  2 1 0  2 0 1  2  8 12 16. Find Eigen values and Eigen vectors A =  1 4 4   0 0 1  Ans:  = 0, 1, 2; 14 ,  04 , 12        0  1  0  3 1 4 17. Find the Eigen values and Eigen vectors of 0 2 6. 0 0 5  Ans:  = 2, 3, 5; 11 , 01 , 23        0 0 1  18. Verify Cayley-Hamilton theorem and find A–1. 1 1 2  1 1  2  (a) 3 1 1  (b) 1 2 1 2 3 1  0 1  1   3 1 4  8  8  2 (c) 0 2 6 (d) 4  3 3 0 0 5 3  4 1 CU IDOL SELF LEARNING MATERIAL (SLM)

 195  Matrices 4 19. Use Cayley-Hamilton theorem and find A–1.  7 2  2 A =  6 1 2  6 2  1 20. Verify Cayley-Hamilton theorem and find A–1, A4, A5 use it.  11  6 2 A =  6 10  4  2  4 6 21. Find minimal polynomial of the matrix and check it is derogatory or non-derogatory  5  6  6 1 0 0 (b) 1 1 0 (a) 1 4 2  3  6  4 0 0  1 22. Find modal matrix P for the following matrix A  7  2 1 A =  2 10  2  1  2 7 23. Show that A and B have same minimal polynomial A = 01 1 00 , B = 20 0 02 2 2     0 0 1 0 0 1  B. Multiple Choice/Objective Type Questions 2 1 . 1. Transpose of matrix 0 2 is  2 0  (b) 1 2 (d) None of above  2 0  (a)   1 2  2 0  (c)   2 1 CU IDOL SELF LEARNING MATERIAL (SLM)