Questions and Answers #3 Matrices

Questions and Answers Sheet 3 Matrices Question #1 Find the characteristic polynomial of the matrices  4 2  3 3   Answer: Step 1 Solution: Consider the given matrix A  4 2 A=   3 3 We know the characteristic polynomial of a A =  11  a21 a12  is a22  a ( A −  I ) =  11  a21 a12  1 0  −    a22  0 1   4 2 Therefore for the given A =    3 3 Characteristic Polynomial is  4 2 1 0  (A − I ) =  −     3 3 0 1  2  4 −  (A − I ) =  3 −    3 ( A −  I ) = (4 −  )(3 −  ) − 6 ( A −  I ) = 12 − 4 − 3 +  2 − 6 ( A −  I ) =  2 − 7 + 6 Step 2 This is the required characteristic polynomial of the given matrix. Question #2 Use elementary matrices to find the inverse of 1 0 0 A =  0 1 0  , c  0  a b c  Answer: Step 1 Given 1 0 0   The matrix is 0 1 0 , c  0    a b c  Step 2 To find inverse of matrix [𝐴|𝐼] = [𝐼|𝐴−1 ] 1 0 0 1 0 0 [ A | I ] =  0 1 0 | 0 1 0  R3  a b c 0 0 1  1 0 0 1 0 0   R3 − aR1[ A | I ] = 0 1 0 | 0 1 0  R3 0 b c −a 0 1  0 0 1 0 0 1   R3 − bR2 [ A | I ] = 0 1 0 | 0 1 0  R3 0 0 c −a −b 1   R3 c  1 0 0 1  [ A | I ] = 0 1 0 | 0  a 0 0 1 − c   0  0  = [ I | A−1 ] 1  c    1 0 0   −1 Therefore Inverse of A is A =  0 1 0  a b 1 − −  c c  c 0 1 b − c Question #3 Let  a11 a12  −6 −4 A=  a31 a32   a41 a42 a d −1 A = g  k a14  6  a34   a44  −7 b c  8 e f  h i j  2 l m a13 8 a33 a43 find ''h'' for which 𝐴−1 is the inverse of A Answer: Step 1 Given:  a11 a12  −6 −4 A=  a31 a32   a41 a42 a d A−1 =  g  k a14  6  a34   a44  −7 b c  8 e f  h i j  2 l m a13 8 a33 a43 Step 2 Given that, A and 𝐴−1 are inverses of each other. ⇒ 𝐴𝐴−1 = 𝐴−1 𝐴 = 𝐼 Where I is an identity matrix of order 4. Consider, 𝐴𝐴−1 = 𝐼  a11 a12 a13  −6 −4 8   a31 a32 a33   a41 a42 a43 a14   a 6   d  a34   g  a44   k −7 b c  1 8 e f  0 = h i j  0   2 l m  0 0 1 0 0 0 0 1 0 0 0  0  1 by equality of matrices, the second-row second column element of the product matrix 𝐴𝐴−1 will be equal to the corresponding second-row second column element of an identity matrix I. For product matrix 𝐴𝐴−1 , second-row second column element is obtained by multiplying the second-row elements of the matrix A with the corresponding column elements of the matrix 𝐴−1 Thus, (-6)(-7)+(-4)(8)+(8)(h)+(6)(2)=1 42-32+8h+12=1 8h+22=1 8h=1-22 8h=-21 −21 ℎ= 8 Therefore, ℎ = −21 8 Question #4 Perform the indicated multiplication if possible.  4 −4 −8 7  −5 −6 1 −7 10 5   4 2     Answer: Step 1 Solution: From the given information, the number of rows of first matrix is 2 and number of columns of first matrix is 4. The number of rows and number of columns of the second matrix is 2. Step 2 Since the number of columns in the first matrix is not equal to the number of rows in the second matrix the multiplication of matrices is not possible. Thus,  NONE  NONE   NONE   NONE NONE NONE NONE NONE NONE NONE NONE NONE NONE  NONE  NONE   NONE  Question #5 Use determinants to determine whether the following 2 × 2 matrices are nonsingular 3 5  2 4   Answer: Step 1 Let , the given matrix 3 5 A=   2 4 Step 2 A matrix A is said to non-singular if the determinant of A is non-zero that is |𝐴| ≠ 0 Step 3 Here, | A |= 3 5 2 4 |𝐴| = (3)(4) − (5)(2) |𝐴| = 12 − 10 |𝐴| = 2 ≠ 0 Therefore, the matrix 3 5 A=  is non-singular.  2 4 Question #6 Find a sequence of elementary matrices whose product is the given nonsingular matrix.  4 −1  3 −1   Answer: Step 1 Solution Consider 𝐼𝐴 = 𝐴 1 0   4 −1 0 1 A =  3 −1     Reducing matrix A on RHS by applying elementary row operations on both side. Applying 𝑅1 ⇒ 𝑅1 4 1 1    4 0  A = 1 − 4       0 1 3 −1  1  0  Let E1 = 4    0 1 Applying 𝑅2 ⇒ 𝑅2 − 3𝑅1 1  1 −    1 0 4   −3 1  E1 A =    0 − 1   4   1 0 Let E2 =    −3 1  Step 2 Applying 𝑅2 ⇒ −4𝑅2 1  1 −  1 0   4 0 −4  E2 E1 A =     0 1  1 0  Let E3 =   0 −4  Applying 𝑅1 ⇒ 𝑅1 + 𝑅2 4 1  1 4  E E E A = 1 0  0 1    3 2 1   0 1   1  1  Let E4 = 4   0 1  Therefore , 𝐴 = 𝐸1−1 𝐸2−1 𝐸3−1 𝐸4−1 Calculating 𝐸1−1 , 𝐸2−1 , 𝐸3−1 , 𝐸4−1 1  0  4 E1−1 =   0 1   16  1 0  E2−1 =   3 1  1 0  −1  E1 =  0 − 1   4 1  1 − −1 E1 =  4   0 1  Step 3 Therefore, 1   4 0  1 0 1 0  1 − 1   A=  4 1     1 3 1 0 −  0  0 1   4    16  Question #7 give the rank and the nullity of the matrices in the given exercises. 1 1 −3 A = 0 2 1  1 −1 −4  Answer: Given: The matrix 1 1 −3 A = 0 2 1  1 −1 −4  Rank of matrix A: The rank of a matrix: The rank of matrix rows (columns) is the maximum number of linearly independent rows (columns) of this matrix. That is the number of non zero rows in its row echelon form. Now reduce matrix A to reduced row-echelon form. Eliminate elements in the first column under the first element. Using the row operation 𝑅3 − 𝑅1 𝑟𝑖𝑔ℎ𝑡 ⇒ 𝑅3 1 1 −3 0 2 1    0 −2 −1 Now perform the operation 𝑅3 + 𝑅2 𝑟𝑖𝑔ℎ𝑡 ⇒ 𝑅3 1 1 −3 0 2 1    0 0 0  As there are two non zero rows in the resulting matrix. Therefore, rank(A) =2. Nullity of matrix A: As rank + nullity = number of columns. Therefore, 2+ nullity =3 Thus nullity(A) = 1. Question #8 Compute the determinants of the following matrices in 𝑀2 × 2(𝑅) 6 a)  2  −5 b)  6 −3 4  2 1  8 0  c)   3 −1 Answer: Step 1 a b   , then its determinant is c d  If a 2 × 2 matrix M with entries from the set of real numbers is M =  a b  | M |=   c d  =ad-bc Step 2 a) The given matrix is: 6 −3 M1 =   2 4  The determinant of the given matrix is: | M1 |= 6 −3 2 4 =(6)(4)-(-3)(2) =24-(-6) =24+6 =30 The determinant of the given matrix is 30. Step 3 b) The given matrix is:  −5 2 M2 =    6 1 The determinant of the given matrix is: | M 2 |= −5 2 6 1 =(-5)(1)-(2)(6) =-5-12 =-17 The determinant of the given matrix is -17. Step 4 c) The given matrix is: 8 0  M2 =   3 −1 The determinant of the given matrix is: | M 2 |= 8 0 3 −1 =(8)(-1)-(0)(3) =-8+0 =-8 The determinant of the given matrix is -8. Step 5  6 −3  2 4  is 30    −5 2  (b) The determinant of the matrix   is -17  6 1 8 0  (c) The determinant of the matrix   is -8 3 −1 Answer: (a) The determinant of the matrix Question #9 a H = { b 1 Let B = { 0 Let b , a, b, c  } which is a subspace of the vector space of all 2 × 2 matrices. c  0  0 1  0 0  , , } which is a basis for H. 0 1 0 0 1 2 1 1. Give the coordinate vector of   relative to B. 1 2 2. TRUE or FALSE : Any set of four matrices from H must be linearly dependent. Please explain. Answer: a H = { b Let 1 B = { 0 b , a, b, c  } c  is basis for H. 0  0 1  0 0  , , } 0  1 0  0 1  2 1 1 0   0 1   0 0 1) Let A =  = 2 + 1 + 2     1 2  0 0  1 0 0 1 2   Co-ordinate vector relative to B is 1    2  2) False as dim(𝐻) = 3 ⇒ any set of 4 matrices are linearly dependent Question #10 Given the matrices A and B, find the matrix product AB 1 3 −2  A= , 3 0 4   3 0 B =  −2 1   0 4  Find AB Answer: Given, 1 3 −2  A= , 3 0 4   3 0 B =  −2 1   0 4   3 0 1 3 −2    = 1(3) + 3(−2) + (−2)(0) 1(0) + 3(1) + (−2)(4)  =  −3 −5 AB =  − 2 1     3(0) + 0(1) + 4(4)   9 16  3 0 4   0 4   3(3) + 0(−2) + 4(0)   Question #11 Find the product of the following matrices  3 0  −1 1  −3 4  2 −1    Answer: Step 1 Product of matrix Suppose A is an m-by-n matrix and C is an n-by-p matrix. Then Ac is defuned to be the m-by-p matrix whose entry in row j, column k, is given by the following equation. (𝐴𝐶)𝑗,𝑘 = ∑𝑛𝑟=1 𝐴𝑗,𝑟 𝐶𝑟,𝑘 Step 2  3 0  −1 1   3  −1 + 2  0 3 1 + 0  −1  −3 3  −3 4  2 −1 =  −3  −1 + 4  2 −3 1 + 4  −1 = 11 −7         Question #12 Find the product of the following two matrices.  0 −5  3 1 −2 3   −2 0    Answer: Step 1 Given,  0 −5  3 1 −2 3   −2 0    Step 2 Now, 0(1) + (−5)(0)   0 −5  3 1   0(3) + (−5)(−2)  =      −2 3   −2 0  (−2)(3) + (3)(−2) (−2)(1) + (3)(0)   0 + 10 0 + 0  =   −6 − 6 −2 + 0  0  10 =   −12 −2  0  0 −5  3 1   10  =      −2 3   −2 0   −12 −2  Question #13 Use the matrix P to show that the matrices A and A' are similar. 1 −12 P = A = A =   0 1  Answer: Step 1 Definition: If A and B are similar matrices then then there exist a invertible matrix P such that 𝐴 = 𝑃𝐵𝑃 −1 1 −12 P = A = A =   0 1  1 −12 −1 Substitute P = A = A =   in 𝐴 = 𝑃𝐵𝑃 and check as follows. 0 1   Given that Step 2 𝐴 = 𝑃𝐵𝑃−1 −1 1 −12  1 −12  1 −12  1 −12  1 −12  0 1  = 0 1  0 1  0 1  0 1          1 −12  1 0  1 −12  1 −12  =     0 1  0 1  0 1  0 1  satisfied Thus, A and A' are similar. Question #14 Use the matrix P to show that the matrices A and A' are similar. 7  1 −1  12 1 −2 P= ,A= , A =     1 2   −20 −11 4 0  Answer: Step 1 Given 7  1 −1  12 1 −2 P= ,A= and A =     1 2  −20 −11 4 0  Step 2 Using Two matrices A and A' are similar if and only if there exist an invertible matrix P, such that 𝐴 = 𝑃𝐴′ 𝑃 −1 or 𝐴𝑃 = 𝑃𝐴′ Now 7  1 −1  12 AP =     −20 −11 1 2  12(−1) + 7(2)   12(−1) + 7(1) =   −20(−1) + (−11)(1) −20(−1) + (−11)(2)   −12 + 7 −12 + 14  =   20 − 11 20 − 22   −5 2  =   9 −2  And 1 −1 1 −2  PA =    1 2   4 0   −1(1) + (−1)(4) −1(−2) + (−1)(0)  = 1(−2) + 2(0)   1(1) + 2(4)  −1 − 4 2 + 0  =   1 + 8 −2 + 0   −5 2  =   9 −2  Here AP=PA' ⇒ 𝐴 and A' are similar. Question #15 Find values for x, y, and z so that the following matrices are equal: 2 x z  y + 7   −10 13 = 4   6 4  Answer: Step 1 It is given that, 2 x z  y + 7   −10 13 = 4   6 4  We have to find the values of x,y and z. Step 2 We have, 2 x z  y + 7   −10 13 = 4   6 4  So, 2𝑥 = −10 ⋯ (1) 𝑦 + 7 = 13 ⋯ (2) and 𝑧 = 6 ⋯ (3) On solving eq. (1),we get −10 ⇒𝑥= 2 ⇒ 𝑥 = −5 On solving eq. (2) ,we get ⇒ 𝑦 + 7 = 13 ⇒ 𝑦 = 13 − 7 ⇒𝑦=6 Hence, x=-5 y=6 and z =6 Question #16 The cipher text text YIFZMA was encrypted by a Hill cipher with a 2 × 2 matrix  9 13 2 3    Find the plaintext Answer: We have to solve given problem: Consider 𝑎 = 0, 𝑏 = 1, 𝑐 = 2, ⋯ , 𝑧 = 25 𝐴 = 0, 𝐵 = 1, 𝐶 = 2, ⋯ , 𝑍 = 25 given cipher text is YIFZMA ⇒ 𝑌 = 24 I=8 F=5 Z=25 M=12 a b  if A =   then inverse of A is c d   d −b  A −1 =    −c a  note: ad-bc Let 9 13 A=  2 3   3 −13 1 (27 − 26)  −2 9   3 −13 A−1 =    −2 9  A−1 = Concept: a  b  if   is cipher text then plaintext will be a  = A−1   mod 26 b  ∴ 𝑌 = 24 I=8  24   3 −13  24   A−1   =     mod 26  8   −2 9   8   72 − 104  =  mod 26  −46 + 72   −32  =  mod 26  24   20  =   24   24   20   A−1   =    8   24  ∴ 𝑈 = 20, 24 = 𝑌 Now, F=5 Z=25  5   3 −13  5  A−1   =     mod 26  25  −2 9   25  15 − 325  =  mod 26  −10 + 225  −310  =  mod 26  215   5   2 A−1   =    25 8  i.e. C=2 , H=8 Now, M=12 A=0 12   3 −13 12  A−1   =     mod 26  0   −2 9   0  36 − 0  =  mod 26  −24   36  =  mod 26  −24  12  10 A−1   =   0 2 K=10, C=2 Hence , required plaintext is UYCHKC Question #17 Find the eigenvalues and eigenvectors of the following matrices  2 −1   −4 2  a)  Answer: Step 1 Consider the given matrix,  2 −1 A=   −4 2  To find the eigen values, set |𝐴 − 𝜆𝐼| = 0 |𝐴 − 𝜆𝐼| = 0 2 −  −1 =0 −4 2 −  (2 − 𝜆)(2 − 𝜆) − (−4)(−1) = 0 (2 − 𝜆)2 − 4 = 0 4 − 4𝜆 + 𝜆2 − 4 = 0 𝜆(𝜆 − 4) = 0 𝜆 = 0,4 Therefore, the eigen values are 0 and 4. Step 2 Eigen vectors: |𝐴 − 𝜆𝐼|𝑥 = 0 For 𝜆 = 0  2 − 0 −1   x1   −4 2 − 0   x  = 0   2  2 −1  x1   −4 2   x  = 0   2  2 x1 − x2   −4 x + 2 x  = 0  1 2 2𝑥1 − 𝑥2 = 0 ⇒ 𝑥2 = 2𝑥1 −4𝑥1 + 2𝑥2 = 0 ⇒ 𝑥2 = 2𝑥1 If 𝑥1 = 1, 𝑥2 = 2 1  2 Therefore, x =   For 𝜆 = 4  2 − 4 −1   x1   −4 2 − 4   x  = 0   2  −2 −1  x1   −4 −2   x  = 0   2  −2 x1 − x2   −4 x − 2 x  = 0  1 2 −2𝑥1 − 𝑥2 = 0 ⇒ 𝑥2 = −2𝑥1 −4𝑥1 − 2𝑥2 = 0 ⇒ 𝑥2 = −2𝑥1 If 𝑥1 = 1, 𝑥2 = −2 1  1 1 . Hence, the eigen vectors of the given matrix are and    −2   2    −2 Therefore , x =  Question #18 Suppose that A, B, C and D are n \times n matrices satisfying the equation ABC=D Show that if DA is invertible , then B is invertible. Write and upload your solution Answer: Step 1 Let us show that if DA is invertible, then B is invertible Given: Suppose that A,B,C and D are 𝑛 × 𝑛 matrices ABC=D Invertible matrix: A square matrix is invertible if and only if the determinant is not equal to zero |𝐴| ≠ 0 Step 2 Let 1 0  1 0  1 0  1 0  A= , B = , C = and D =  0 1  0 1  0 1  0 1        ABC=D L.H.S 1 0  1 0  1 0  1 0  1 0  1 0  0 1   0 1    0 1  =  0 1    0 1  =  0 1              =D =R.H.S ABC=D Step 3 1 DA =  0 1 DA =  0 0  1 0   1  0 1  0 | 1  1 0  DA |=   0 1  |𝐷𝐴| = 1 ≠ 0 DA is invertible 1 0  1 0  B= | B |=    0 1  0 1  |𝐵| = 1 ≠ 0 B is invertible Step 4 Hence proved If DA is invertible, then B is invertible Question #19 Prove the triple scalar product identity 𝐴 ⋅ (𝐵 × 𝐶) = (𝐴 × 𝐵) ⋅ 𝐶 Answer: Step 1 A rectangular array of numbers (or other mathematical objects) for those the operations such as addition and multiplication are defined is called matrices. Matrix are used for plotting graph, statistics and also to do scientific studies and research in almost different fields. Step 2 Given: To prove the scalar identity 𝐴 ⋅ (𝐵 × 𝐶) = (𝐴 × 𝐵) ⋅ 𝐶 Step 3 Suppose, 𝐴⃗ = 𝑎1 𝑖̂ + 𝑎2 𝑗̂ + 𝑎3 𝑘̂ ⃗⃗ = 𝑏1 𝑖̂ + 𝑏2 𝑗̂ + 𝑏3 𝑘̂ 𝐵 𝐶⃗ = 𝑐1 𝑖̂ + 𝑐2 𝑗̂ + 𝑐3 𝑘̂ Now, i j k A( BC ) = (a1iˆ + a2 ˆj + a3kˆ) b1 b2 b3 c1 c2 c3 (𝑎1 𝑖̂ + 𝑎2 𝑗̂ + 𝑎3 𝑘̂){(𝑏2 𝑐3 − 𝑏1 𝑐2 )𝑖̂ − (𝑏1 𝑐3 − 𝑏3 𝑐2 )𝑗̂ + (𝑏1 𝑐2 − 𝑐1 𝑏2 )} a1 a2 b1 b2 c1 c2 Similarly, a3 b3 c3 i j ( A  B )C = a1 a2 b1 b2 k a3 b3 ( A  B )C = c1 (a2b3 − b2 a3 ) + c2 (a3b1 − a1b3 ) + c3 (a1b2 − a2b1 ) c1 c2 ( A  B )C = a1 a2 b1 b2 c3 a3 b3 a1 a2 ( A  B )C = b1 b2 c1 c2 a3 b3 c3 therefore RHS=LHS Hence proved. Question #20 Decide if the following matrices are positive definite? 3 2 1 A =  2 5 3  1 3 4  2 1 1 B = 1 2 1  1 1 2  5 2 2 C =  2 6 1  1 2 7  Answer: Step 1 Definition Method to test positive definiteness: Determinant of all possible k*k upper left sub matrices should be positive. Step 2 Explanation A) |3|=3 3 2 1 3 2 = 15 − 4 = 11 2 5 3 2 5 1 3 4 = 3(20 − 9) − 2(8 − 3) + 1(6 − 5) = 24 B) |2|=2 2 1 = 4 −1 1 2 2 1 1 =31 2 1 1 1 2 = 2(4 − 1) -1(2-1)+1(1-2)=4 C) |5|=5 5 2 2 5 2 = 30 − 4 = 26 2 6 1 2 6 1 2 7 = 5(42 − 2) -2(14-1)+2(4-6)=170 Step 3 All three matrix, A,B,C are satisfying positive definiteness test. Thus are positive definite matrix.