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# Sect3 4

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### Sect3 4

1. 1. SECTION 3.4 MATRIX OPERATIONS The objective of this section is simple to state. It is not merely knowledge of, but complete mastery of matrix addition and multiplication (particularly the latter). Matrix multiplication must be practiced until it is carried out not only accurately but quickly and with confidence — until you can hardly look at two matrices A and B without thinking of "pouring" the ith row of A down the jth column of B.  3 −5   −1 0  9 −15  −4 0   5 −15 1. 3  + 4  3 −4 =  6 21  + 12 −16 = 18 5  2 7           2 0 − 3  − 2 3 1  10 0 −15  6 −9 −3   16 −9 −18 2. 5  − 3  7 1 5 =  −5 25 30  +  −21 −3 −15 =  −26 22 15   −1 5 6          5 0   −4 5   −10 0   −16 20   −26 20  3. 0 7  + 4  3 2  =  0 −14 +  12 −2  8  =  12 −6          3 −1    7 4    −6  2   28 16      22 18     2 −1 0   6 −3 − 4  4.  4 0 −3 + 5 5 2 −1 7     5 −2 7    0 7 9    14 −7 0  30 −15 −20   44 −22 −20   28 0 −21 +  25 10 =  −5  =  53 10 −26      35 −14 49   0 35 45      35 21 94     2 −1  −4 2   −9 1   −4 2   2 −1  −2 8  5.  3 2   1 3  =  −10 12  ;  1 3   3 2  =  11 5            1 0 −3   7 −4 3   7 −13 −24  6.  3 2 4  1 5 −2  =  23 10 41        2 −3 5  0 3 9     11 −8 57     7 −4 3   1 0 −3   1 −17 −22  1 5 −2  3 2 4  = 12 16 7        0 3 9   2 −3 5      27 −21 57   
2. 2.  3 3 3 6 9  7. [1 2 3]  4 = [26];    4  1 2 3 =  4 8 12   [ ]   6   6     5 10 15    3 0  3 0 3 0 9 1 0 3    21 15  −1 4  1 0 3  =  7 −20 13  8.  2 −5 4   −1 4  =   ;    2 −5 4       6 5 35 0   6 5   16 −25 38        0 −2   4   0 −2   3 1   3  =  7  but the product  3   3 1  is not defined. 9.    −2     −2     −4 5     −22     −4 5         2 1   −1 0 4  1 −2 13 10. AB =    3 −2 5  = 5 −6 31 but the product BA is not defined.  4 3      2 7 5 6 11. AB = [3 −5]   = [11 1 5 3] but the product BA is not defined.  −1 4 2 3 12. Neither product matrix AB or BA is defined.  3 1    2 5  0 1   3 1  10 17   32 51  13. A(BC) =     −3 1  2 3  =  −1 4  2 0  =  −2 −17   −1 4             3 1   2 5   0 1  3 16   0 1  32 51  ( AB)C =     −3 1   2 3 =  −14 −1  2 3 =  −2 −17    −1 4            2 5  6    −13 14. A ( BC ) = [2 −1]     −5  = [2 −1]  −23 = [−3]   −3 1         2 5   6  6 ( AB ) C =  [2 −1]     −5 = [7 9]  −5 = [−3]   −3 1       2 0  3   3  12 15  15. A ( BC ) =    [1 −1 2] 0 3   =   [4 5] =     2  1 4   2   8 10    
3. 3. 2 0  2 0    3   =  3 −3 6  0 3  = 12 15 ( AB ) C =    [1 −1 2] 0 3      8 10    2  1 4   2 −2 4   1 4        2 0 0  1 −1 1 0 −1 2   16. A ( BC ) =  3    3 −2  3 2 0 1   1  4      2 0  −4 −4 −2 2  = 0   −2 −2 −1 1  =  −9 −12 −9 12  3   −3 −4 −3 4    4  1   −14 −18 −13 17       2 0    1 −1  1 0 −1 2  ( AB ) C =  0 3   3 −2   3 2 0 1  1  4      2 −2   −4 −4 −2 2  1 0 −1 2 = 9 −6   =  −9 −12 −9 12    3 2 0 1   −9   13   −14 −18 −13 17      Each of the homogeneous linear systems in Problems 17-22 is already in echelon form, so it remains only to write (by back substitution) the solution, first in parametric form and then in vector form. 17. x3 = s, x4 = t , x1 = 5s − 4t , x2 = − 2 s + 7t x = s (5, −2,1, 0 ) + t ( −4, 7, 0,1) 18. x2 = s, x4 = t , x1 = 3s − 6t , x3 = − 9t x = s (3,1, 0, 0 ) + t ( −6, 0, −9,1) 19. x4 = s, x5 = t , x1 = − 3s + t , x2 = 2 s − 6t , x3 = − s + 8t x = s ( −3, 2, −1,1, 0 ) + t (1, −6,8, 0,1) 20. x2 = s, x5 = t , x1 = 3s − 7t , x3 = 2t , x4 = 10t x = s (3,1, 0, 0, 0 ) + t ( −7, 0, 2,10, 0 )
4. 4. 21. x3 = r , x4 = s, x5 = t , x1 = r − 2 s − 7t , x2 = − 2r + 3s − 4t x = r (1, −2,1, 0, 0 ) + s ( −2,3, 0,1, 0 ) + t ( −7, −4, 0, 0,1) 22. x2 = r , x4 = s, x5 = t , x1 = r − 7 s − 3t , x3 = s + 2t x = r (1,1, 0, 0,0 ) + s ( −7, 0,1,1, 0 ) + t ( −3, 0, 2, 0,1)  2 1   a b  1 0  23. The matrix equation   =  entails the four scalar equations  3 2   c d  0 1  2a + c = 1 2b + d = 0 3a + 2c = 0 3b + 2d = 1 that we readily solve for a = 2, b = −1, c = −3, d = 2. Hence the apparent inverse  2 −1 matrix of A, such that AB = I, is B =   . Indeed, we find that BA = I  −3 2  as well. 3 4   a b  1 0  24. The matrix equation   =  entails the four scalar equations 5 7   c d   0 1  3a + 4c = 1 3b + 4d = 0 5a + 7c = 0 5b + 7 d = 1 that we readily solve for a = 7, b = −4, c = −5, d = 3. Hence the apparent inverse  7 −4  matrix of A, such that AB = I, is B =   . Indeed, we find that BA = I  −5 3  as well.  5 7   a b  1 0  25. The matrix equation   =  entails the four scalar equations  2 3   c d  0 1  5a + 7c = 1 5b + 7 d = 0 2a + 3c = 0 2b + 3d = 1 that we readily solve for a = 3, b = −7, c = −2, d = 5. Hence the apparent inverse  3 −7  matrix of A, such that AB = I, is B =   . Indeed, we find that BA = I  −2 5  as well.
5. 5.  1 −2   a b  1 0  26. The matrix equation   =  entails the four scalar equations  −2 4   c d   0 1  a − 2c = 1 b − 2d = 0 −2a + 4c = 0 − 2b + 4d = 1. But the two equations in a and c obviously are inconsistent, because ( −1)(1) ≠ 0, and the two equations in b and d are similarly inconsistent. Therefore the given matrix A has no inverse matrix.  a1 0 0 0 b1 0 0 0  a1b1 0 0 0  0 a2 0 0 0 b 0 0   0 ab 0 0     2  2 2  27. 0 0 a3 0 0 0 b3 0 =  0 0 a3b3 0              0  0 0 an   0 0  0  bn   0  0 0 anbn   Thus the product of two diagonal matrices of the same size is obtained simply by multiplying corresponding diagonal elements. Then the commutativity of scalar multiplication immediately implies that AB = BA for diagonal matrices. 28. The matrix power A n is simply the product AAA A of n copies of A. It follows (by associativity) that parentheses don't matter: A r A s = ( AAA A) ( AAA A) = ( AAA A) = A r + s , r copies s copies r + s copies the product of r + s copies of A in either case. a b  1 0  29. (a + d ) A − (ad − bc )I = (a + d )   − (ad − bc ) 0 1  c d    (a + ad ) − (ad − bc) 2 ab + bd   a 2 + bc ab + bd  =  =    ac + cd (ad + d 2 ) − (ad − bc)   ac + cd bc + d 2  a b  a b  =  c d  = A 2 c d    2 1 30. If A =   then a + d = 4 and ad − bc = 3. Hence 1 2  2 1  1 0  5 4 A 2 = 4 A − 3I = 4   −3  =  ; 1 2 0 1  4 5
6. 6. 5 4 2 1  14 13 A 3 = 4 A 2 − 3A = 4   − 3 1 2  = 13 14  ; 4 5      14 13  5 4   41 40  A 4 = 4A3 − 3A 2 = 4   − 3  4 5  =  40 41 ; 13 14       41 40  14 13 122 121 A5 = 4A5 − 3A3 = 4   − 3 13 14  = 121 122  .  40 41      2 −1 1 5  31. (a) If A =   and B = 3 7  then  −4 3     3 4   1 −6   −25 −34  ( A + B )( A − B ) =    −7 −4  =  −71 −34   −1 10      but  8 −5  16 40   −8 −45 A 2 − B2 =   −  24 64  =  −44 −51 .  −20 13      (b) If AB = BA then ( A + B )( A − B ) = A ( A − B) + B( A − B) = A 2 − AB + BA − B 2 = A 2 − B 2 .  2 −1 1 5  32. (a) If A =   and B = 3 7  then  −4 3     3 4  3 4   5 52  ( A + B)2 =    −1 10 =  −13 96  −1 10     but  8 −5   2 −1 1 5  16 40  A 2 + 2 AB + B 2 =   + 2  −4 3  3 7  +  24 64   −20 13        8 −5   −1 3 16 40   22 41 =   + 2  5 1 +  24 64  = 14 79  .  −20 13        (b) If AB = BA then ( A + B )( A − B ) = A ( A − B) + B( A − B) = A 2 − AB + BA − B 2 = A 2 − B 2 . 33. Four different 2 × 2 matrices A with A2 = I are 1 0   −1 0  1 0   −1 0  0 1  ,  0 1  , 0 −1 , and  0 −1 .        
7. 7. 1 −1 If A =  ≠ 0 then A 2 = (0) A − (0)I = 0. 1 −1 34.    2 −1 35. If A =   ≠ 0 then A = (1) A − (0)I = A. 2  2 −1 0 1 If A =  ≠ 0 then A 2 = (0) A − ( −1)I = I. 1 0 36.    0 1 37. If A =   ≠ 0 then A = (0) A − (1)I = − I. 2  −1 0  0 1  0 1 38. If A =   ≠ 0 is the matrix of Problem 36 and B =  −1 0  ≠ 0 is the matrix 1 0   of Problem 37, then A + B = (I ) + (−I ) = 0. 2 2 39. If Ax1 = Ax2 = 0, then A ( c1x1 + c2 x 2 ) = c1 ( Ax1 ) + c2 ( Ax 2 ) = c1 (0 ) + c2 ( 0 ) = 0. 40. (a) If Ax0 = 0 and Ax1 = b, then A ( x 0 + x1 ) = Ax0 + Ax1 = 0 + b = b. (b) If Ax1 = b and Ax2 = b, then A ( x1 − x 2 ) = Ax1 − Ax 2 = b − b = 0. 41. If AB = BA then (A + B) 3 ( = ( A + B )( A + B ) = ( A + B ) A 2 + 2AB + B 2 2 ) ( ) ( = A A 2 + 2AB + B 2 + B A 2 + 2AB + B 2 ) = ( A + 2A B + AB ) + ( A B + 2AB 3 2 2 2 2 +B 3 ) = A + 3A B + 3AB + B . 3 2 2 3 (A + B) ( A + B ) = ( A + B )( A + B ) 4 4 3 To compute , write and proceed similarly, ( A + B ) just obtained. 3 substituting the expansion of
8. 8. 0 0 4  0 0 0  42. (a) 20 0 0  and N3 = 0 0 0  . Matrix multiplication gives N =     0 0 0    0 0 0    1 0 0  0 2 0  0 0 4 1 4 4 (b) 2 20 1 0 + 2  0 0 2  + 0 0 0  =  0 1 4  A = I + 2N + N =         0 0 1   0 0 0  0 0 0     0 0 1    1 0 0  0 2 0  0 0 4  1 6 12  3  0 1 0 + 3  0 0 2  + 3 0 0 0  =  0 1 6  A = I + 3N + 3N =  2        0 0 1    0 0 0    0 0 0    0 0 1    1 0 0 0 2 0 0 0 4 1 8 24 4 0 1 0 + 4  0 0 2  + 6  0 0 0  = 0 1 8  A = I + 4N + 6N =  2        0 0 1    0 0 0    0 0 0    0 0 1     2 −1 −1  6 −3 −3  43. 2  −1 2 −1 =  −3 6 −3 = 3A. Then First, matrix multiplication gives A =      −1 −1 2     −3 −3 6    A 3 = A 2 ⋅ A = 3A ⋅ A = 3A 2 = 3 ⋅ 3A = 9 A, A 4 = A 3 ⋅ A = 9 A ⋅ A = 9 A 2 = 9 ⋅ 3A = 27 A, and so forth.