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# Sect3 5

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### Sect3 5

1. 1. SECTION 3.5 INVERSES OF MATRICES The computational objective of this section is clearcut — to find the inverse of a given invertible matrix. From a more general viewpoint, Theorem 7 on the properties of nonsingular matrices summarizes most of the basic theory of this chapter. In Problems 1-8 we first give the inverse matrix A −1 and then calculate the solution vector x.  3 −2   3 −2   5  3 1. A −1 =  ; x =    6  =  −2   −4 3   −4 3       5 −7   5 −7   −1  −26  2. A −1 =  ; x =    3  =  11   −2 3   −2 3       6 −7   6 −7   2   33  3. A −1 =  ; x =    −3 =  −28  −5 6   −5 6      17 −12  17 −12  5  25  4. A −1 =  ; x =   5 =  −10   −7 5   −7 5      1  4 −2  1  4 −2   5  1  8  5. A −1 =  −5 3  ; x =  −5 3  6  = 2  −7  2   2      1  6 −7  1  6 −7  10  1  25  A −1 = x = = 3  −3 4  3  −3 4   5  3  −10  6. ;        1  7 −9  1  7 −9   3 1  3  7. A −1 =  −5 7  ; x =  −5 7   2  = 4  −1 4   4      1 10 −15 1 10 −15 7  1  25  A −1 = x = = 5  −5 8  5  −5 8   3 5  −11 8. ;        In Problems 9-22 we give at least the first few steps in the reduction of the augmented matrix whose right half is the identity matrix of appropriate size. We wind up with its echelon form, whose left half is an identity matrix and whose right half is the desired inverse matrix. 5 6 1 0 R1− R 2  1 1 1 −1 R 2− 4 R1 1 1 1 −1 9. 4 5 0 1  → 4 5 0 1  → 0 1 −4 5       
2. 2. R1− R 2 1 0 5 −6   5 −6  →  0 1 −4 5  ; thus A −1 =      −4 5  5 7 1 0 R1− R 2 1 1 1 −1 R 2− 4 R1 1 1 1 −1 10. 4 6 0 1  → 4 6 0 1  → 0 2 −4 5        (1/ 2) R 2 1 1 1 −1 R1− R 2 1 0 3 − 7  1  6 −7  → 0 1 −2 5  →  2 ; thus A −1 =  −4 5   2   0 1 −2 5  2 2   1 5 1 1 0 0  R 2− 2 R1 1 5 1 1 0 0  11. 2 5 0 0 1 0 → 0 −5 −2 −2 1 0      2 7 1 0 0 1    2 7 1 0 0 1    R 3−2 R1 1 5 1 1 0 0  1 2 0 −1 0 1  R1+ R3 → 0 −5 −2 −2 1 0  → 0 −5 −2 −2 1 0       0 −3 −1 − 2 0 1     0 −3 −1 − 2 0 1    R 2− 2 R 3 1 2 0 −1 0 1  1 2 0 −1 0 1  R 3+3 R 2 →  0 1 0 2 1 −2  →  0 1 0 2 1 −2      0 −3 −1 −2 0 1    0 0 −1 4 3 −5    ( −1) R 3 R1−2 R 2  1 0 0 −5 −2 5   − 5 −2 5  → →  0 1 0 2 1 −2  ; thus A −1 =  2 1 −2       0 0 1 −4 −3 5     −4 −3 5    1 3 2 1 0 0  R 2− 2 R1 1 3 2 1 0 0  12. 2 8 3 0 1 0  → 0 2 −1 −2 1 0       3 10 6 0 0 1     3 10 6 0 0 1    R 3−2 R1 1 3 2 1 0 0  1 3 2 1 0 0  R 2− R 3 → 0 2 −1 −2 1 0  → 0 1 −1 1 1 −1     0 1 0 −3 0 1     0 1 0 −3 0 1    1 3 2 1 0 0  R1−3 R 2 1 0 0 18 2 −7  R 3− R 2 → 0 1 −1 1 1 −1 → → 0 1 0 −3 0 1  ;     0 0 1 −4 −1 2    0 0 1 −4 −2 2    18 2 −7  thus A −1 =  −3 0 1     −4 −1 2   
3. 3. 2 7 3 1 0 0 SWAP ( R1, R 2) 1 3 2 0 1 0  13. 1 3 2 0 1 0 → 2 7 3 1 0 0     3 7 9 0 0 1    3 7 9 0 0 1    R 2− 2 R1 1 3 2 0 1 0  1 3 2 0 1 0  R 3−3 R1 →  0 1 −1 1 −2 0  →  0 1 −1 1 −2 0      3 7 9 0 0 1    0 −2 3 0 −3 1    1 0 0 −13 42 −5  −13 42 −5  R 3+ 2 R 2 → →  0 1 0 3 −9 1  ;   thus A −1 =  3 −9 1    0 0 1 2 −7 1     2 −7 1    3 5 6 1 0 0   1 1 3 1 −1 0  R1− R 2 14. 2 4 3 0 1 0  → 2 4 3 0 1 0     2 3 5 0 0 1    2 3 5 0 0 1     1 1 3 1 −1 0   1 1 3 1 −1 0  R 2− R 3 R 3− 2 R1 →  0 1 −2 0 1 −1 → 0 1 −2 0 1 −1     2 3 5 0 0 1     0 1 −1 −2 2 1    1 0 0 11 −7 −9   11 −7 −9  R 3− R 2 → →  0 1 0 −4 3 3  ;   thus A −1 =  −4 3 3     0 0 1 −2 1  2   −2 1  2 1 1 5 1 0 0  1 1 5 1 0 0  R 2− R1 15. 1 4 13 0 1 0  →  0 3 8 −1 1 0      3 2 12 0 0 1    3 2 12 0 0 1    R 3−3 R1 1 1 5 1 0 0  1 1 5 1 0 0  R 2+ 2 R 3 →  0 3 8 −1 1 0  → 0 1 2 −7 1 2      0 −1 −3 −3 0 1    0 −1 −3 −3 0 1    1 0 0 −22 2 7   −22 2 7  R 3+ R 2 → → 0 1 0 −27 3 8  ;   thus A −1 =  −27 3 8    0 0 1 10 −1 −3    10 −1 −3  
4. 4.  1 −3 −3 1 0 0  R 2+ R1 1 −3 −3 1 0 0  16.  −1 1 2 0 1 0  →  0 −2 −1 1 1 0       2 −3 −3 0 0 1     2 −3 −3 0 0 1    R 3−2 R1  1 −3 −3 1 0 0  1 −3 −3 1 0 0  R 2+ R 3 → 0 −2 −1 1 1 0 →  0 1 2 −1 1 1      0 3 3 −2 0 1    0 3 3 −2 0 1    R 3−3 R 2  1 −3 − 3 1 0 0  ( −1/3( R 3)  1 −3 − 3 1 0 0  0 1 2 −1 1 1    →   →  0 1 2 −1 1 1   0 0 −3 1 −3 −2    0 0 1 − 1 1 3   3 2  R1+3 R 2  1 0 0 −1 0 1   −3 0 3    =  −1 −3 −1 1 → →  0 1 0 − 1 −1 − 1  ; 3 3 thus A −1  3 0 0 1 − 1 1  3 2  3   −1 3 2     1 −3 0 1 0 0   1 −3 0 1 0 0  R 2+ R1   →  0 −1 −1 1 1 0  17.  −1 2 −1 0 1 0     0 −2 2 0 0 1     0 −2 2 0 0 1    ( −1) R 2  1 −3 0 1 0 0  R 3+ 2 R 2  1 −3 0 1 0 0 → 0 1 1 −1 −1 0  → 0 1 1 −1 −1 0      0 −2 2 0 0 1    0 0 4 −2 −2 1    (1/ 4) R 3 1 0 0 − 1 2 −2 3 − 4 3  −2 −6 −3 → 0 1 0 − 1 − 1; 1  →  2 −1 2 4 thus A −1 =  −2 −2 −1 4 0 0 1 − 2  1 −2 1 1  4   −2 −2 1    1 −2 2 1 0 0  R 2−3 R1  1 −2 2 1 0 0    →  0 6 −5 −3 1 0  18. 3 0 1 0 1 0    1 −1 2 0 0 1     1 −1 2 0 0 1     1 −2 2 1 0 0  R 2−5 R 3  1 −2 2 1 0 0  R 3− R1 → 0 6 −5 −3 1 0  → 0 1 −5 2 1 −5     0 1 0 −1 0 1    0 1 0 −1 0 1     1 −2 2 1 0 0  1 0 0 1 2 − 5 2 R 3− R 2 5 5 0 1 −5 2 1 −5 (1/5) R 3  1 ; →   → → 0 1 0 −1 0   0 0 5 −3 −1 6    0 0 1 − 3 − 1  5 5 6  5 
5. 5.  1 2 −2  =  −5 0 5  . −1 1 thus A  5  −3 −1 6    1 4 3 1 0 0  1 4 3 1 0 0 R 2− R1 19. 1 4 5 0 1 0  →  0 0 2 −1 1 0      2 5 1 0 0 1    2 5 1 0 0 1    SWAP ( R 2, R 3) 1 4 3 1 0 0  R 2− 2 R1 1 4 3 1 0 0  → 2 5 1 0 0 1  → 0 −3 −5 −2 0 1       0 0 2 −1 1 0     0 0 2 −1 1 0    ( −1/3) R 2 1 4 3 1 0 0  1 0 0 − 72 11 6 4 3  − ; (1/ 2) R 3 → 0 1 5 2 0 − 1  → → 0 1 0 23 − 5 1  3 3 3 6  3  0 0 2 −1 1 0    0 0 1 − 1  2 1 2 0   −21 11 8  =  9 −5 −2  −1 1 thus A  6  −3 3 0     2 0 −1 1 0 0  1 0 −4 1 −1 0  R1− R 2 20. 1 0 3 0 1 0  → 1 0 3 0 1 0      1 1 1 0 0 1    1 1 1 0 0 1     1 0 −4 1 −1 0   1 0 −4 1 −1 0  R 3− R1 SWAP ( R 2, R 3) → 1 0 3 0 1 0 →  0 1 5 −1 1 1      0 1 5 −1 1 0    1 0 3 0 1 0     1 0 −4 1 − 1 0  1 0 0 73 1 0 R 3− R1 7   (1/7) R 3   → 0 1 5 −1 1 1  → → 0 1 0 − 72 − 3 7 1 ; 0 0 7 −1 2 0    0 0 1 − 1  7 2 7 0   3 1 0 =  −2 −3 7  1 thus A −1  7  −1 2 0   
6. 6. 0 0 1 0 1 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 SWAP ( R1, R 2) 0 0 1 0 1 0 0 0 21.   →   0 1 2 0 0 0 1 0 0 1 2 0 0 0 1 0     3 0 0 1 0 0 0 1 3 0 0 1 0 0 0 1 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 SWAP ( R 2, R 3) 0 1 2 0 0 0 1 0 R 4 −3 R1 0 1 2 0 0 0 1 0 →   →   0 0 1 0 1 0 0 0 0 0 1 0 1 0 0 0     3 0 0 1 0 0 0 1 0 0 0 1 0 −3 0 1 1 0 0 0 0 1 0 0 0 1 0 0 R 2− 2 R 3 0 1 0 0 −2 0 1 0  −2 0 1 0 →  ; thus A −1 =   0 0 1 0 1 0 0 0 1 0 0 0     0 0 0 1 0 −3 0 1  0 −3 0 1 4 0 1 1 1 0 0 0 1 −1 −2 0 1 −1 0 0 3 1 3 1 0 1 0 0 R1− R 2 3 1 3 1 0 1 0 0 22.   →   0 1 2 0 0 0 1 0 0 1 2 0 0 0 1 0     3 2 4 1 0 0 0 1 3 2 4 1 0 0 0 1 1 −1 −2 0 1 −1 0 0 1 −1 −2 0 1 −1 0 0 R 2−3 R1 0 4 9 1 −3 4 0 0 R 4−3 R1 0 4 9 1 −3 4 0 0 →   →   0 1 2 0 0 0 1 0 0 1 2 0 0 0 1 0     3 2 4 1 0 0 0 1  0 5 10 1 −3 3 0 1 1 −1 −2 0 1 −1 0 0 1 −1 −2 0 1 −1 0 0 R 2−3 R3 0 1 3 1 −3 4 −3 0 R3− R 2 0 1 3 1 −3 4 −3 0 →   →   0 1 2 0 0 0 1 0 0 0 −1 −1 3 −4 4 0     0 5 10 1 −3 3 0 1 0 5 10 1 −3 3 0 1 1 0 0 0 1 −1 1 0   1 −1 1 0  R 4−5 R 2 0 1 0  0 0 −2 −1 2   0 −2 −1 2  → →  ; thus A =  −1  0 0 1 0 0 1 1 −1  0 1 1 −1     0 0 0 1 −3 3 −5 1   −3 3 −5 1  In Problems 23-28 we first give the inverse matrix A −1 and then calculate the solution matrix X.  4 −3   4 −3  1 3 −5  7 18 −35 23. A −1 =  ; X =    −1 −2 5  =  −9 −23 45   −5 4   −5 4     
7. 7.  7 −6   7 −6   2 0 4   14 −30 46  24. A −1 =  ; X =   0 5 −3 =  −16 35 −53  −8 7   −8 7       11 −9 4   11 −9 4   1 0 3   7 −14 15  25. A −1 =  −2 2 −1 ;    −2 2 −1  0 2 2  =  −1 3 −2  X =       −2 1 0     −2 1 0   −1 1 0      −2 2 −4     −16 3 11   −16 3 11   2 0 1   −21 9 6  26. A −1 =  6 −1 −4  ;    6 −1 −4   0 3 0  =  8 −3 −2 X =       −13 2 9     −13 2 9  1 0 2      −17 6 5     7 −20 17   7 −20 17  0 0 1 1  17 −20 24 −13 27. A −1 0 =  −1 1 ; X =  0  0 1 0 1  =  1 −1 1   −1 1 −1       −2 6 −5     −2 6 −5   1 0 1 0      −5 6 −7 4     −5 5 10   −5 5 10   2 1 0 2   −5 5 10 1  28.  −8 8 A =−1  ; X =  −8 8 15    −1 3 5 0  =  −8 8 15   15 7       24 −23 −45    24 −23 −45  1 1 0 5   24 −23 −45 −13      29. (a) The fact that A–1 is the inverse of A means that AA −1 = A −1A = I. That is, that when A–1 is multiplied either on the right or on the left by A, the result is the identity matrix I. By the same token, this means that A is the inverse of A–1. (b) A n ( A −1 ) n = A n −1 ⋅ AA −1 ⋅ ( A −1 ) n −1 = A n −1 ⋅ I ⋅ ( A −1 ) n −1 = = I. Similarly, ( A −1 )n A n = I , so it follows that ( A −1 ) n is the inverse of A n . 30. ABC ⋅ C−1B −1A −1 = AB ⋅ I ⋅ B −1A −1 = A ⋅ I ⋅ A −1 = I, and we see is a similar way that C−1B −1A −1 ⋅ ABC = I. 31. Let p = − r 0, q = − s 0, and B = A −1. Then A r A s = A − p A − q = ( A −1 ) p ( A −1 ) q = B pBq = B p+q (because p, q 0) = ( A −1 ) p + q = A − p − q = A r + s as desired, and ( A r )s = ( A − p ) − q = (B p ) − q = B − pq = A pq = A rs similarly. 32. Multiplication of AB = AC on the left by A–1 yields B = C.
8. 8. 33. In particular, Ae j = e j where e j denotes the jth column vector of the identity matrix I. Hence it follows from Fact 2 that AI = I, and therefore A = I–1 = I. 34. The invertibility of a diagonal matrix with nonzero diagonal elements follows immediately from the rule for multiplying diagonal matrices (Problem 27 in Section 3.4). The inverse of such a diagonal matrix is gotten simply by inverting each diagonal element. 35. If the jth column of A is all zeros and B is any n × n matrix, then the jth column of BA is all zeros, so BA ≠ I. Hence A has no inverse matrix. Similarly, if the ith row of A is all zeros, then so is the ith row of AB. 36. If ad – bc = 0, then it follows easily that one row of A is a multiple of the other. Hence the * *  reduced echelon form of A is of the form   rather than the 2 × 2 identity matrix. 0 0  Therefore A is not invertible. 37. Direct multiplication shows that AA −1 = A −1A = I. 3 0   a b  3a 3b  38. EA =   c d  =  c d  0 1      1 0 0   a11 a12 a13   a11 a12 a13  39. EA = 0 1 0   a21   a22  =  a a23   21 a22 a23    2 0 1   a31   a32 a33    a31 + 2a11  a32 + a12 a33 + a13   0 1 0   a11 a12 a13   a21 a22 a23  40. EA = 1 0 0   a21   a22  = a a23   11 a12 a13   0 0 1   a31   a32 a33    a31  a32 a33   41. This follows immediately from the fact that the ijth element of AB is the product of the ith row of A and the jth column of B. 42. Let ei denote the ith row of I. Then ei B = B i , the ith row of B. Hence the result in Problem 41 yields  e1   e1B   B1  e  e B  B  IB =   B =  2  2 =  2  = B.            e m    e m B    B m   
9. 9. 43. Let E1 , E2 , , Ek be the elementary matrices corresponding to the elementary row operations that reduce A to B. Then Theorem 5 gives B = Ek Ek −1 E2 E1A = GA where G = E k E k −1 E2 E1. 44. This follows immediately from the result in Problem 43, because an invertible matrix is row-equivalent to the identity matrix. 45. One can simply photocopy the portion of the proof of Theorem 7 that follows Equation (20). Starting only with the assumption that A and B are square matrices with AB = I, it is proved there that A and B are then invertible. 46. If C = AB is invertible, so C–1 exists, then A(BC−1 ) = I and (C−1 A)B = I. Hence the fact that A and B are invertible follows immediately from Problem 45.