Unformatted text preview: Math 330 Homework 2.2 (Pages 126127)
(4) Find the inverse of the matrix: 3 4 7 8 SOLUTION: It is certainly okay to use the formula where we divide by 3(8)  (4)(7) = 4 and we swap the diagonal entries and negate the off diagonal entries to get 3 4 7 8
1 = 1 4 8 4 7 3 = 2 1 7 3 4 4 However, we should also be able to do this by the augmented matrix method: 3 4 1 0 7 8 0 1
R1R1+R2
7 R2R2 3 R1 = 3 4 1 0 4 0 3 7 1 3 = R23R2 = 3 4 1 0 0 4 7 3 = 3 0 6 3 0 4 7 3 R1 R1 & R2 R2 3 4 1 0 2 1 7 0 1 4 3 4 Which confirms our original answer. (6) Use the inverse found in exercise 3 to solve: 8x1 + 5x2 = 9 7x1  5x2 = 11 SOLUTION: Note that the system is equivalent to the matrix equation Ax = b where Ax = The solution is thus x = A1 b or 8 5 7 5 x1 x2 = 9 11 =b x1 x2 = 8 5 7 5 1 1 7 5 8 5 1 9 11 9 11 = 1 8(5)  (5)(7) 1 7 5 1 8 5 5 5 7 8 2  9 11 = 1 5 2  25 5 5 5 7 8 2 5 9 11 = = 9 + 11 = 63 5 88 5 = = Notice that since A is invertible the answer always exists and is unique no matter what the value of b might be. (14) Suppose that (B  C)D = 0 where B and C are m n matrices and D is invertible. Show that B = C. SOLUTION: Since D1 exists, we may multiply on both sides by it: (B  C)D = 0 = (B  C)DD1 = 0D1 = 0 = (B  C)I = B  C = 0 = B = C 1 (32) Find the inverse of the matrix, if it exists: SOLUTION: We augment using I3 and proceed towards reduced Echelon form: 1 2 1 4 7 3 2 6 4 1 2 1 1 0 0 1 2 1 1 0 0 R2R24R1 & R3R3+2R1 = 4 7 3 0 1 0 0 1 1 4 1 0 2 6 4 0 0 1 0 2 2 2 0 1 1 2 1 1 0 0 R3R32R2 = 0 1 1 4 1 0 0 0 0 10 2 1 At this point we can see that it will not be possible to get I3 on the left since we do not have three pivot columns within the original 3 3 matrix. Therefore, the inverse does not exist. (38) Construct a 4 2 matrix D using only 1 and 0 as entries, such that AD = I2. Is it possible that CA = I4 for some 4 2 matrix C? Why or why not? A= 1 1 1 0 0 1 1 1 SOLUTION: The 4 2 matrix D = [d1 d2 ] will have two columns. These columns each have 4 entries which 1 indicate the weights in a linear combination of the columns of A which produce (with d1 ) and 0 1 0 0 (with d2 ). Since the first column of A is and the last column of A is it should be 0 1 1 clear that since 1 and 0 we have that 1 0 +0 1 1 +0 1 1 +0 0 1 = 1 0 1 0 +0 1 1 +0 1 1 +1 0 1 = 0 1 D= 2 1 0 0 0 0 0 0 1 ...
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This note was uploaded on 09/19/2009 for the course MATH 330 taught by Professor Johnson,j during the Spring '08 term at Nevada.
 Spring '08
 Johnson,J
 Math, Linear Algebra, Algebra

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