# H22-1 - Math 330 Homework 2.2(Pages 126-127(4 Find the...

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Unformatted text preview: Math 330 Homework 2.2 (Pages 126-127) (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 = A-1 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 D-1 exists, we may multiply on both sides by it: (B - C)D = 0 = (B - C)DD-1 = 0D-1 = 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 R2R2-4R1 & 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 R3R3-2R2 = 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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