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MATH 330 Exam 2 Solution

# MATH 330 Exam 2 Solution - Math 330 Exam 2 March 2 2009 No...

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Math 330 Exam 2, March 2, 2009 No calculators are allowed on the following. Be sure to show all your work, especially any row operations. Matrix Operations - 20 points Consider the following matrices: A = bracketleftBigg 1 2 3 0 - 1 - 1 bracketrightBigg , B = 1 5 1 0 0 10 Compute each of the following and answer the indicated questions (5 points each part) (a) What is A T ? What is A T + B ? SOLUTION: bracketleftBigg 1 2 3 0 - 1 - 1 bracketrightBigg T + 1 5 1 0 0 10 = 1 0 2 - 1 3 - 1 + 1 5 1 0 0 10 = 2 5 3 - 1 3 9 (b) Find a matrix C so that C + 3 A is the appropriate sized matrix with all entries zero. SOLUTION: C + 3 A = 0 means that C = - 3 A . So C = - 3 A = - 3 bracketleftBigg 1 2 3 0 - 1 - 1 bracketrightBigg = bracketleftBigg - 3 - 6 - 9 0 3 3 bracketrightBigg (c) Calculate BA . Does AB = BA ? Explain and/or calculate. SOLUTION: 1 5 1 0 0 10 bracketleftBigg 1 2 3 0 - 1 - 1 bracketrightBigg = 1 - 3 - 2 1 2 3 0 - 10 - 10 Since AB is a 2 × 2 matrix and BA is a 3 × 3 matrix they cannot be equal. (d) Specify any matrix D (with all entries non-zero), so that DB = bracketleftBigg 0 0 0 0 bracketrightBigg . We should make D a 2 × 3 matrix so that DB is 2 × 2. The two columns of B give weights for linear combinations of the columns of D which produce the zero vector. Therefore, with D = [ vector d 1 vector d 2 vector d 3 ] then we need (from the first column of B ) that vector d 1 + vector d 2 = vector 0. From the second column of B we need that 5 vector d 1 + 10 vector d 3 = vector 0. There are many possibilities, one of the simpler ones D = bracketleftBigg 2 - 2 1 2 - 2 1 bracketrightBigg 1

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Inverse Matrices - 20 points Find the inverses (or explain why these do not exist) of each of the following matrices: (10 pts each) (a) A = - 1 0 0 0 0 1 0 0 0 0 0 5 0 0 1 0 SOLUTION: - 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 5 0 0 1 0 0 0 1 0 0 0 0 1 R 1 ←- R 1 & R 3 R 3 5 = 1 0 0 0 - 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 0 1 5 0 0 0 1 0 0 0 0 1 R 3 R 4 = 1 0 0 0 - 1 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 1 5 0 So the inverse matrix is A - 1 = - 1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 5 0 (b) B = 1 1 0 2 3 2 - 1 0 1 SOLUTION: 1 1 0 1 0 0 2 3 2 0 1 0 - 1 0 1 0 0 1 R 2 R 2 - 2 R 1 & R 3 R 3+ R 1 = 1 1 0 1 0 0 0 1 2 - 2 1 0 0 1 1
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