ProblemSet5 Solutions

# ProblemSet5 Solutions - Department of Economics Ryerson...

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Department of Economics Ryerson University ECN230-Mathematics For Economics Problem Set #5 - Solutions EXERCISE 4.6 Problem 4: We consider all possible matrix product combinations of the four matri- ces and check which ones yield the identity matrix. It turns out that: DF = 10 01 ¸ and EG = ¸ . Hence, D and F are inverses of each other, as well as E and G . Problem 5: Let D = AB . Then: ( ABC ) 1 =( DC ) 1 = C 1 D 1 = C 1 ( AB ) 1 = C 1 ¡ B 1 A 1 ¢ = C 1 B 1 A 1 . Problem 6: (a) A and X 0 X must be square, say of dimension n × n .Ma t r ix X only needs to be of dimension n × m ,where m is not necessarily equal to n . (b) Recall that I is itself an idempotent matrix. In other words, II = I . The objective in this exercise is to show that AA = A .U s ingthede f nition of A : AA = h I X ¡ X 0 X ¢ 1 X 0 ih I X ¡ X 0 X ¢ 1 X 0 i . Using the result from Exercise 4.4-Problem 6, we can write: AA = IX ¡ X 0 X ¢ 1 X 0 X ¡ X 0 X ¢ 1 X 0 I + X ¡ X 0 X ¢ 1 X 0 X ¡ X 0 X ¢ 1 X 0 . 1

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Note that X 0 X ( X 0 X ) 1 = I . Therefore, AA = II IX ¡ X 0 X ¢ 1 X 0 X ¡ X 0 X ¢ 1 X 0 I + X ¡ X 0 X ¢ 1 0 . Using the properties of the identity matrix, the previous expression is equiv- alent to: AA = I X ¡ X 0 X ¢ 1 X 0 X ¡ X 0 X ¢ 1 X 0 + X ¡ X 0 X ¢ 1 X 0 = I X ¡ X 0 X ¢ 1 X 0 = A .
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ProblemSet5 Solutions - Department of Economics Ryerson...

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