081st_tut3sol

# 081st_tut3sol - MATH1111/2008-09/Tutorial III Solution 1...

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Unformatted text preview: MATH1111/2008-09/Tutorial III Solution 1 Tutorial III Suggested Solution 1. Let A and B be square matrices of the same order. Show that AB = I if and only if BA = I . In other words, B is the inverse of A if either AB = I or BA = I . [ Remark . Recall that by definition, B is the inverse of A if both AB = I and BA = I . This exercise tells that actually one of them can guarantee B to be the inverse.] Ans . “only if” part: Suppose AB = I . Then det( AB ) = det I = 1. As A and B are of the same size, det( AB ) = (det A )(det B ). So (det A )(det B ) = 1. ∴ det A 6 = 0 and hence A is nonsingular. Multiply A- 1 from the left, we have A- 1 AB = A- 1 I IB = A- 1 B = A- 1 . As B is the inverse of A , BA = I in view of the definition. “if” part: The proof is similar. Leave to you! MATH1111/2008-09/Tutorial III Solution 2 2. (a) Justify that det( I- A ) 6 = det I- det A in general. (b) Let A be a square matrix such that A m = 0 for some m ≥ 1....
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081st_tut3sol - MATH1111/2008-09/Tutorial III Solution 1...

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