Solq4 - Solutions to quiz 4(September 30 1 Computing the inverse we obtain 1 2 3 1 0 −→ 0 0 1 −1 01 1a 01 |100 1 −→ 0 |010 |001 0 | 1 00

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Unformatted text preview: Solutions to quiz # 4 (September 30) 1. Computing the inverse, we obtain 1 2 3 1 0 −→ 0 0 1 −1 01 1a 01 |100 1 −→ 0 |010 |001 0 | 1 00 10 −→ 0 1 | −2 1 0 | −5 1 1 00 1 a+2 4−a 0 1 −1 0| 0| 1| 0 0 1 6 −1 −1 5a − 2 1 − a −a . −5 1 1 1 |1 a | −2 1 − a | −3 0 1 0 Answer: The inverse matrix is 6 5a − 2 −5 −1 1−a 1 −1 −a . 1 2. Solving for X , we obtain ABXC = D =⇒ A−1 (ABXC )C −1 = A−1 DC −1 =⇒ A−1 A B X C C −1 = A−1 DC −1 =⇒ IBXI = A−1 DC −1 =⇒BX = A−1 DC −1 =⇒ B −1 (BX ) = B −1 A−1 DC −1 =⇒ B −1 B X = B −1 A−1 DC −1 =⇒ IX = B −1 A−1 DC −1 =⇒X = B −1 A−1 DC −1 . Answer: X = B −1 A−1 DC −1 . 1 ...
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This note was uploaded on 12/26/2011 for the course MATH 214 taught by Professor Conger during the Fall '08 term at University of Michigan.

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