Unformatted text preview: ±1 ² and rotating ± 1 ² by 3 π 2 radians yields ± 1 ² . Thus, T ( e 1 ) = T ³± 1 ²´ = ±1 ² T ( e 2 ) = T ³± 1 ²´ = ± 1 ² Hence, the standard matrix for T is [ T ( e 1 ) T ( e 2 ) ] = ± 11 ² . 3. (2.2.24) Suppose A is n × n and the equation A x = b has a solution for each b in n . Explain why A must be invertible. [ Hint: Is A row equivalent to I n ?] Since A x = b has a solution for each b in n , we can solve each of the n matrix equations A u 1 = e 1 A u 2 = e 2 ··· ··· A u n = e n Thus, A [ u 1 u 2 ··· u n ] = [ A u 1 A u 2 ··· A u n ] = [ e 1 e 2 ··· e n ] = I . This means that the matrix B = [ u 1 u 2 ··· u n ] satisFes AB = I , which implies that A is invertible....
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 Spring '03
 BUSS
 Linear Algebra, Algebra, Vectors, homogeneous matrix equation, radians yields

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