T6 - ERG2018 Advanced Engineering Mathematics Tutorial...

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ERG2018 Advanced Engineering Mathematics Tutorial Notes 6 October 28, 2008 Examples 1. “Discuss” the rank of the matrix A , where A = x 1 2 1 x 1 1 1 1 . Solution : exchange the Row 1 and Row 3, by Guass elimination, we get 1 1 1 0 x - 1 0 0 0 2 - x . Then the determinant of matrix A det ( A ) = - ( x - 1)(2 - x ) . Case 1: x 6 = 1 and x 6 = 2, rank ( A )=3; Case 2: x = 1, 1 1 1 0 0 0 0 0 1 . There is a submatrix 1 1 0 1 , whose determinant is non-zero, hence rank A =2; Case 2: x = 2, 1 1 1 0 1 0 0 0 0 . 1
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There is a submatrix 1 1 0 1 , whose determinant is non-zero, hence rank A =2. 2. Let A be m × n and B be n × k , prove that rank( AB ) min { rank A , rank B } (hint: rank M + nullity M = #of the columns of M ). Solution : rank ( AB ) = dimension of the column vector space of matrix AB = dimension { y R m : y = ABx, for any x R k } dimension { y R m : y = Ax * , for any x * R n } = rank
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This note was uploaded on 11/13/2010 for the course SEEM SEEM2018 taught by Professor Chan during the Spring '10 term at CUHK.

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T6 - ERG2018 Advanced Engineering Mathematics Tutorial...

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