HW3 Sol

HW3 Sol - EE 103, Winter 09, Prof SEJ: HW 3 Sol. Applied...

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EE 103, Winter ’09, Prof SEJ: HW 3 Sol. Page 1 / 12 Applied Numerical Computing Instructor: Prof. S. E. Jacobsen HW3 Solution Students: Distributed HW solutions are a component of the course and should be fully understood. Problem 1) We have included “arrows” to remind you that the components, especially the “ i b ”, are vectors. We know that we can add, transpose, and multiply block matrices if the corresponding blocks have the correct sizes (and if we pay attention to the order of multiplication as we multiply the blocks). We write A and B as follows: 1 12 2 ... , and n mn nk n b b Aa a a B b ×× ⎡⎤ == ⎣⎦ G G GG G # G , where i a G are 1 m × column vectors and j b G are 1 k × row vectors. A and B have the form of two block matrices now and we can treat each block as if it were a number, and as long as the dimensions match, we can proceed to multiply and sum. Normally we have: [] 1 2 1 1 nn n n y y x xx x yx y y ⎢⎥ =+ + "" # so, in the case of block matrices A and B, we can write: 1 21 2 2 1 ... . n j nj j n b b AB a a a a b a b a b a b b = ⇒= = + + + = G G G G G G G G G " # G the dimensions of each pair of matrices (row and column vectors) work for multiplications and summation. Each i j ab G G produces an mk × matrix and the result of the sum will also be × as we expect. Alternatively, the results can be shown as follows:
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EE 103, Winter ’09, Prof SEJ: HW 3 Sol. Page 2 / 12 1 2 jj j j nj j ab ⎛⎞ ⎜⎟ = ⎝⎠ G G G G # G The ( , ) th uv element of j j G G is ,, uj jv . Therefore, summing over j , the ( , ) th element of 11 nn ju j j v u a b == ∑∑ G G G G , The usual inner product of the th u row of A with th v column of B . Problem 2) IGE proceeds by using the main diagonal elements to eliminate the elements below the main diagonal, while leaving the diagonal elements unchanged. By interchanging rows, we pivot on the non-zero element with the largest absolute value. 12 1 0 01 1 1 13 1 2 14 2 4 A ⎡⎤ ⎢⎥ = ⎣⎦ , we proceed my making all the elements below the diagonal zero on the first column. 1 1 1 1 000 0100 01 11 , 1010 01 0 2 1001 02 3 4 PI E AE P A = = −− next, we interchange rows 2 and 4 (since on the second column, the largest element on the diagonal or below is 2 which is in row 4) , 22 1 1000 0001 , 0010 01 0 2 PP A =⇒ = we proceed to make all the elements below the diagonal zero on the second column. 2 2 1 100 0 1 2 10 010 0 0 23 4 0 1 / 210 003 / 2 4 01 / 2 0 1 0 0 1 / 21 EA E P A = = we do not need to interchange in the next step (since on the third column, the largest element on the diagonal or below is 3/2 ). we proceed to make all the elements below the diagonal zero on the third column.
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EE 103, Winter ’09, Prof SEJ: HW 3 Sol.
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This note was uploaded on 03/30/2009 for the course EE 103 taught by Professor Vandenberghe,lieven during the Winter '08 term at UCLA.

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HW3 Sol - EE 103, Winter 09, Prof SEJ: HW 3 Sol. Applied...

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