Solution_2 - THE CHINESE UNIVERSITY OF HONG KONG Department...

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Unformatted text preview: THE CHINESE UNIVERSITY OF HONG KONG Department of Mathematics MAT2310 Linear Algebra and Applications (Fall 2007) Homework 2 Suggested Solutions 1. (a) • (-2) times the first row was added to its second row, • (-3) times the first row was added to its third row, we obtain 1 0 1 0 2 0 0 4 2 (-2) times the second row was added to its third row, we obtain 1 0 1 0 2 0 0 0 2 Hence we obtain the elementary matrices, E 1 = 1 0 0- 2 1 0 0 0 1 E 2 = 1 0 0 0 1 0- 3 0 1 E 3 = 1 0 0 1 0- 2 1 and the corresponding upper matrix is U = 1 0 1 0 2 0 0 0 2 . (b) The inverse of elementary matrices E 1 , E 2 , E 3 are E- 1 1 = 1 0 0 2 1 0 0 0 1 E- 1 2 = 1 0 0 0 1 0 3 0 1 2 E- 1 3 = 1 0 0 0 1 0 0 2 1 respectively. And the lower triangular matrix L is equal to L = 1 0 0 2 1 0 3 2 1 . After some algebraic calculations, we obtain LU = 1 0 0 2 1 0 3 2 1 1 0 1 0 2 0 0 0 2 = 1 0 1 2 2 2 3 4 5 = A 2. Let U be an n × n upper triangular matrix with non-zero diagonal entries. (a) i. First, the elementary row operation (II) can be used to change the diagonal entries, u ii , all to ones. ii. Then, the system ˜ U x = will be in row echelon form, and in triangular form. This implies that the system ˜ U x = will have a unique solution x = . iii. Thus by the list of non-singular equivalence (cf. Lecture Note 1-2, page 43), ˜ U is non-singular, and U is row equivalent to ˜ U . Hence, U is also non-singular. (b) When the augmented matrix form [ U | I ] is done to find the inverse of U , (or U- 1 ), all of the entries which need the elementary row operation (III) to eliminate are above the diagonal. Hence, using the elementary row operation will get the identity matrix I in its upper diagonal values only, which will lead to an upper diagonal matrix for U- 1 ....
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Solution_2 - THE CHINESE UNIVERSITY OF HONG KONG Department...

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