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# HW1-sol - ECE 493 Univ of Illinois HW#1 – Version 1.00...

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Unformatted text preview: ECE 493 Univ. of Illinois HW #1 – Version 1.00 March 15, 2011 Due Thur, Mar 3 Spring 2011 Prof. Allen Topic of this homework: LinearAlgebra(Inverse of matrix, Cramer’s law, Gauss elimination, computing determinant) Deliverables: You best attempt at the questions. It is not in your best interest to answer questions you don’t understand (e.g., don’t copy stuﬀ from Wikipedia). 1. P407 1-(h) Derive the solution set for each of the following systems using Gauss elimination and augmented matrix format. Document each step(e.g. 1nd row → 2nd row → + 5 times 1st row), and classify the result(e.g. unique solution, the system is inconsistent, 3 parameter family of solutions, etc.) x1 + x2 − 2x3 = 3 x1 − x2 − 3x3 = 1 x1 − 3x2 − 4x3 = −1 Solution: 1 1 −2 3 1 −1 −3 1 1 −3 −4 −1 → 1 1 −2 3 0 −2 −1 −2 0 −2 −1 −2 → 1 1 −2 3 0 −2 −1 −2 00 0 0 1 parameter family of solution 2. P493 2-(e) Evaluate the determinant of given matrix using a cofactor expansion about the ﬁrst and last rows, and also about the last column. 123 234 345 Solution: first row 23 24 34 =0 +3 −2 34 35 45 12 13 23 =0 +5 −4 23 24 34 12 12 23 =0 +5 −4 23 34 34 third row 3 third column 3 3. Vandermonde determinant Show that for the real numbers x1 ,x2 , · · · xn , 1 x1 · · · 1 x2 · · · ··· 1 xn · · · = i<j n x1 −1 n−1 x2 n xn−1 (xj − xi ). Solution: Multiply each column by x1 and substract it from the next column on the right, starting from the right hand side. Then we can see that Vn = n (xi − i=2 x1 ) Vn−1 where Vn represents the n×n Vandermonde determinant. Then use an induction. 4. P522 1-(f )Compute the inverse of the given matrix. If it doesn’t exist, explain why. 12 1 A= 2 1 3 0 3 −1 Solution: if we compute the determinant,it is zero [ Det(A) = 1(−1 − 9)+2(3+2)+ 0(6 − 1) = −10 + 10 = 0 ]. Thus the inverse does not exist. 5. P523 5.(a),(d) Solve for x1 , x2 by Cramer’s rule. (a) x1 + 4x2 = 0, 3x1 − x2 = 6 (d) x1 + 2x2 + 3x3 = 9, x1 + 4x2 = 6, x1 − 5x3 = 2 ˛ ˛ ˛0 ˛ ˛ ˛6 ˛ ˛ ˛1 ˛ ˛ ˛3 Solution: (a)x1 = 4˛ ˛ −1˛ ˛ 4˛ ˛ −1˛ ˛ ˛ ˛ ˛ ˛ , x2 = ˛ ˛ ˛1 ˛ ˛ ˛3 ˛ ˛ ˛1 ˛ ˛ ˛3 ˛ 0˛ ˛ 6˛ ˛ 4˛ ˛ −1˛ ˛ ˛ ˛ ˛ ˛ ˛ ˛ (d) x1 = ˛ ˛9 ˛ ˛ ˛6 ˛ ˛ ˛2 ˛ ˛1 ˛ ˛ ˛1 ˛ ˛ ˛1 2 3˛ ˛ ˛ 4 0˛ ˛ ˛ 0 −5˛ ˛ , x2 = 2 3˛ ˛ ˛ 4 0˛ ˛ ˛ 0 −5˛ ˛ ˛ ˛1 ˛ ˛ ˛1 ˛ ˛ ˛1 ˛ ˛1 ˛ ˛ ˛1 ˛ ˛ ˛1 ˛1 2 9˛ 9 3˛ ˛ ˛ ˛ ˛ ˛ ˛ ˛1 4 6˛ 6 0˛ ˛ ˛ ˛ ˛ ˛ ˛ ˛1 0 2˛ 2 −5˛ ˛ , x3 = ˛ ˛ ˛1 2 2 3˛ 3˛ ˛ ˛ ˛ ˛ ˛ ˛ ˛1 4 4 0˛ 0˛ ˛ ˛ ˛ ˛ ˛ ˛ ˛1 0 −5˛ 0 −5˛ Version 1.00 March 15, 2011 ˜ /493/Assignments/HW #1 – Version 1.00 March 15, 2011 ...
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