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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 Spring '11
 JontB.Allen
 Linear Algebra, Determinant

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