CMI_2011_A2[S]

CMI_2011_A2[S] - 242: CMI-Assignment 2 [ A2-Sol ] [ Q1 ] [...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 242: CMI-Assignment 2 [ A2-Sol ] [ Q1 ] [ Total: 17 marks ] (a) Let ( E ) : 3 x + 2 y − z = 10 5 x − y − 4 z = 17 x + 5 y + az = b . (i) Find the values of a and b such ( E ) has infinitely many solutions. [ 4 ] (ii) Hence, find the solution set of ( E ) . [ 3 ] (b) Consider the system of linear equations: ( E ) x + 4 y + ( a + 2 ) z = 2 − 3 x − 12 y + 6 z = b 2 x + 8 y + az = 4 . (i) If ( E ) is consistent, find a and b . [ 4 ] (ii) Hence, find the solution set of ( E ) . [ 6 ] [ Sol ] (a) (i) The augmented matrix of ( E ) : 3 2 − 1 5 − 1 − 4 1 5 a 10 17 b −→ 1 5 a − 13 − 1 − 3 a − 26 − 4 − 5 a b 10 − 3 b 17 − 5 b −→ 1 5 a 0 1 1 + 3 a 13 0 0 a − 2 b 3 b − 10 13 b − 3 ··· ( E ′ ) ( E ) has infinitely many solutions if a − 2 = 0 and b − 3 = 0, i.e., a = 2, b = 3. (ii) ( E ′ ) becomes 1 5 2 0 1 7 13 0 0 3 − 1 13 , i.e., ¤ x + 5 y + 2 z = 3 y + 7 13 z = − 1 13 . Set z = t , we have y = − 1 13 − 7 13 t and x = 3 − 5 y − 2 z = 3 − 5 ( − 1 13 − 7 13 t ) − 2 t = 44 13 + 9 13 t ∴ S . S . = { 44 13 + 9 13 t , − 1 13 − 7 13 t , t | t ∈ R } . AD: College Mathematics I [ 2011-12 ] Louis Lau 1 242: CMI-Assignment 2 [ A2-Sol ] [ Sol ] (b) Reduce the augmented matrix of ( E ) to its row echelon form as follows: 1 4 a + 2 − 3 − 12 6 2 8 a 2 b 4 −→ 1 4 a + 2 0 0 3 a + 12 0 0 − a − 4 2 b + 6 −→ 1 4 a + 2 0 0 a + 4 0 0 2 1 3 ( b + 6 ) ··· ( E ′ ) (i) If ( E ) is consistent, then b = − 6 and a can be any number. (ii) (1) If b = − 6 and a ̸ = − 4, then ( E ′ ) becomes 1 4 a + 2 0 0 a + 4 0 0 2 , i.e., ¤ x + 4 y + ( a + 2 ) z = 2 ( a + 4 ) z = . ∴ z = 0. Let y = t , then x = 2 − 4 t . ∴ S . S . = { ( 2 − 4 t , t ,0 ) : t ∈ R } . (2) If b = − 6 and a = − 4, then ( E ′ ) becomes 1 4 − 2 0 0 0 0 2 , i.e., x + 4 y − 2 z = 2. Let z = t , y = s , then x = 2 − 4 s + 2 t . ∴ S . S . = { ( 2 − 4 s + 2 t , s , t ) : s , t ∈ R } . [ Q2 ] [ Total: 28 marks ] (a) Let A = 2 − 2 0 − 2 1 2 2 5 . (i) Find | A | , | 2 A | , | A 2 | and | A − 1 | . [ 4 ] (ii) Given that 2 − 1 1 and − 1 2 4 are eigenvectors of A , find their corresponding eigenvalues. [ 2 ] (iii) Given that − 1 is the third eigenvalues of A , find a corresponding normalised eigenvector. [ 3 ] (iv) State a matrix P and a diagonal matrix D such that P T AP = D . [ 4 ] (b) Let A = 1 0 2 0 4 0 2 0 1 ....
View Full Document

This note was uploaded on 02/22/2012 for the course CHEM yscn0027 taught by Professor Drtong during the Fall '10 term at HKU.

Page1 / 10

CMI_2011_A2[S] - 242: CMI-Assignment 2 [ A2-Sol ] [ Q1 ] [...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online