ExmSol

# ExmSol - MATH1111/Exam-Outlined Solution 1 MATH1111 Linear...

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MATH1111/Exam-Outlined Solution/ 1 MATH1111 Linear Algebra 2007 Dec Section A . 1. B = E 4 E 3 E 2 E 1 A where E 1 = ± 0 1 1 0 ² ; E 2 = ± 1 0 ± 2 1 ² ; E 3 = ± 1 2 0 1 ² ; E 4 = ± 1 0 0 ± 1 ² : Hence, E = ± 2 ± 3 ± 1 2 ² . 2. (a) A basis for r ( A ) is f ( ± 2 ± 1 1 ± 3) ; (0 0 1 ± 1) g . (b) A basis for c ( A ) is f ( ± 2 2 4) T ; (1 1 1) T g . (c) N ( A ) = 8 > > < > > : 0 B B @ ± t ± s 2 s t t 1 C C A : s;t 2 R 9 > > = > > ; . A basis for c ( A ) is f ( ± 1 0 1 1) T ; ( ± 1 2 1 0 0) T g . 3. (a) The standard matrix repn. is ± 1 1 0 ± 1 0 2 ² . (b) As dim R 3 = 3, it su±ces to show that u 1 ;u 2 ;u 3 are linearly independent, which is equiv- alent to the nonvanishing of ³ ³ ³ ³ ³ ³ 1 1 1 0 1 0 ± 1 1 0 ³ ³ ³ ³ ³ ³ : As the last determinant is 1, the proof is done. (c) As [ T ( u 1 )] F = ( ± 3 1) T , [ T ( u 2 )] F = (1 2) T , [ T ( u 3 )] F = ( ± 1 1) T , the matrix repn relative to E and F is ± ± 3 1 ± 1 1 2 1 ² : 4. The characteristic equation is x 2 ± 1 = 0. The eigenvalues are 1 and ± 1. For ± = 1, an eigenvector is (4 1) T . For ± = ± 1, an eigenvector is (2 1) T . Hence, ± 4 2 1 1 ² ± 1 A ± 4 2 1 1 ² = ± 1 0 0 ± 1 ² . A 2007 = ± 4 2 1 1 ²± 1 0 0 ± 1 ² 2007 ± 4 2 1 1 ² ± 1 = A . 5. Bookwork. (The proof was explained in lecture.) 6. (a) No. Take x 1 = (1 0) T , and x 2 = x 3 = x 4 = (0 1) T . (b) No. Take A = 0 @ 1 0 0 1 1 1 1 A and B = 0 @ 1 0 0 1 0 0 1 A , then A and B are row equivalent but c ( A ) 6 = c ( B ), for (1 0 1) T = 2 c ( B ).

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MATH1111/Exam-Outlined Solution/ 2 (c) Yes. Let A be n ± n . Then dim N ( A ) = n ² rank( A ) = n ² rank( A T ) = dim N ( A T ) : (d) No. Take T (( x y z ) T ) = (2 x + z 2 z ) T and L (( x y z ) T ) = (2 x + y 2 y ) T . T and L are not equal as T ( e 3 ) 6 = L ( e 3 ). Section B . 1. See Assignment or Tutorial. 2. See See Assignment or Tutorial. 3. (a) Bookwork (A complete proof was discussed in lecture.) (b) As rank( A ) = 2, we get dim N ( A ) = 4 ² 2 = 2, so 0 is an eigenvalue of A and its algebraic multiplicity is at least 2. Since the sum of algebraic multiplicities equal 4, this forces that both 1 and ² 1 have algebraic multiplicity 1 and A has eigenvalues 0 ; 1 ; ² 1. The characteristic polynomial of A is p ( x ) = x 2 ( x ² 1)( x + 1) :
MATH1111/Exam-Outlined Solution/ 3 2008 May 1. See See Assignment or Tutorial. 2. See See Assignment or Tutorial. 3. (a) * (det X ) 4 = det( X 4 ) = 0, ) det X = 0 and X is singular. Hence, 0 is an eigenvalue of X . Let ± be an eigenvalue of X with an eigenvector x . Then X 4 x = ± 4 x = 0 . ) ± 4 = 0 ) ± = 0. (b) Assume X is diagonalizable. As X has the eigenvalue 0 only, S ± 1 XS = 0 for some nonsin- gular S . This implies X = 0. Contradict to our assumption. (c) Yes. ( I ± X )( I + X + X 2 + X 3 ) = ( I + X + X 2 + X 3 )( I ± X ) = I , so I ± X is invertible with the inverse I + X + X 2 + X 3 . 4. (a) (i) No. r (1) = 3. The zero polynomial is not belonged to U (ii) Yes. p (1) = 0. Let v 1 ;v 2 2 V . Then v 1 + v 2 2 P 3 and ( v 1 + v 2 )(1) = 0 + 0 = 0. ) v 1 + v 2 2 V . Similarly, ²v 2 V whenever v 2 V and ² 2 R . (b) Let c 1 p + c 2 q + c 3 r = 0. Then we get c 1 + 2 c 2 + c 3 = 0 ; c 2 ± 2 c 3 = 0 ; ± c 1 ± 3 c 2 + 4 c 3 = 0. Or equivalently, 0 @ 1 2 1 0 1 ± 2 ± 1 ± 3 4 1 A 0 @ c 1 c 2 c 3 1 A = 0 .

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