Unformatted text preview: (b)The basis for the null space of A is ï£® ï£¯ ï£¯ ï£° 21 1 ï£¹ ï£º ï£º ï£» , ï£® ï£¯ ï£¯ ï£° 32 1 ï£¹ ï£º ï£º ï£» . 4. Solution: HINT: Since Q T Q = I and det( Q ) = det( Q T ), we have (det( Q )) 2 = 1. 5. Solution: HINT: B T = A (( A T A )1 ) T A T = A (( A T A ) T )1 A T = B . 6. Solution: (a)HINT: The eigenvalues of A are 1 , 2 , 3, and the corresponding eigenvectors are ï£® ï£° 11 1 ï£¹ ï£» , ï£® ï£° 1 1 ï£¹ ï£» , ï£® ï£° 1 1 ï£¹ ï£» , respectively. Hence P = ï£® ï£° 1 1 01 0 1 1 1 1 ï£¹ ï£» . (b) C = ï£® ï£° 46 161642 14 4210 16 40 ï£¹ ï£»...
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 Spring '06
 Dr.JeffChakFuWONG
 Linear Algebra, Algebra, Hong Kong Department, CHINESE UNIVERSITY OF HONG KONG Department of Mathematics, MAT2310 Linear Algebra, Examination Paper Note

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