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Math 21 - 451(4)in If Β 2 0 O A β€”-6 l ~β€”4 β€”3 0 β€”1...

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Unformatted text preview: 451(4)in If Β» 2 0 O A: β€”-6 l ~β€”4 , β€”3 0 β€”1 then pat) = β€”β€”det(A β€”β€” AI) 2β€”)Β» 0 0 =β€”- β€”6 1~A β€”4 ~3 0 β€”-1β€”A Expanding across the first row, 170Β») = β€˜(2 β€” 1) 1β€”11 β€”-4 0 β€”-1β€”)t = (A -2)(1 ~ A)(β€”β€”l β€”~ A) = (A β€”β€” 2)()Β» ~ 1)(}t +1). Thus, the eigenvalues are 2, 1, and β€”l, respectively. Because A β€” 21 reduces, 0 O 0 1 0 1 A~β€”21= β€”6 β€”1 - β€”β€”> 0 1 β€”2 , β€”-3 0 β€”3 0 O 0 it is easily seen that the nullspace of A β€” 21 is generβ€” ated by the eigenvector v1 = (β€”1,2, 1)T. In a sim- ilar manner, we arrive at the following eigenvalue- eigenvector pairs. 0 0 1 β€”> 1 ,, and ~ 1 β€”-> 2 0 1 Because β€”1 0 0 det 2 1 2 : β€”1, 1 O 1 the eigenvectors are independent. . If 1 O 0 A = 3 β€”-Z 1 , 5 β€”5 2 then β€˜ pot) 2: β€”-det(A β€” M) 1 β€” A 0 0 = β€” 3 β€”β€”2 β€” it 1 . 5 β€”5 2 -β€” A Expanding across the first row, V β€”2 β€” A 1 : (A β€” 1)((~2 β€” m2 ~ A) + 5) = (A ~ 1)(A2 +1) Thus, the eigenvalues are 1, i, and β€”i, respectively. Because A ~ 1'] reduces, β€˜ - lβ€”i 0 0 Aβ€”il: 3 β€”2β€”β€”i 1 5 ~5 Zβ€”i l 0 0 β€”> 0 1 ~2/5+1/5i , 0 0 0 it is easily seen that the nullspaces of A ~ i I is gener- ated by the eigenvector v z (0, 2 ~ 1', 5)T. In a simβ€” ilar manner, we arrive at the following eigenvalueβ€” eigemvector pairs. /0 1 fii_+5\2+i} and 1β€”β€”-> 1 5 . Because the eigenvectors are independent. ...
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