Consider the matrix
3
whose en vales
are 1. -1 and 3. Then the trace of (A - 3A²) is
Answers
Answer:
2L AB0;8 A;CG08 ?@52KH5=8O <0:A8<0;L=>9 AB02:8 B>?-10 10=:>2 1>;55 G5< =0 2 ?. ?. ! 22 45:01@O 2014 3>40 @53C;OB>@ @5:><5=4>20; 10=:0< =5 ?@52KH0BL C@>25=L A@54=59 <0:A8<0;L=>9 AB02:8 B>?-10 :@548B=KE >@30=870F89 1>;55 G5< =0 3,5 ?. ?., 0 4> :>=F0 2014-3> 4>?CA:0;>AL ?@52KH5=85 <0:A8<C< =0 2 ?. ?.
! :0:8<8 8A?KB0=8O<8 <>3CB AB>;:=CBLAO @>AA89A:85 D8=0=AK 8 @K=:8 2 =0ABC?0NI5< 3>4C, @5H8;8 2KOA=8BL "5=L38".
! A2O7K20NB A;01>ABL ?>:070B5;O A B@5<O >A=>2=K<8 D0:B>@0<8: A8;L=>9 107>9 =>O1@O 2016 3>40, A45;:>9 ?> A>:@0I5=8N 4>1KG8 + 8 1>;55 B5?;>9 ?>3>4>9 2 MB>< 3>4C. ;02=K9 M:>=><8AB "! ;>10; 0@:5BA" ;048<8@ "8E><8@>2 >B<5G05B, GB> A=865=85 ?@><KH;5==>3> ?@>872>4AB20 2 =>O1@5 2 3>4>2>< 2K@065=88 =0 3,6% AB0;> ?5@2K< A?04>< A O=20@O 2016 3>40. "%>BO 70<54;5=85 B5<?>2 @>AB0 ?@>8AE>48;> A 8N=O B5:CI53> 3>40", - A:070; >= 035=BAB2C "@09<".
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