Question

Could you please help me to solve these three questions. Thanks!

Diagonalize the matrix A, if possible. That is, find an invertible matrix P and a diagonal matrix D such that A- PDP-1

Answer 1

Ti 147 A=10 -4 o s -1-8] TO(A)=-11 A = 32 128= -8 +20=12 A33= -4 Air= 40 1A1= 32- (80) = 32-80 =-48 characteristic equation is ginen by 3 + 11 d² +40 x +48=0 - (+3)(x+4)²=0 d=-3,-4-4 Now eigen vector for d=-4 5a + b + 4c=0 5 1 47 55 147 Cal b 0 0 0 ~ Ooo || Let b=OP CE-S 6=4 L-3--ul loo olol lo Let c=0 & 6=-5 U=(4, 0 =(1, -5,0 g a=1 For d=-3 a+c=0 14 147 Ti 0 -1 10 0 0 0 let c=-1 b=0 L-5 -1 -5 Loo a=1 w=(1,0,-1) Hence transition matrix p is given by 4 1 1 1-4 oo p= 1o-5 o 8 D= o-uo 1-5 0-1 Too-3 J We know that eigen vectors are not unique therefore correct answer is (B). 16 oo A= 6o Too6J since A is lower triangelar matrix therefore eigenvalues of A are 6, 6, 6. EJE

for d=6 let us find the eigen vector rio07 Toon oool Looo oool! looo7 a = 0,26&c are free variable. => Geometric multiplicity of (d=6)=2 # Algebric multiplicity=6) Hence A is not sliceguenziable.diagonalizable 3 78 0007 A= To-8 oo 11 -4 8 o W 2 0 8 Again A is lower triangular matrix therefore, eigen values of A = -8,-8, 8,8 How eigen vector for d=-8 To o o o7 T-4 1607 ro-4 16 0707 To ooola To-2 16 16.0 6-8-8162 0 0 0 0 0 0 0 0 0 L-1 2 0 16 lo o o o o o o alla aubt 160 = 0 6-86-8d=0 Let c=0 & d=1 6=8 a-32=0 9=32 u= (32,8,0,1) et d=0 e=1 6=8 a-32 +16=0 a=16 For L = (16,8,1,0) d=8 . 1-16 ooo y azo b=0 1 0 1 10-16 0 0 0 0 16-O 8 c, d are free variable 11 -4 oo let C=0 &d=1 L-1 20 w=(0,0,0,1) Let e=;,d=0 x=(0,0,1,0) 32 16ool 8 ooo7 p=8 8 ool D-o-ool o I 01 0 1 0 0 Looo8J OOOO reence To o 8o (B)