Question

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Diagonalize the matrix A, if possible. That is, find an invertible matrix Pand a diagonal matrix D such that A = PDP-1. A = -11 0 6 3 -5 -3 -91 0 4 12 A = 1 LO 0 0 2 0 0 2 0 0 0 9 A= 9 0 -16 0 0 0 16 9 4 1 0 0

Answer 1

SOLUTIONO 3 -9 : 6 Its eigen valve equation: 1 A-Il=0 -11-1 - 9 3 - 5-1 6 - 3 4-1 O => => 1-11-1) [C-5-1) (0-1)] – 9( 30+61] = (5+1) [(1+1)(4-1) - que] =0 (5+1) [-12-74 +44 - 59] =0 (5+2) [+²+70+10] =0 (5+1) ++s) C+2) = 0 - - 2, - - 5 4 3 = -5 Eigen vector corresponding 1,5-2 => [A+24]x=0 T-/+2 3 - 9 -5+2 -3 4+2 61-63 6 3 -9 x JE [8] -3 23

R, R, + R2 + R₂ R3-R2 9 -9 - 3 22 6 6 X3 - R3 R3 + 6 RI 9 - 9 ² -3 : 42 +1:] x3 here xz is free Cariable choose x=1 = -92, ~983=0>) 2,5-1 - 3x₂=0 X2=0 Hence eigen vector corresponding 4,==2 X, - (7) Now eigen vector corresponding =-5 [A+5]]= 0 = ) 6 3 -9 24 x2 13 o xz 6 -3 9 R₂ R₂ + R, 6 9 xz

Here x2 + x both are free varrabe first Choose R2 = x3=0 - 6x + 3x2 - 9x3=0 - 6x + 3 =0 2 = 1/2 Hence Y2 Now Choose - 6x + 372-923=0 = 3/2 x=9 Hence x3 = 3/2 D since all three eigen vectors are dinearly Independent. Hence Given matrix is diagonalizible. The diogonalizing matix P: [x, x2 *3] e is given by -1 Y2 3/2 Hence D = P'AP= -2 o C A, 2 o OO 13 1: -5

Since questions are too lenghty and time consuming I solved only one. Do the rest same.