![0.4. biven- 21, -i, + iz = 5.9 + fa- -i = 0.7 -3i + 2 12 = - 3.1 matrix- make an @gym argumented Let US 5.9 o et ! iz -3.1 -](//img.homeworklib.com/questions/5e96db00-2c30-11eb-9a1c-e1a8dd69701a.png?x-oss-process=image/resize,w_560)
![Hence, Restoring the original mataise - 5.9 -2.25 3 - 574 2.375 0 Hence, le 2 + z = 5.9 3 13 = -2.25 pa 11 slur 2.375 0,6 2.3](//img.homeworklib.com/questions/630a56a0-2c30-11eb-8eef-2fc966ef88dc.png?x-oss-process=image/resize,w_560)
![3 O 1-1 O 1 0 -> I-n 0 1-3) R1 about -- Opening -> (1-9) ((1-0)2-0) -- (-1) + 0 = 0 => 11-093 +1.= 0](//img.homeworklib.com/questions/63eeaa30-2c30-11eb-956e-bdb1a553cfa5.png?x-oss-process=image/resize,w_560)
![=> 1-03_ 32 11-9).+!=0 - 1-23- 37 +372 +1 = 0 73-372 +39-2=0 2=2 n=1 10 Hence, the eigen values of A are 7=2, 3 and est](//img.homeworklib.com/questions/6503a720-2c30-11eb-97ed-c9c9945b34b5.png?x-oss-process=image/resize,w_560)
![b) Now, For eigen vectoas cones ponding to the eigen values calculated in part A- For 2= 2 - 1 x2 223 1-2 Xz=2 -xet x = 0 - X](//img.homeworklib.com/questions/65ea3d10-2c30-11eb-b0fa-df39d699a6e5.png?x-oss-process=image/resize,w_560)
![5 Xz -22 = -> x + x = 0 221 Ka = 22 +2₂=0 x = k x3=0 X, + Hence, the coresponding eigen vector, for n = 1/2 X3 K X2= -> X2 =](//img.homeworklib.com/questions/66e8ce30-2c30-11eb-a97c-3b1ec98e456e.png?x-oss-process=image/resize,w_560)
0.4. biven- 21, -i, + iz = 5.9 + fa- -i = 0.7 -3i + 2 12 = - 3.1 matrix- make an @gym argumented Let US 5.9 o et ! iz -3.1 - 3 R3 - R₂ + 3 R Rg - Ra elor 5.9 ia -2.25 O 13 5.75 3 (R3 R3 + 3 Rue) 5.9 i. 2. -2.25 ia iz 5 2.375 0 0
Hence, Restoring the original mataise - 5.9 -2.25 3 - 574 2.375 0 Hence, le 2 + z = 5.9 3 13 = -2.25 pa 11 slur 2.375 0,6 2.3 1}z=log 0_2. briven matrix - 9) A = Guven, AX= 2x → (A-TI) x = 0
3 O 1-1 O 1 0 -> I-n 0 1-3) R1 about -- Opening -> (1-9) ((1-0)2-0) -- (-1) + 0 = 0 => 11-093 +1.= 0
=> 1-03_ 32 11-9).+!=0 - 1-23- 37 +372 +1 = 0 73-372 +39-2=0 2=2 n=1 10 Hence, the eigen values of A are 7=2, 3 and est
b) Now, For eigen vectoas cones ponding to the eigen values calculated in part A- For 2= 2 - 1 x2 223 1-2 Xz=2 -xet x = 0 - Xq + 2z=0 xe, - &z=0 Hence, the conespondinding eigen vector, For n=2 1: X, = constant where K is a For n = 's I-5 22 dz
5 Xz -22 = -> x + x = 0 221 Ka = 22 +2₂=0 x = k x3=0 X, + Hence, the coresponding eigen vector, for n = 1/2 X3 K X2= -> X2 = K -ako -ak Hence since the the last eigen value is the eigen vector (X2 = X2) Hence, x = R · Yg= ées