Question

Let E be the matrix given by -3 E= -12 -1 -1] 0 5 -2 -1] Determine if E is diagonalizable. If it is, find a basis of C consisting of eigenvectors of E.

Answer 1

-- 3 0 5 sol @ To Find use is diagonalizable. D, P-AP (gt Ps stalishes this, then 81 is Dragonalizable Eigen values for it Now we hind that is det (€ - 1I) -1 z -3-1 - -19 ord 5 29 -|- - 1 f-1/2-) ---- [ - _-] *13-1) Ed)(-+-)+10)+1(812) 6-1-0)-20] -[auth] z (-3-1) 1+12410] + [18418d-20)-24-ud 23-312-30 ---73-307. 19+ 180-20-24-yt z-/3-312+7) -96. so eigen values are 7,2-2, dzz-2V5-1 and &2 2V5-1 Call eigen values are No w Eigen veche conosponding to each distino -/ I for diz-2 12 U 32 3V5+7 -485 +7 Azz 2891 22V5-1 for dat eigen value. 2 for 2 34567 1 UN+7 4

Vo 8:01 PM 0.OKB/s na 51 Х Eigenvalues and Eigenv... https://matrixcalc.org Clean 1. Find eigenvalues from the characteristic polynomial -3-1 -1 -1 -12 0-2 -5 4 -2 -1-2 = 11 -23-4*2 2 + 15 x 2 + 38 (?) -(2 + 2) * (22+ 2 x - 19) (?) -(2 + 2) * (1 + (2 + (2 X V5 + 1)) } (1-(2x V5 - 1)) Details (Triangle's rule) 1.11=-2 2. 12 = -2 * V5 - 1 3.13 = 2 * 15-1 2. For every 2 we find its own vector(s): 1. 11 = -2 -1 -1 -1 A-211 = -12 2 -5 4 -2 1

8:01 PM 0.6KB/s Vo LTE 51 1 1.11--2 2. 12 = -2 * V5 - 1 3.13 = 2 V5 - 1 2. For every 2 we find its own vector(s): 1. 11 = -2 -1 -1 -1 A-211 = | -12 2 -5 4 -2 1 TIL Av = av * (A - 2)v = 0 So we have a homogeneous system of linear equations, we solve it by Gaussian Elimination: -1 -1 -1 0 1*(-1) -12 2 -50 4 -2 0 1 o) III ? R1/(-1) - R1 1 0 -12 2 -50 1 ܠܙ ]*(12) 4 -2 1 0 =

Vo 8:01 PM 7.8KB/s na 51 ? 1 1 0 R1/(-1) - R1 1 1 8) x(12) 2 -50 -12 4 -2 1 = ? R2-(-12) *R1 - R2 1 1 10 0 14 70 4 -2 10 *(-4) ? R3 R3-4x R 1 1 1 1 0 (14 0 -6 -30 ? R2/(14) - R2 o 1 1 1 (11 10

Vo 8:02 PM 1.1KB/s 51 ? R2/(14) - R 2 1 1 0 (1 1 o x(6) 0 -6 -30 ? *R 2 R3 R3 -(-6) 1 1 1 0 1 *(-1) 0 (1 0 0 0 0 1 1 0 0 ? 2 R1-1*R 2 - R1 1 0 1 0 2 0 0 0 0 = X 1 + ***X3 х = 0 { (1) x2 +*xx. = 0 = . Find the variable x, from

Vo 8:02 PM 0.0KB/s 51 = Find the variable x2 from the equation 2 of the system (1): x2 = {xx3 2 • Find the variable x, from the equation 1 of the system (1): x1 = xx3 Answer: 3 Х = 1 XX3 2 X2 = XX3 - 1 x X3 = X3 Ex X 3 General Solution : X= 1;1xx3 -1 2 2 X3 -1 2 The solution set: {X3* | -1 fo 2 1

Vo 8:02 PM 0.0KB/s QA 51 -1 xX3 2. General Solution 2:X=11xx3 2. X 3 = 2. -1 The solution set: X3 * 2. 1 2. Let x 3 = 1, V1= | -1 2. 1 2. 2. - 2 × V5 - 1 A-L= 2 x V5 - 2. -12 2 x 5 +1 4 -2. - 1 15 - 5 2 × 15 = Av = AV (A - JJv = 0 So we have a homogeneous system of linear equations we

8:02 PM 0.1KB/s na Vo LTE 51 Αν = λν (A - 21)v = 0 So we have a homogeneous system of linear equations, we solve it by Gaussian Elimination: (15+ 2 x 15 - 2 -1 -1 0 8 -12 2 x 5 + 1 -5 0 2 x 150 4 -2 E ? R 1/(2x V5 – 2) - R1 1 -15 -1 -15 -1 8 0. 8 ]*(12) -12 2 x 15 + 1 -5 0 4 -2 2 150 = ? 0 R2-(-12) Ⓡ R 1 - R2 -15 - 1 -15 - 1 (1 8 8 V5 - 1 -3x V5 - 13 0 2 2 4 -2 2 x 15 *(-4) 0 0

Vo 8:03 PM 35.4KB/s 02 51 ? 0 R3-4×R → R3 -5 - 1 -V5 - 1 1 8 8. V5 - 1 -3 × V5 - 13 0 2 2. 5 - 3 5 x V5 + 1 2. 2. V5 + 1) 0 x (S) 0 0 III ? ~ R/ PS) - R2 V5 - 1 2. -/5 - 1 8 1 1 -5 - 1 0 8 - 4 x V5 - 7 10 5 x V5 + 1 0 2. 0 -/5 +3) X V5 - 3 (15) 0 이 2. ||| ? R3 (V5 - 3 2. -5 - 1 (12) x R2 → R3 -5 - 1 1 0 8 8 (V5 + 1 8 0 1 - 4 × V5 - 7 10

8:03 PM 75.2KB/s na Vo LTE 51 15 - 3 R3 - 2 -V5 - 1 1 xR2 - R3 -V5 - 1 0 8 -4* V5 - 70 0 0 (15 + 1 8 1-15 0 1 0 0 ? R (**)* R - R R 1 1 -3 V5 - 7 1 0 0 2 0 1 -4 * V5 - 70 0 0 0 o II X 1 0 3* V5 +7 X2 0 (1) (4 x V5 + 7)* X3 = • Find the variable x 2 from the equation 2 of the system

Vo 8:03 PM 1.3KB/s AA 51 • Find the variable x 2 from the equation 2 of the system (1): x2 = (4 * v5 + 7)*xg • Find the variable x 1 from the equation 1 of the system (1): 3 x 15 + 7 X 3 Answer: 3 x V5 + 7 X13 X 2 = (4 * v5 + 7) XXg X 3 = X3 General Solution : 3 x V5 + 7 X 1 2 XX3 2 X X3 2 X= (4 x V5 + 7) *X 3 X3 The solution set: 3 x 15 + 7 2 X 3 х 4xV5 + 7 1

8:03 PM 1.2KB/s na Vo LTE 51 X X 3 General Solution 3 x V5 + 7 2 X= (4 x V5 + 7) * x3 X wo The solution set: 3 x 15 + 7 2 X3 Х 4* V5 + 7 1 3 x 5 + 7 2 Let x 3 = 1, V2 = 4* 5 + 7 1 TIL 3. 23 = 2 * V5 - 1 A-131 = -2 x V5 - 2 -1 -1 -12 -2 x V5 + 1 -5 4 -2 -2 x V5 = Αν = λν (A - 2I)v = 0 So we have a homogeneous

8:03 PM 0.OKB/s na Vo LTE 51 Αν = λν (A - 21) v = 0 So we have a homogeneous system of linear equations, we solve it by Gaussian Elimination: -2 * 15 2 -1 -1 0 V-V5 + 1 -2 x V5 + -12 -5 0 -2 4 15 Х 8 1 -2 ? R1/(-2xv5-2) - R1 1) 15 - 1 15 - 1 8. 0 8 (12) -12 -2 * 15 + 1 -5 0 4 -2 -2 x 150 ? R2-(-12) Ⓡ R1 - R2 15 - 1 1 15 - 1 8 0 8

Vo 8:03 PM 58.2KB/s 2 A 51 0 R2-(-12) Ⓡ R1 - R2 15 - 1 15 - 1 1 8 8 -V5 - 1 315 - 13 2 2 4. -2 -2 x 15 1 *(-4) o 0 0 ? R3-4*R1 - R3 15 - 1 15 - 1 1 0 8 8 -V5 - 1 3 x V5 - 13 0 0 X -V5 + 1 2 2 2 -V5 - 3 -5 x 15 + 1 0 0 2 2 ? R 2 15 - 1 2 15 - 1 R 2 15 - 1 1 0 8 8 0 1 4* V5 - 70 -V5-3 -5 x V5 + 1 0 2 2 14(457.35 ) 0 III

8:03 PM 61.2KB/s 2 A Vo LTE 51 ?. R3 -(45= 3)* R2 - R$ 15 - 1 15 - 1 1 0 15 + 1 8 8 x 8 0 1 4x V5 - 70 0 0 0 0 E ? ~ R1-(* ) *R2 + R, 0 3 x V5 - 7 1 0 2 0 1 4x V5 - 7 0 0 0 o 0 = X 1 + = 0 3x V5 - 7 X 2 + = 0(1) (4 x V5 – 7) * X 3 III • Find the variable x 2 from

8:03 PM 61.2KB/s 2 A Vo LTE 51 ?. R3 -(45= 3)* R2 - R$ 15 - 1 15 - 1 1 0 15 + 1 8 8 x 8 0 1 4x V5 - 70 0 0 0 0 E ? ~ R1-(* ) *R2 + R, 0 3 x V5 - 7 1 0 2 0 1 4x V5 - 7 0 0 0 o 0 = X 1 + = 0 3x V5 - 7 X 2 + = 0(1) (4 x V5 – 7) * X 3 III • Find the variable x 2 from

Vo 8:03 PM 0.0KB/s 51 • Find the variable x2 from the equation 2 of the system (1): x2 = (-4x V5 + 7) * x3 • Find the variable x 1 from the equation 1 of the system (1): -3x V5 + 7 Answer: X 1 = -3 *V7+2x x3 x2 = (-4x V5 + 7) * x3 X3 = X 3 General Solution : -3 V5 + 7 2 X= (-4x V5 + 7) XX3 XX3 X3 The solution set: -3x V5 + 7 2 X 3 -4 * V5 + 7 1

12 3V567 2 - 1/2 UV5+7 a 1 13 -3547 -u 5+1 2 14/19 16 19 -375 +8 8V5-15 19 95 31578-8V5-15 95 - 715725 190 -9V5715 715 +25 190 - 1619 ----14-12 346547 - Best Daptap 6/19 14/19 3 -- -3V5+8 8V5-15 -7V5+25 19 95 -12 12 05 190 1/2 41541 3V5t8 - 8V5-15 7V5 +25 4-8 19 95 190 255.7560 - 18.2984 +25.0130 1289.846429 - 9003434 - 127.80097) 70:36656 - 7.6louz t is Lontcim Toen valu D- -4-5896 not diagonazable on the notion value

Vo 8:11 PM 0.0KB/s 51 Insert in A Insert in B Clean V 3 x 15 + 7 (-1) -1 2 -1 - 3x √5 + 7 2 -34 35 2 = 4x V5 + 7 -4* V5 + 7 2 1 1 1 = 6 -16 19 19 8 x 15 - 15 -3 V5 + 8 14 19 -7 V5 + 25 190 7 x V5 + 25 190 19 95 3 x 5 + 8 -8 * V5 - 15 19 95 = Details (Montante's method (Bareiss algorithm)) Details (Gauss-Jordan elimination) Details (Using the adjugate matrix) Insert in A Insert in B Clean