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. xy =0, X= %=2c, , , , , , , , For A ==1, the eigen vectors is x, =(1 2 0)", , , , B ttl, Li soodel matrix is P=(X, X, 102, , 220, , —+~+ 2 2, , Pray 22 3, , ss 2, , 2 #0 -l, , -t 0 0, , B=P"'AP=|0 -1 0, , . 0 03, , Hamilton Theorem a, , P4207 43245=0_, , istics equations, then according to this theorem the above equation is satisfied by A., , d A-I from above equation. On premultiplying A", we get, P+24+37+5A4'=0, f' =-1/5 (A? + 24431, , , , 12 1, 39: Find the characteristics of the matrix A. Verify Cayley-Hamilton theorem 4-10 1-1, d ; 3 -1 1, fhe characteristics equation of above matrix 4 is | A —A/|=0, Ha 2° 1, A=| 0 1-4 --1 |=0, 3 -I I-a, +307+2-9 =0, By Cayley-Hamilton theorem |, A +347+A-91=0, - A -3A?-A+9I=0, Mathematics - ; 55°, , Scanned with CamScanner
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aif lau 7). *; of, , 4u 1) (43 0)(1.2 1 1 0:0, &-34-A+91=|-9 -2 -7|-3/-3 2 -2]-/0 1 te, ; au. 7) (6 4 5)\3-1 Ty oot, , (4 We 1) fiz 6 0) (t 2 1) (9 0, A&-34-At91=|-9 -2 -7|-|-9 6 -6|-}0 1 -1}+/0 9, “(2p ue 7) 8 12 15) (3 -1 1). (0 0-9):, , = Hence Cayley-Hamilton theorem is verified., , 2-11 * oy, , A=|-1-2°-1 :, 1-12, , Solution:, , The characteristics equation of above matrix A is] A—A7|=0, , , , 2-42 -1 1, Me As| 1) 2-4 Jeo, i I =P 2-4)., ie. V-O7 4927-4 =0.. °, ‘2. ° By Cayley-Hamilton theorem, , 4-64? +9A—41=0, , _ Verification: ; . *, , , , , , Scanned with CamScanner
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' (22-22 W) (6 “5 5) |, f-64-94- 41 =|-21 22 ee ie re, a -2ii-6|-5 6 -5|-9/-1 2 -t|-4]0 1 0}=0, 2 2) ls 5 6 1-12) loot, , Hence Cayley-Hamilton theorem is verted, , Now, to find A-t, > 4-644-944-4100, , Multiply by A’, we get, > A -6A4+9-44'=0, , = ae :, A "= G(4"-6A+91), , 6 -5 5 2 -1 1 100 3 1-1, , wats 6 !, =5 -s|-6J-1 2 -1]+9]0 1 ol|=a}1 3}, 5 5 6) (1 -1 2) lo 01} 13, , me, , Example 41: Use Cayley-Hamilton theorem to find the matrix represented by, , , , , , , , (541+ TAS abs ae 5A} + 8A?—2A + 1=0, wh 23, - where 4-10 1 0, 2, Solntion:, “he characteristics equation of above matrix A is |A— -ul-0 a /, 2-4 1 1 i :, Les Az] 0 I-A: 0 |=O- ! 1 2-4, ie | 4-507 + TA-3 = 0, Ort pe x, Baga Bie PPP! ea ge (1), ig ii ‘, —5A'+7A 6 _ 345+ A'-5A 34 gA?—-24 +1 =0, fn st via _ 3 + A= 5A + TA 3h+ A+ Atl, SR+ATN, 211/21!, {lo 1-0/0 10, 11, 2b 4 2, Engineering Mathematics “Il, , Scanned with CamScanner
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UNIT —1, , Hamilton theorem and hence *, Example 42: Find the characteristics of the matrix A. Verify Cayley ! feds, , 1 oio3 < ;, de} loo3 -3, 2 +4 +4, Solution:, , + fie] AAI] =0, The characteristics equation of above matrix 4 is | A Ml |, , I-A 1 3 ., te ae] 1 o3-a 33 Jeo, —2 -+ -+4-2, ic, F-20048 =0 ., By Cayley-Hamilton theorem, , A*-204+8/7=0, Verification:, , 4-8 -12//1 1 3 12, , 20 60, #=A44=(10 2 6H 1-3 -3|=| 29 52-60, 2 2 2f-2 ~4 -4) |-40'~20 'g¢ :, 12 20° 60 11 3 10, 4-20A+81=| 20 52 _60|~20, , 0, 1 3. 314810 1 oleo, 40-80, 38 i, , 2 4 4) |o 9, Hence Cayley-Hamilton theorem is verified,, Now, to find A-!, , > = A-204+87=0, Multiply by A”, we get, > A#&-204+8 4H, , A‘ =F0201~ 4%), , 1,0 0) (-4 -8 ~12 9, af. 48, Atel 2010 1 O}-l10 _ | 12 12,4 6, 8 22 6 ios -10 -2 gle, 0041 22 2» 8 S(.75 <1 0-3, , , , , , i ry seal, : Engineering Mathematics!, Scanned with CamScanner