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M4

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52 views23 pages

M4

Uploaded by

chetansawant8112003
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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theorem is verified, by actual division, we see that : © p86A5 +904 + 403-1202 + 2A-1=(A9-7A2 44 =O+2A-1 te A8-7A2 + 16A— periom 51 (1 010 5 20 10 =2|-2 -3 -4|- 1 O0|=|-4 -7 -8 my 0 Siu 7 0 6 10 13 +0 Example 6 : Find the characteristic equation of the matrix A given below and hence, find matrix represented by A°— 5A” + TAS — 34° + A* -— 5A° + BA? — 2A + I, where @ : re bh A=|0 10 (MU. 2000, 03, 19) 1.1.2) Sol. : The characteristic equation is j Den Ghat 0 1-02 O |=0 4 1-2-2 s (2-A)[(1 =A) (2-2)-0]= 10-0] +1 [0-1 (1-41 =0 3. (4-40 +22) (1 =2)= (1-2) =0 . 640402 — 40 + 402-209-140 =0 asl 2 8-512 + 74-3 = 0. ; This equation is satisfied by A. Now dividing 28 — 527 +798 - 3° +04 - 51° + 87 Quotient 25+ 2. and the remainder 2+ A+ 1. In terms of the matrix A this means AP -5AT + 70° = 3A5 + A 5A? + 8A? —2A4 7 = (AP = 542 +TA-3(AS 4A) + (A HA HD 2 +1 by 93-542 + 72-3, we get the, But (4°54? +7A-31)=0 + LHS. A+ AT ctivetth nort words |? 0 Nik Mee 10 0 e 45 L © p nit walt Seanned with CamScanner ic equation of A is 2: =0 | A)-4=0 -~4=0 Lats sat pa ont one Seanned with CamSeanner ‘Seanned with CamScanner i is — 10. The column headed by — 4 Re cen ieee ee ea ead below this column and put ae Petia anon ive the elements in the column of R.H.S. by the corresponding elma in the key column and obtain the column of ratios. Neglecting ree ValUes, We fn jy minimum positive ratio. It is 5. The row having the element 5 is the key row or Pivot row. Wen ‘an arrow head on the right and also put this row in a rectangular box. The element common foi, key-column and the key-row is the key-element or pivot element. It is 4. The element on the ty key rowis the outgoing element. Thus, s2 leaves. The element, at the top of the key columnisty incoming element. Thus, x; enters. Using the key element and by using elementary row operations, we bring zeros atallatie Places in the key column. For this we divide the elements of the key row by 4 and wil resulting row in another table. Thus, we get the following row. (See the third row of the next atl) x A 114 3/4 0 114 5 Using this row, we bring zeros at all other Of this row by 10 and id them to the corres; above the dotted line. (See Places in the key-column. So multiply these elem Pondiing elements of the first row. Thus, we getnel™ the fist row of the next table) z 0 3/2 13/2 0 S/2 50 Now multiply the elements of th Seen secon ow THs, wa gel ene Suet them fom the came th new row. (See the second row of the next) ~ 18/4 4 1/4 Scanned with CamSeanner ’ YZ ey Jequation, weget » XQ = 3. Re Xp) = x1 + x2-4 = 0 a s z- 1, om = 1 and all other partial derivatives are zero, x; # (x4, X2) = 4x4 + 8X9 — x12 — x2 af Baier eee! os Ox, ax? bro 9X19x2 9x1 OX. ah Ff ath of _, oh 9X2 9X2 x, 9X2 Ox, ax,? ax.” Ons 1. =|1 -2 o}=-1)) i + [228 0) -2 i 2 » 4 Is positive, Xy is a maxima, Seanned with CamScanner _oj_2t-t (1-8) 01-212 01=0 (MAB 4a tM 42 (1A 2 eye Rem eet s2- 2.4 = De rayoend oH FOR jee itn theorem states that this equation is satisfied by ne matrix A, je. AX-5A?+9A-1=0 « Ae 2 -2]fet 2-2 PA Ricrd Nowmeeetersy O|/=1, 2, 0|=|-4-°%. 2 Seer iio -2 1], 2-8 1 Peete 4)[_4.-2-~2]-[-13 42-2 and AP=|-4 7 2//-1 3 Ol=|-11 9 10 2-8 4], 0-2-1 10-22 -3 a easily seen that Sees peat i -11 9 10|-5)-4 7 2|+9|-1 10 -22 -3 ee 1 0 -2 Seanned with CamScanner 414i : ample 8 : Evaluate J ( az] is(1/a)>|z[>a J 2G) + 2 age [Gl Gis P = azn ez |: 4+ i i sa Bese | 32 eae tee a ~~) 2 ee bye= (2: eT (SZ) BS. 3 Bao) 3z 3z feet ee b

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