Thus,
We can use either
đ _1â đ _3
đ _1â đ _2 + đ _1
đ _3â 2đ _3
OR
đ¶_1â" " đ¶_2 + 9đ¶_2
đ¶_1â 5/2 đ¶_1
đ¶_2â đ¶_3â3đ¶_1
But not
đ _1â đ _3
đ _1â đ _2 + ă2đ ă_1
đȘ_đâ đȘ_đ + ăđđȘă_đ
This is wrong
Find inverse of [â 8(đ&đ@đ&đ)]
Let A = [â 8(3&2@1&4)]
We know that
A = A I
This becomes Aâ1
Convert to I using elementary transformation
A = A I
[â 8(3&2@1&4)] = A [â 8(1&0@0&1)]
Making 3 to 1
đ _1 âđ _1â 2đ _2
[â 8(3â2 (1) &2â2(4)@1&4)] = A [â 8(1â2 (0) &0â2(1)@0&1)]
[â 8(3â2 &2â8@1&4)] = A [â 8(1&â2@0&1)]
[â 8(1&â6@1&4)] = A [â 8(1&â2@0&1)]
Making 1 to 0
đ _2 âđ _2â đ _1
[â 8(1&â6@1â1&4â(â6))] = A [â 8(1 &â2@0â1&1â(â2))]
[â 8(1&â6@0&4+6)] = A [â 8(1&â2@â1&1+2)]
[â 8(1&â6@0&10)] = A [â 8(1&â2@â1&3)]
Making 10 to 1
đ _2 âđ _2/10
[â 8(1&â6@0/10&10/10)] = A [â 8(1 &â2@(â1)/10&3/10)]
[â 8(1&â6@0&1)] = A [â 8(1 &â2@(â1)/10&3/10)]
Making â6 to 0
đ _1 âđ _1+ 6đ _2
[â 8(1+6(0)&â6+6(1)@0&1)] = A [â 8(1+6((â1)/10) &â2+6(3/10)@(â1)/10&3/10)]
[â 8(1&â6+6@0&1)] = A [â 8(1â6/10&â2+18/10@(â1)/10&3/10)]
[â 8(1&0@0&1)] = A [â 8(4/10&(â2)/10@(â1)/10&3/10)]
This is similar to
I = AAâ1
Therefore,
Aâ1 = [â 8(4/10&(â2)/10@(â1)/10&3/10)]
Find inverse of [â 8(9&2&1@5&â1&6@4&0&â2)]
Let A = [â 8(9&2&1@5&â1&6@4&0&â2)]
We know that
A = AI
[â 8(9&2&1@5&â1&6@4&0&â2)] = A [â 8(1&0&0@0&1&0@0&0&1)]
Making 9 to 1
đ _1 âđ _1â 2đ _3
[â 8(9â2(4)&2â2(0)&1â2(â2)@5&â1&6@4&0&â2)] = A [â 8(1â2(0)&0â2(0)&0â2(1)@0&1&0@0&0&1)]
[â 8(9â8&2â0&1+4@5&â1&6@4&0&â2)] = A [â 8(1&0&â2@0&1&0@0&0&1)]
[â 8(1&2&5@5&â1&6@4&0&â2)] = A [â 8(1&0&â2@0&1&0@0&0&1)]
Making 5 to 0
đ _2 âđ _2â 5đ _1
[â 8(1&2&5@5â5(1)&â1â5(2)&6â5(5)@4&0&â2)] = A [â 8(1&0&â2@0â5(1)&1â5(0)&0â5(â2)@0&0&1)]
[â 8(1&2&5@5â5&â1â10&6â25@4&0&â2)] = A [â 8(1&0&â2@â5&1&10@0&0&1)]
[â 8(1&2&5@0&â11&â19@4&0&â2)] = A [â 8(1&0&â2@â5&1&10@0&0&1)]
Making 4 to 0
đ _3 âđ _3â 4đ _1
[â 8(1&2&5@0&â11&â19@4â4(1)&0â4(2)&â2â4(5))] = A [â 8(1&0&â2@â5&1&10@0â4(1)&0â4(0)&1â4(â2))]
[â 8(1&2&5@0&â11&â19@4â4&â8&â2â20)] = A [â 8(1&0&â2@â5&1&10@â4&0&1+8)]
[â 8(1&2&5@0&âđđ&â19@4&â8&â22)] = A [â 8(1&0&â2@â5&1&10@â4&0&9)]
Making â11 to 1
đ _2 âđ _2/(â11)
[â 8(1&2&5@0/(â11)&(â11)/(â11)&(â19)/(â11)@0&â8&â22)] = A [â 8(1&0&â2@(â5)/(â11)&1/(â11)&10/(â11)@â4&0&9)]
[â 8(1&đ&5@0&1&19/11@0&â8&â22)] = A [â 8(1&0&â2@5/11&(â1)/11&(â10)/11@â4&0&9)]
Making 2 as 0
đ _1 âđ _1â 2đ _2
[â 8(1â2(0)&2â2(1)&5â2(19/11)@0&1&19/11@0&â8&â22)] = A [â 8(1â2(5/11)&0â2((â1)/11)&â2â2((â10)/11)@5/11&(â1)/11&(â10)/11@â4&0&9)]
[â 8(1&0&5â38/11@0&1&19/11@0&â8&â22)] = A [â 8(1â10/11&2/11&â2+20/11@5/11&(â1)/11&(â10)/11@â4&0&9)]
[â 8(1&0&17/11@0&1&19/11@0&âđ&â22)] = A [â 8(1/11&2/11&(â2)/11@5/11&(â1)/11&(â10)/11@â4&0&9)]
Making â8 as 0
đ _3 âđ _3 + 8đ _2
[â 8(1&0&17/11@0&1&19/11@0+8(0)&â8+8(1)&â22+8(19/11) )] = A [â 8(1/11&2/11&(â2)/11@5/11&(â1)/11&(â10)/11@â4+8(5/11)&0+8((â1)/11)&9+8((â10)/11) )]
[â 8(1&0&17/11@0&1&19/11@0&â8+8&â22+152/11)] = A [â 8(1/11&2/11&(â2)/11@5/11&(â1)/11&(â10)/11@â4+40/11&(â8)/11&9â80/11)]
[â 8(1&0&17/11@0&1&19/11@0&0&(âđđ)/đđ)] = A [â 8(1/11&2/11&(â2)/11@5/11&(â1)/11&(â10)/11@(â4)/11&(â8)/11&19/11)]
We Make (â90)/11 to 1
đ _3 âđ _3 Ă (â11)/90
[â 8(1&0&17/11@0&1&19/11@0Ă(â11)/90&0Ă(â11)/90&(â90)/11Ă(â11)/90)] = A [â 8(1/11&2/11&(â2)/11@5/11&(â1)/11&(â10)/11@â4Ă(â11)/90&(â8)/11Ă(â11)/90&19/11Ă(â11)/90)]
[â 8(1&0&17/11@0&1&19/11@0&0&1)] = A [â 8(1/11&2/11&(â2)/11@5/11&(â1)/11&(â10)/11@2/45&4/45&(â19)/90)]
We Make 17/11 to 0
đ _1 âđ _1 â 17/11 đ _3
[â 8(1â17/11(0)&0â17/11(0)&17/11â17/11(1)@0&1&19/11@0&0&1)] = A [â 8(1/11â17/11 (2/45)&2/11â17/11 (4/45)&(â2)/11â17/11 ((â19)/90)@5/11&(â1)/11&(â10)/11@2/45&4/45&(â19)/90)]
[â 8(1â17/11(0)&0â17/11(0)&17/11â17/11(1)@0&1&19/11@0&0&1)] = A [â 8(1/11 (1â34/45) &2/11 (1â34/35)&1/11 (â2+323/90)@5/11&(â1)/11&(â10)/11@2/45&4/45&(â19)/90)]
[â 8(1&0&0@0&1&19/11@0&0&1)] = A [â 8(1/11Ă11/45&2/11Ă11/45&1/11Ă143/90@5/11&(â1)/11&(â10)/11@2/45&4/45&(â19)/90)]
[â 8(1&0&0@0&1&19/11@0&0&1)] = A [â 8(1/45&2/45&13/90@5/11&(â1)/11&(â10)/11@2/45&4/45&(â19)/90)]
We Make 19/11 as 1
đ _2 âđ _2 â 19/11 đ _3
[â 8(1&0&0@0â19/11(0)&1â19/11(0)&19/11â19/11(1)@0&0&1)] = A [â 8(1/45&2/11&13/90@5/11â19/11 (2/45)&(â1)/11â19/11 (4/45)&(â10)/11â19/11 ((â19)/11)@2/45&4/45&(â19)/90)]
[â 8(1&0&0@0&1&0@0&0&1)] = A [â 8(1/45&2/45&13/90@1/11 (5â38/45)&(â1)/11Ă121/45&1/11Ă((â539)/90)@2/45&4/45&(â19)/90)]
[â 8(1&0&0@0&1&0@0&0&1)] = A [â 8(1/45&2/45&13/90@17/45&(â11)/45&(â49)/90@2/45&4/45&(â19)/90)]
I = AAâ1
Therefore,
Aâ1 = [â 8(1/45&2/45&13/90@17/45&(â11)/45&(â49)/90@2/45&4/45&(â19)/90)]
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.
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