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Question

If A= 235324112, then find A1.
Use it to solve the system of equations
2x3y+5z=113x+2y4z=5x+y2z=3

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Solution

A = 235324112

|A|=2(4+4)+3(6+4)+5(32)=06+5=10

A is non-singular so its inverse exists.
Cij are the cofactors of the determinant,

C11=(1)2 [2412]=0

C12=(1)3 [3412]=2

C13=(1)4 [3211]=1

C21=(1)3 [3512]=1

C22=(1)4 [2512]=9

C23=(1)5 [2311]=5

C31=(1)4 [3524]=2

C32=(1)5 [2534]=23

C33=(1)6 [2332]=13

adj.A=02119522313T=01229231513

A1=1|A|adj.A=1101229231513

= 01229231513

Also, 2x3y+5z=113x+2y4z=5 x+y2z=3

The given system of equations can be written as AX=B, where

A= 235324112,X=xyzand B=1153

So, X = A1B

xyz=01229231513×1153

xyz= 05+62245+691125+39=123

Hence, x=1,y=2 and z=3.

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