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Question

Let A=121231115

[adjA]1=adj(A1)
 


Solution

Here, adj(A)=141151143531=B(let)

 |B|=|adj A|=141151143531=14(49)11(1115)5(33+20)=182+286+65=1690
Cofactors of B are

B11=(49)=13,B12=(1115)=26,B13=(33+20)=13B21=(1115)=26,B22=1425=39,B23=(42+55)=13B31=(33+20)=13,B32=(42+55)=13,B33=56121=65

adj(B)=adj(adj A)=132613263913131365=132613263913131365
 B1=[adj A]1=1|adj A(adj(adj A))[adjA]1=1169132613263913131365=113121231115   (i)
Cofactors of A1 are
A11=13169,A12=26169,A13=13169,A21=26169,A22=39169,A23=13169A31=13169,A32=13169,A33=65169

Now, adj(A1)=⎢ ⎢ ⎢131692616913169261693916913169131691316965169⎥ ⎥ ⎥T=⎢ ⎢ ⎢131692616913169261693916913169131691316965169⎥ ⎥ ⎥
adj(A1)=113121231115    (ii)
From Eqs. (i) and (ii), we get that adj(A1)=(adjA)1


Mathematics

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