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Question

Find the adjoint of 112302103

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Solution

Let,
A=∣ ∣112302103∣ ∣
Let the minors,
M11=0,M12=11,M13=0
M21=3,M22=1,M23=1
M31=2,M32=4,M33=4
Therefore,
adjA=M11M21M31M12M22M32M13M23M33
=0321114014

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