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Question

If A = diag (a, b, c) = ⎢ ⎢a000b000c⎥ ⎥ such that abc0
then A1=diag(1a,1b,1c)=⎢ ⎢ ⎢ ⎢1a0001b0001c⎥ ⎥ ⎥ ⎥

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Solution

|A| = abc 0 and hence A is non-singular whose inverse will exit. Now A1=adj.(A)|A|
If C be the matrix formed by cofactors of the elements of A then C = ⎢ ⎢bc000ca000ab⎥ ⎥
adj. A = Transpose of C = ⎢ ⎢bc000ca000ab⎥ ⎥
Rowscol and Colrows
A1=1|A|Adj.A=1abc⎢ ⎢bc000ca000ab⎥ ⎥
=⎢ ⎢ ⎢ ⎢1a0001b0001c⎥ ⎥ ⎥ ⎥

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