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Suppose that D is an n × n diagonal matrix with entries dii. Show that D-1 is an n × n diagonal matrix with entries 1/dii.

User Charlietfl
by
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1 Answer

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Answer:

For matrix D=[d11 0 0 ....... 0

0 d22 0 ...... 0

0 0 d33 0 0 ....0

.

.

0 0 0 0 0 . . . dnn]

Adj D =[d22d33..dnn 0 0 0 ....0

0 allvaluesexceptd22 0 0 ...0

0 0 allvaluesexpectd33

.

.

..........allvaluesexceptdnn]

|D|=d11d22d33....dnn

So D-1 = Adj D/|D|

So dividing we get:

D-1=[1/d11 0 0 .... 0

0 0 1/d22 ...0

0 0 0 ......1/dnn]

Explanation:

as we know that D-1=AdjD/|D|

and for adj more than 2cross2 we use cofactor method

to find cofactors of matrix we can get

leaving first row and first col

D11=d22d33d44...dnn (as its a diagonal matrix and determinant can be found by multiplying diagonal values)

D12=0 same as all upto D1n=0

D21=0 D22=d11d33d44...dnn all other D22 to D2n=0

and upto so on we can get cofactor matrix as :

[d22d33..dnn 0 0 0 ....0

0 allvaluesexceptd22 0 0 ...0

0 0 allvaluesexpectd33

.

.

..........allvaluesexceptdnn]

Adj D is transpose of cofactor matrix which comes out to be the same :

[d22d33..dnn 0 0 0 ....0

0 allvaluesexceptd22 0 0 ...0

0 0 allvaluesexpectd33

.

.

..........allvaluesexceptdnn]

Now comes determinant:

determinant of diagonal matrix can be calculated by multiplying all diagonal elements

So

|D|=d11d22d33d44...dnn

Solving for D-1 we get :

AdjD/|D|=[1/d11 0 0 .... 0

0 0 1/d22 ...0

0 0 0 ......1/dnn]

User Ross Sheppard
by
4.2k points
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