If x, y, z are non-zero real numbers, then the inverse of matrix `A=[(x,0, 0) ,(0,y,0),( 0, 0,z)]`is
(A) `[[x^(-1),0,0],[0,y^(-1),0],[0,0,z^(-1)]]`
(B) `xyz[[x^(-1),0,0],[0,y^(-1),0],[0,0,z^(-1)]]`
(C) `(1)/(xyz)[[x,0,0],[0,y,0],[0,0,z]]]`
(D) `(1)/(xyz)[[1,0,0],[0,1,0],[0,0,1]]`

Explanation

Given `A=[(x,0, 0) ,(0,y,0),( 0, 0,z)]`

We have to find `A^(-1)`
We know that
`A^(-1)=1/|A|adj(A), " " " " ` exists if `|A|ne0`

Calculating `|A|`

`|A|=|(x,0, 0) ,(0,y,0),( 0, 0,z)|`

` " " " " =x|[y,0],[0,z]|-0|[0,0],[0,z]|+0|[0,y],[0,0]|`

` " " " " =x(yz-0)-0+0=xyz`

Since `|A|ne0`
Thus, `A^(-1)` exist

Now,
`adj A=[[A_11,A_21,A_31],[A_12,A_22,A_32],[A_13,A_23,A_33]]`

`A=[(x,0, 0) ,(0,y,0),( 0, 0,z)]`

`M_11=|[y,0],[0,z]|=yz-0=yz, " " " ` `M_12=|[0,0],[0,z]|= 0 , " " " ` `M_13=|[0,y],[0,0]|= 0 `

`M_21=|[0,0],[0,z]|=0, " " " ` `M_22=|[x,0],[0,z]|=xz, " " " ` `M_23=|[x,0],[0,0]|= 0 `

`M_31=|[0,0],[0,z]|=0, " " " ` `M_32=|[x,0],[0,0]|=0, " " " ` `M_33=|[x,0],[0,y]|= xy `


`A_11=(-1)^(1+1)M_11=(-1)^2yz=yz, " " " ` `A_12=(-1)^(1+2)M_12=(-1)^3 .0=0, " " " ` `A_13=(-1)^(1+3)M_13=(-1)^4 .0=0`

`A_21=(-1)^(2+1)M_21=(-1)^3 .0=0, " " " ` `A_22=(-1)^(2+2)M_22=(-1)^4 .xz=xz, " " " ` `A_23=(-1)^(2+3)M_23=(-1)^5 .0=0`

`A_31=(-1)^(3+1)M_31=(-1)^4 .0=0, " " " ` `A_32=(-1)^(3+2)M_32=(-1)^5 .0=0, " " " ` `A_33=(-1)^(3+3)M_33=(-1)^6 .xy=xy`

Thus, `adj A=[[xy,0,0],[0,zy,0],[0,0,xy]]`

Now, `A^(-1)=1/|A|adj(A)`

` " " " " = 1/(xyz)[[xy,0,0],[0,zy,0],[0,0,xy]]`

` " " " " =[[(xy)/(xyz),0,0],[0,(zy)/(xyz),0],[0,0,(xy)/(xyz)]]`

` " " " " = [[(1)/(x),0,0],[0,(1)/(y),0],[0,0,(1)/(z)]]`

`A^(-1)=[[x^(-1),0,0],[0,y^(-1),0],[0,0,z^(-1)]]`

Thus, the correct option is (A).