# Let A = [[1, 2, 0],[2, 5, 0],[k, 2k + 1, 2k + 4]] Find the inverse of A when A is invertible.

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Given matrix is `A=[[1,2,0,1,0,0],[2,5,0,0,1,0],[k,2k+1,2k+4,0,0,1]]` `~~[[1,2,0,1,0,0],[0,-1,0,2,-1,0],[0,-1,-2k-4,k,0,-1]]` `~~[[1,2,0,1,0,0],[0,1,0,-2,1,0],[0,1,2k+4,-k,0,1]]` `~~[[1,0,0,5,-2,0],[0,1,0,-2,1,0],[0,0,-2k-4,-2+k,1,-1]]` `~~[[1,0,0,5,-2,0],[0,1,0,-2,1,0],[0,0,1,(2-k)/(2k+4),-1/(2k+4),1/(2k+4)]]`

So, `A^-1=[[5,-2,0],[-2,1,0],[(2-k)/(2k+4),-1/(2k+4),1/(2k+4)]]` . This is the required inverse of the given matrix A.

`A=[[1,2,0],[2,5,0],[k,2k+1,2k+4]]`

we have to find for wich K Det A=0

Running along last column:

`DetA=2(k+2)` so `A` is invertible if `k!=-2`

Since :

`Det A=2(k+2) Det [[1,2],[2,5]]` its enough to find inverse of `[[1,2],[2,5]]`

`[[1,2],[2,5]]^-1=` `[[5,-2],[-2,1]]`

so:

`A^(-1)=[[5,-2,0],[-2,1,0],[r_1,r_2,r_3]]`

For the third row `(r_1,r_2,r_3)` we have to(by Product third row for 1st, 2nd 3rd column):

`5k-2(2k+1)+2 r_1 (k+2)=0`

`-2k+2k+1+2r_2 (k+2)=0`

`2 r_3 (k+2)=1`

Then:

`r_1=(2-k)/(2(k+2))` `r_2=-1/(2(k+2))` `r_3=1/(2(k+2))`

So that the inverse of `A` is:

`A^(-1)=` `[[5,-2,0],[-2,1,0],[(2-k)/(2(k+2)),-1/(2(k+2)),1/(2(k+2))]]`

Reading the last row it draws home how needs condition `k!=-2`

is to be accomplisched.

A = [[1, 2, 0],[2, 5, 0],[k, 2k + 1, 2k + 4]]

`A=[[1,2,0],[2,5,0],[k,2k+1,2k+4]]`

`|A|=(2k+4)(5-4)`

`=2(k+2)`

`A_11=5(2k+4)``=10(k+2),`

`A_12=-2(2k+4)``=-4(k+2),`

`` `A_13=2(2k+4)-5k=8-k,`

`A_21=-(2(2k+4))=-4(k+2),`

`A_22=(2k+4),`

`A23=-(2k+1-2k)=-1,`

`A_31=0,`

`A_32=0,`

`A_33=(5-4)=1`

`A^(-1)=1/(2k+2)[[10(k+2),-4(k+2),0],[-4(k+2),2(k+2),0],[8-k,0,1]]`

Ans.