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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rakesh05 | High School Teacher | (Level 1) Assistant Educator

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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.

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oldnick | (Level 1) Valedictorian

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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]]`




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`



`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.

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pramodpandey | College Teacher | (Level 3) Valedictorian

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






`` `A_13=2(2k+4)-5k=8-k,`