# If A = [[1,-1,0,0],[2,-3,0,0],[0,0,1,-1],[0,0,-2,3]] then A^-1 = ?

### 2 Answers | Add Yours

Given matrix is `A=[[1,-1,0,0],[2,-3,0,0],[0,0,1,-1],[0,0,-2,3]]` .

Now to find the inverse of the matrix A we write the matrix equation as `AI=A.` where I is the `4xx4` identity matrix. Now by applying elementary row operations reduce the matrix A on the right hand side to identity matrix I.

i.e. `AB=I` . Here the matrix B will be the inverse of the matrix A.

So, `A[[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1]]=[[1,-1,0,0],[2,-3,0,0],[0,0,1,-1],[0,0,-2,3]]`

by `R_2->2R_1-R_2` we get

`A[[1,0,0,0],[2,-1,0,0],[0,0,1,0],[0,0,0,1]]=[[1,-1,0,0],[0,1,0,0],[0,0,1,-1],[0,0,-2,3]]`

by `R_1->R_1+R_2` and `R_4->2R_3+R_4` we get

`A[[3,-1,0,0],[2,-1,0,0],[0,0,1,0],[0,0,2,1]]=[[1,0,0,0],[0,1,0,0],[0,0,1,-1],[0,0,0,1]]`

by `R_3->R_3+R_4` we get

`A[[3,-1,0,0],[2,-1,0,0],[0,0,3,1],[0,0,2,1]]=[[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1]]` .

Now we see that the matrix on right hand side is the identity. So the matrix `B=[[3,-1,0,0],[2,-1,0,0],[0,0,3,1],[0,0,2,1]]` is the required inverse of the matrix A.

So `A^-1=[[3,-1,0,0],[2,-1,0,0],[0,0,3,1],[0,0,2,1]]` .

A = [[1,-1,0,0],[2,-3,0,0],[0,0,1,-1],[0,0,-2,3]]

`A=[[1,-1,0,0],[2,-3,0,0],[0,0,1,-1],[0,0,-2,3]]`

`A=[[X,0],[0,Y]]`

`X=[[1,-1],[2,-3]],Y=[[1,-1],[-2,3]]`

`AB=[[X,0],[0,Y]][[E,G],[F,H]]=[[I,0],[0,I]]`

`XE=I, XG=0,YF=0,YH=I`

`E=-[[-3,1],[-2,1]]=[[3,-1],[2,-1]]`

`H=[[3,1],[2,1]]`

`B=[[E,0],[0,H]]=A^(-1)=[[3,-1,0,0],[2,-1,0,0],[0,0,3,1],[0,0,2,1]]`

``

``

``