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

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rakesh05's profile pic

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

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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]]` .

 

pramodpandey's profile pic

pramodpandey | College Teacher | (Level 3) Valedictorian

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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]]`

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