1 Answer | Add Yours
Given matrix is `A=[[c,1,0],[1,c,1],[0,1,c]]` .
Inverse of the matrix A will exist only when `det A!=0` .
Now det A=`c^3-2c!=0` .
i.e. `c!=0` and `c!=+-sqrt2` .
Now for all real values of `c!=0,+-sqrt2` we calculate the inverse of the given matrix A as under-
The minors for the matrix A are `a_11=c^2-1, a_12=-c, a_13=1,a_21=-c, a_22=c^2, a_23=-c, a_31=1,a_32=-c, a_33=c^2-1`
` A^-1=(Adj A)/det A.`
We’ve answered 333,896 questions. We can answer yours, too.Ask a question