Given matrix B is the inverse matrix of `3 xx 3` matrix A, where `B= [{:(5, 2alpha, 1),(0, 2, 1),(alpha, 3, -1):}]`
We know that,
`"det"(A) = (1)/("det"(B)) " "[because "det"(A^(-1)) = (1)/("det"(A))]`
Since, det(A) +1 `=0 " " ("given")`
`(1)/("det"(B))+1 =0`
`rArr ` det (B) =-1
`rArr 5(-2-3)-2alpha(0-alpha) +1(0-2alpha) =-1`
`rArr -25 + 2alpha^(2)-2alpha =-1`
`rArr 2alpha^(2)-2alpha-24 =0`
`rArr alpha^(2) -alpha -12 =0`
`rArr (alpha-4)(alpha +3) = 0`
`rArr alpha = -3, 4`
So, required sum of all values of `alpha` is 4-3 =1