First row of a matrix A is `[1,3,2]`. If
adj `A=[(-2,4,alpha),(-1,2,1),(3alpha,-5,-2)]`, then a det (A) is

Options

A. `-2`
B. `-1`
C. 0
D. 1

Explanation

We know that
det `(A)=a_(11)A_(11)+a_(12)A_(21)+a_(13)A_(31)`
`=(1) (-2)+(3) (-1)+(2) (3alpha)`
`= 6 alpha-5`
Also, `("det. (A)")^(2)=` det. (adj. A) `=-10+17 alpha-6 alpha^(2)`
`:. (6 alpha-5)^(2)=-10+17 alpha-6 alpha^(2)`
`implies 42 alpha^(2)-77alpha+35=0`
`implies 6alpha^(2)-11alpha+5=0`
`implies (alpha-1) (6alpha-5)=0`
`implies alpha=1, 5//6`
For `alpha=1`, det (A) `=6-5=1`
For `alpha=5//6`, det `(A)=0`