First row of a matrix A is 1    3    2. If adj⁡A=−24α−1213α−5−2 then a possible value of det(A) is

We know thatdet⁡(A)=a11A11+a12A12+a13A13=(1)(−2)+(3)(−1)+(2)(3α)=6α−5and  (det⁡(A))2=det⁡(adj⁡A)=11α−10Thus, (6α−5)2=11α−10⇒ 36α2−71α+35=0⇒ α=1,35/36Therefore, a possible value of det(A) is 1.