| Simplify the determinant by the column operation C1⟶C1−C2 : ∆=|
k2−1
2k+1
1
k−1
k+2
1
0
3
1
| Expanding along the third row, ∆=−3|
k2−1
1
k−1
1
|+|
k2−1
2k+1
k−1
k+2
|=(k−1)3. Thus ∆=(k−1)3. Sign analysis - k>0 : if 0<k<1,∆<0; if k>1,∆>0⇒ Statement I is false. - k<0:∆<0⇒ Statement II is true. - k=0:∆=(−1)3=−1≠0⇒ Statement III is false. ∴ Only Statement II is correct ⇒ exactly one statement is true.