Engineering Board Exam Reviewer · Cramer's Rule and Linear Systems · hard
For what value of k does the system x+y+z=3, x+2y+3z=6, x+3y+kz=10 have no solution?
- 0
- 3
- -5
- 5
Answer
D. 5
Solution
The coefficient determinant is D= vmatrix1&1&1\1&2&3\1&3&k vmatrix=k-5. For k=5, R3-2R2+R1 gives 0x+0y+0z=10-12+3=1, a contradiction. So the system has no solution when k=5.
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