For what value(s) of k, if any, will the system have no solution, a unique solution, and infinitely many solutions? (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)
x − 2y + 8z = 6
x + y + z = k
2x − y + 9z = k2
1
Expert's answer
2021-04-19T07:26:37-0400
Consider the system of linear equations
x-2y+8z=6
x +y +z=k
2x-y+9z=k2
The augmented matrix for the above linear system:
A = ⎣⎡112−21−18∣1∣9∣6kk2⎦⎤
The row reduction process as follows:
R2⟹R2−R1
A = ⎣⎡102−23−18−79∣∣∣6k−6k2⎦⎤
R3⟹R3−2R1
A = ⎣⎡100−2338−7−7∣∣∣6k−6k2−12⎦⎤
R3⟹R3−R2
A = ⎣⎡100−2308−70∣∣∣6k−6k2−k−6⎦⎤ ........ (1)
The system has no solution if the the rank of the matrix ⎣⎡100−2308−70⎦⎤ is not equal to rank of matrix ⎣⎡100−2308−70∣∣∣6k−6k2−k−6⎦⎤
Let k2−k−6=0
(k+2)(k−3)=0
k=−2,3
Thus, the rank of matrix ⎣⎡100−2308−70⎦⎤ is 2.
Thus, the rank of matrix ⎣⎡100−2308−70∣∣∣6k−6k2−k−6⎦⎤ is 3.
Hence the system has no solution if k=−2,3
The system has unique solution if the rank of matrix ⎣⎡100−2308−70⎦⎤ is equal to rank of matrix ⎣⎡100−2308−70∣∣∣6k−6k2−k−6⎦⎤=3
This is not possible.
So, the system has no unique solution.
The system has infinitely many solutions if the rank of matrix ⎣⎡100−2308−70⎦⎤ is equal to rank of matrix ⎣⎡100−2308−70∣∣∣6k−6k2−k−6⎦⎤ less than 3.
Let k2−k−6=0
(k+2)(k−3)=0
k=−2,3
Thus, the rank of matrix ⎣⎡100−2308−70⎦⎤ is 2.
Thus, the rank of matrix ⎣⎡100−2308−70∣∣∣6k−6k2−k−6⎦⎤ is 2.
Hence, the system has infinitely many solutions if k=−2,3 .
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