For what value of k does the system below not have a unique solution?
-1x + 7y + 2z = 6
2x + 1y + kz = -11
3x - 6y - 2z = -25
A.5
B.0
C.-2
D.-8
A system of equations does not have a unique solution when the equations are not linearly independent. The equations are linearly independent if none of the left-hand side of the equations can be represented as a linear combination of the others.
If we add equations (1) and (3)
-1x + 7y + 2z = 6 ...(1)
3x - 6y - 2z = -25 ...(3)
we get
2x +1y + 0z = -19 ...(2A)
Examine the left-hand side of equation (2A) carefully.
What value of k in equation (2) would make its left-hand side identical to a linear combination of (1) and (3), namely equation (2A)?
To determine the value of k that would make the system not have a unique solution, we need to check if the system is consistent or inconsistent.
A system of linear equations has a unique solution if it is consistent and has no free variables. If a system is consistent but has one or more free variables, it has infinitely many solutions. On the other hand, if a system is inconsistent, it has no solution or a unique solution.
To check whether the system is consistent or inconsistent, we can use Gaussian elimination or row reduction.
The given system:
-1x + 7y + 2z = 6 (equation 1)
2x + 1y + kz = -11 (equation 2)
3x - 6y - 2z = -25 (equation 3)
We can start by performing row reduction on the augmented matrix of the system:
[ -1 7 2 | 6 ]
[ 2 1 k | -11 ]
[ 3 -6 -2 | -25 ]
Performing row operations:
1. Multiply equation 1 by 2 and add it to equation 2:
[ -1 7 2 | 6 ]
[ 0 15 k | 1 ]
[ 3 -6 -2 | -25 ]
2. Multiply equation 1 by 3 and subtract it from equation 3:
[ -1 7 2 | 6 ]
[ 0 15 k | 1 ]
[ 0 -27 -8 | -43 ]
3. Multiply equation 2 by -27/15 and add it to equation 3:
[ -1 7 2 | 6 ]
[ 0 15 k | 1 ]
[ 0 0 -8+(27k/15) | (-43 + k) ]
Now, to have a unique solution, the third row should not have any free variable. In other words, the coefficient of the variable z in the third row should not be zero.
Looking at the third row, we see that for it to have a unique solution, the coefficient of z should not be zero:
-8 + (27k/15) ≠ 0
To find the value of k for which the coefficient becomes zero, we solve the equation:
-8 + (27k/15) = 0
Multiplying both sides by 15:
-120 + 27k = 0
27k = 120
k = 120/27
Simplifying the fraction:
k ≈ 4.444
Therefore, for the given system not to have a unique solution, the value of k should be approximately 4.444.
None of the answer choices (A, B, C, D) is equal to approximately 4.444. Hence, the correct answer cannot be determined from the given options.