Question 1209084: For a certain value of k, the system
x + y + 3z = 10,
4x + 5y + 6z = 7,
kx - 3y + 2z = 3
has no solutions. What is this value of k? Answer by math_tutor2020(3817) (Show Source):
I placed the matrix in a table or grid format to separate out each entry.
From here we can apply row operations to get the matrix into Row Echelon Form (REF).
This is where we turn each pivot into a 1, and zero out each item below the pivots.
The pivots are along the northwest diagonal.
Here's what the steps look like.
1
1
3
10
4
5
6
7
k
-3
2
3
1
1
3
10
0
1
-6
-33
R2 - 4*R1 --> R2
k
-3
2
3
1
1
3
10
0
1
-6
-33
0
-3-k
2-3k
3-10k
R3 - k*R1 --> R3
1
1
3
10
0
1
-6
-33
0
0
-16-9k
-96-43k
R3 - (-3-k)*R2 --> R3
1
1
3
10
0
1
-6
-33
0
0
1
(-96-43k)/(-16-9k)
(1/(-16-9k))*R3 --> R3
At this point we can stop since the pivots are 1 and each item below the pivot is 0.
The third row leads to the equation or
Set the denominator equal to zero and isolate k.
-16-9k = 0
-9k = 16
k = -16/9
If k = -16/9, then the denominator of is zero, and it would mean there are no solutions to the original system.
For any other value of k, that fraction is some real number; and we can use that value of z to find y, and then find x using back substitution.
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