Consider the system
x-2y=k-1
4x+(1-k^2)y=8
For what values of k does the system have
a. unique solution?
b. infinitely many solutions?
c. no solution?
rearranging things a bit, you have
y = x/2 + (1-k)/2
y = 4x/(k^2-1) + 8/(1-k^2)
for infinitely many solution, we need
y = mx+b
y = mx+c
where b=c
same slope:
1/2 = 4/(k^2-1)
k^2-1 = 8
k^2 = 9
k = ±3
b=c:
(1-k)/2 = 8/(1-k^2)
(1-k)(1-k^2) = 16
k=3
So, if k=3 there are infinitely many solutions.
If k = -3, there are no solutions.
For any other value of k, there is one solution.
Can you perform using matrix? Thank you very much
To determine for what values of k the given system has a unique solution, infinitely many solutions, or no solution, we can use the method of solving linear equations: elimination or substitution.
a. For the system to have a unique solution, the two equations must intersect at a single point, meaning they are not parallel and not coincident.
To check if they are parallel, we can calculate the ratio of the coefficients of x and y in both equations. If the ratios are equal, they are parallel. If not, they may intersect at a single point.
The ratio of the coefficients of x in both equations is `1/4`.
The ratio of the coefficients of y in both equations is `(1-k^2)/(-2)`.
If the two ratios are not equal, the system will have a unique solution. Thus, we need to set up an equation to check if `1/4` is equal to `(1-k^2)/(-2)`.
Solving `(1-k^2)/(-2) = 1/4` for k:
(1-k^2)/(-2) = 1/4
Cross-multiply:
-8(1-k^2) = 2
Distribute:
-8 + 8k^2 = 2
Move constant term to the left side:
8k^2 = 2 + 8
Simplify:
8k^2 = 10
Divide by 8:
k^2 = 10/8
k^2 = 5/4
Taking the square root of both sides:
k = ± sqrt(5/4)
The system will have a unique solution for `k = sqrt(5/4)` or `k = -sqrt(5/4)`.
(Note: The square root of a positive number has both a positive and a negative value.)
b. For the system to have infinitely many solutions, the equations must be coincident, meaning they are multiples of each other.
To check if they are multiples of each other, we can calculate the ratios of the coefficients of both equations. If the ratios are equal, they are multiples of each other.
Considering the ratio of the coefficients of x, `1/4`, and the ratio of the coefficients of y, `(1-k^2)/(-2)`, we need to find a value of k for which these ratios are equal to each other.
Setting up the equation:
(1-k^2)/(-2) = 1/4
Cross-multiply:
4(1-k^2) = -2
Distribute:
4 - 4k^2 = -2
Move constant term to the left side:
4k^2 = 4 + 2
Simplify:
4k^2 = 6
Divide by 4:
k^2 = 6/4
k^2 = 3/2
Taking the square root of both sides:
k = ± sqrt(3/2)
The system will have infinitely many solutions for `k = sqrt(3/2)` or `k = -sqrt(3/2)`.
c. If neither condition (a) nor (b) is satisfied, the system will have no solution.
Therefore, for `k ≠ sqrt(5/4)`, `k ≠ -sqrt(5/4)`, `k ≠ sqrt(3/2)`, and `k ≠ -sqrt(3/2)`, the given system will have no solution.