Determine all values of h and k for which the linear system has no solution.
-6x+5y-6z=6
-8x+9y+3z=4
-26x+31y+hz=k
The linear system has no solution if k≠ ?
and h = ?
no solution if the determinant is zero.
|-6 5 -6|
|-8 9 3| = 0
|-26 31 h|
To determine the values of h and k for which the linear system has no solution, we need to consider the coefficients of the variables in each equation.
First, we can rewrite the system of equations in matrix form:
A * X = B,
where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix:
| -6 5 -6 | | x | | 6 |
| -8 9 3 | * | y | = | 4 |
| -26 31 h | | z | | k |
For the system to have no solution, the coefficient matrix A must be non-invertible, which means its determinant must be zero.
Det(A) = -6(9h - 3) - 5(-8h - 3) + 6(-8*31 + 9*26)
= -54h + 18 + 40h + 15 - 38h + 39h
= -13h + 33
To find the value of k for no solution, we need to determine the range of h that makes the determinant zero:
-13h + 33 = 0
-13h = -33
h = 33/13
h ≈ 2.54
So, k ≠ 2.54 for the linear system to have no solution. Therefore, the values of k for no solution are any real number except 2.54.
To recap:
1. For the linear system to have no solution, k ≠ 2.54.
2. The value of h for which the linear system has no solution is h ≈ 2.54.