Final answer:
The basis for the solution space of the given homogeneous system of linear equations is the vector (1, 4, -3), and the dimension of this solution space is 1, as there is one linearly independent solution derived from one free variable.
Step-by-step explanation:
To find a basis for and the dimension of the solution space of the homogeneous system of linear equations given by:
- -x + y + z = 0
- 4x - y = 0
- 2x - 5y - 6z = 0
we must perform Gaussian elimination to reduce the system to row-echelon form. This process will identify any free variables, which are necessary to express the solution space.
First, let's combine the equations to eliminate y by using the second equation 4x - y = 0 to get y in terms of x, which gives y = 4x.
Now, substituting y into the first and third equation, we get:
- -x + 4x + z = 0 ⇒ 3x + z = 0 ⇒ z = -3x
- 2x - 5(4x) - 6z = 0 ⇒ -18x - 6z = 0 ⇒ But since z = -3x, this equation is now redundant.
Hence, we are left with y = 4x and z = -3x. This means x is the free variable, and we can express the solution space in terms of x as:
t(1, 4, -3) , where ℝ is the set of all real numbers, and t is a scalar multiple of the vector (1, 4, -3).
Therefore, the basis for the solution space is the single vector (1, 4, -3), and the dimension of this solution space is 1 as there is one free variable that can take any real value.