Homework SourseIn Exercises II and 12 find a basis for the null space of ea
In Exercises II and 12, find a basis for the null space of each given matrix A. A = [1 2 3 -1 2 3 2 0 3 4 1 1 1 1 -1 1]
![In Exercises II and 12, find a basis for the null space of each given matrix A. A = [1 2 3 -1 2 3 2 0 3 4 1 1 1 1 -1 1]Solutioncan be transformed by a sequence](/WebImages/13/in-exercises-ii-and-12-find-a-basis-for-the-null-space-of-ea-1017587-1761525943-0.webp "In Exercises II and 12, find a basis for the null space of each given matrix A. A = [1 2 3 -1 2 3 2 0 3 4 1 1 1 1 -1 1]Solutioncan be transformed by a sequence")
Solution
can be transformed by a sequence of elementary row operations to the matrix
The reduced row echelon form of the augmented matrix is
which corresponds to the system
The leading entries in the matrix have been highlighted in yellow.
A leading entry on the (i,j) position indicates that the j-th unknown will be determined using the i-th equation.
Those columns in the coefficient part of the matrix that do not contain leading entries, correspond to unknowns that will be arbitrary.
The system has infinitely many solutions:
The rank of the matrix is 2. It equals the number of leading entries.
The nullity of the matrix is 2. This is the dimension of the null space. It equals the number of columns without leading entries.
| Transform the matrix to the reduced row echelon form (Show details) |
