Linear mapping/Kernel/2 3 0 -1/4 2 2 5/Exercise/Solution

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We determine the solution space of the system of linear equations

We get

Thus we have an equivalent system in row echelon form.

We choose first and . Then according to III and according to I we get . Hence

is a solution.

We choose now and . Then according to III and according to I we get

Therefore

is another solution, which is linearly independent of the first solution. Since the rank of the matrix is , the kernel is two-dimensional, hence the kernel equals