Question

Determine the null space of each of the following matrices:
\(\begin{pmatrix}1 & 3 &-4 \\ 2 & -1 & -1 \\ -1 & -3 &4 \end{pmatrix}\)

Answer 1

Step 1
Consider the given matrix:
\(\begin{pmatrix}1 & 3 &-4 \\ 2 & -1 & -1 \\ -1 & -3 &4 \end{pmatrix}\)
Now solve the system of equations below to find the null spaces:
\(\begin{pmatrix}1 & 3 &-4 \\ 2 & -1 & -1 \\ -1 & -3 &4 \end{pmatrix}\begin{pmatrix}x_1 \\ x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0 \end{pmatrix}\)
Solve the above matrix using Gauss Elimination method . Find the row echelon form: \(R_2=R_2-2R_1\)
\(\begin{pmatrix}1 & 3 &-4 \\ 0& -7 & 7\\ -1 & -3 &4 \end{pmatrix}\begin{pmatrix}x_1 \\ x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0 \end{pmatrix}\)
\(R_3=R_3+R_1\)
\(\begin{pmatrix}1 & 3 &-4 \\ 0& -7 & 7\\ 0 & 0 &0 \end{pmatrix}\begin{pmatrix}x_1 \\ x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0 \end{pmatrix}\)
Step 2
we get equations:
\(-7x_2+7x_3=0\)
\(7x_2=7x_3\)
\(x_2=x_3\) And \(x_1+3x_2-4x_3=0\)
\(x_1+3x_2-4x_2=0\)
\(x_1=x_2\)
Therefore null space is given by:
\(\begin{pmatrix}x_2 \\ x_2 \\ x_2 \end{pmatrix}=x_2\begin{pmatrix}1 \\ 1 \\ 1 \end{pmatrix}\)