(1) Let $A=\left(\begin{array}{ccc}1 & 0 & -1 \\ 3 & 1 & -1 \\ 1 & 1 & 1\end{array}\right)$. (a) Find the RREF of $A$. (b) Find a basis for null $(A)$, and determine if null $(A)$ is a point, line, plane or all of $\mathbb{R}^{3}$. (c) Find a basis for $\operatorname{im}(A)$, and determine if $\operatorname{im}(A)$ is a point, line, plane or all of $\mathbb{R}^{3}$.
Solution
Step 1 ：First, we subtract 3 times the first row from the second row and subtract the first row from the third row to get the matrix \(\left(\begin{array}{ccc}1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 1 & 2\end{array}\right)\).
Step 2 ：Then, we subtract the second row from the third row to get the Reduced Row Echelon Form (RREF) of the matrix \(A\), which is \(\left(\begin{array}{ccc}1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 0\end{array}\right)\).
Step 3 ：The null space of \(A\) (null \(A\)) is the set of all vectors \(x\) such that \(Ax = 0\). From the RREF of \(A\), we can see that the system of equations \(Ax = 0\) has a free variable, which corresponds to the third column of \(A\). Therefore, a basis for null \(A\) is \(\left\{\left(\begin{array}{c}1 \\ -2 \\ 1\end{array}\right)\right\}\). Since there is one free variable, null \(A\) is a line in \(\mathbb{R}^{3}\).
Step 4 ：The image of \(A\) (im \(A\)) is the span of the column vectors of \(A\). From the RREF of \(A\), we can see that the first two columns are linearly independent, and the third column is a linear combination of the first two. Therefore, a basis for im \(A\) is \(\left\{\left(\begin{array}{c}1 \\ 0 \\ 1\end{array}\right), \left(\begin{array}{c}0 \\ 1 \\ 1\end{array}\right)\right\}\). Since there are two basis vectors, im \(A\) is a plane in \(\mathbb{R}^{3}\).
