MHB 14.2 find a basis for NS(A) and dim{NS(A)}

[karush]

$\tiny{370.14.2.}$
For the matrix
$A=\left[
\begin{array}{rrrr}
1&0&1\\0&1&3
\end{array}\right]$
find a basis for NS(A) and $\dim{NS(A)}$
-----------------------------------------------------------
altho it didn't say I assume the notation means Null Space of A
Reducing the augmented matrix for the system $$AX=0$$ to reduced row-echelon form.
$\left[
\begin{array}{ccc}
1 & 0 & 1 \\ 0 & 1 & 3
\end{array} \right]
\left[ \begin{array}{c}
x_{1} \\ x_{2} \\ x_{3}
\end{array} \right]
=\left[ \begin{array}{c}
0 \\ 0
\end{array}
\right]$
OK just seeing if I am going in the right direction

 

[HallsofIvy]

$\tiny{370.14.2.}$
For the matrix
$A=\left[
\begin{array}{rrrr}
1&0&1\\0&1&3
\end{array}\right]$
find a basis for NS(A) and $\dim{NS(A)}$
-----------------------------------------------------------
altho it didn't say I assume the notation means Null Space of A
Reducing the augmented matrix for the system $$AX=0$$ to reduced row-echelon form.
$\left[
\begin{array}{ccc}
1 & 0 & 1 \\ 0 & 1 & 3
\end{array} \right]
\left[ \begin{array}{c}
x_{1} \\ x_{2} \\ x_{3}
\end{array} \right]
=\left[ \begin{array}{c}
0 \\ 0
\end{array}
\right]$
OK just seeing if I am going in the right direction

Yes, that is correct. Do you see that the matrix equation is equivalent to the two equations x_1+ x_3= 0 and x_2+ 3x_3= 0? Do you see that x_1 and x_2 can be written in terms of x_3 so this null space is one dimensional?