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[SOLVED] 14.2 find a basis for NS(A) and dim{NS(A)}

karush

Well-known member
Jan 31, 2012
2,657
$\tiny{370.14.2.}$
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
 
Last edited:

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
$\tiny{370.14.2.}$
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 [tex]x_1+ x_3= 0[/tex] and [tex]x_2+ 3x_3= 0[/tex]? Do you see that [tex]x_1[/tex] and [tex]x_2[/tex] can be written in terms of [tex]x_3[/tex] so this null space is one dimensional?