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

karush

Well-known member
$\tiny{370.14.2.}$
For the matrix
$A=\left[ \begin{array}{rrrr} 1&0&1\\0&1&3 \end{array}\right]$
find a basis for NS(A) and $\dim{NS(A)}$
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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
$\tiny{370.14.2.}$
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?