Find a basis for the null space of the matrix.
-$$A = \left[ \begin{array} { r r r r r }
1 & 0 & - 4 & 0 & - 4 \
0 & 1 & 2 & 0 & 2 \
0 & 0 & 0 & 1 & 1 \
0 & 0 & 0 & 0 & 0
\end{array} \right]$$
A)
$ \left\{\left[\begin{array}{r}-4 \ 2 \ 1 \ 0 \ 0\end{array}\right],\left[\begin{array}{r}-4 \ 2 \ 0 \ -1 \ 1\end{array}\right]\right\} $

B)
$$\left\{ \left[ \begin{array} { r }
- 4 \
2 \
1 \
0 \
0
\end{array} \right] , \left[ \begin{array} { r }
- 4 \
2 \
0 \
- 1 \
1
\end{array} \right] \right\}$$
C)
$\left\{\left[\begin{array}{r}
1 \
0 \
-4 \
0 \
-4
\end{array}\right],\left[\begin{array}{l}
0 \
1 \
2 \
0 \
2
\end{array}\right]\right\}$

D)
$\left\{\left[\begin{array}{l}
1 \
0 \
0 \
0
\end{array}\right],\left[\begin{array}{l}
0 \
1 \
0 \
0
\end{array}\right],\left[\begin{array}{l}
0 \
0 \
1 \
0
\end{array}\right]\right\}$

Question

Find a basis for the null space of the matrix.
-$$A = \left[ \begin{array} { r r r r r } 
1 & 0 & - 4 & 0 & - 4 \
0 & 1 & 2 & 0 & 2 \
0 & 0 & 0 & 1 & 1 \
0 & 0 & 0 & 0 & 0
\end{array} \right]$$
A)
$ \left\{\left[\begin{array}{r}-4 \ 2 \ 1 \ 0 \ 0\end{array}\right],\left[\begin{array}{r}-4 \ 2 \ 0 \ -1 \ 1\end{array}\right]\right\} $

B)
$$\left\{ \left[ \begin{array} { r } 
- 4 \
2 \
1 \
0 \
0
\end{array} \right] , \left[ \begin{array} { r } 
- 4 \
2 \
0 \
- 1 \
1
\end{array} \right] \right\}$$
C)
$\left\{\left[\begin{array}{r}
1 \
0 \
-4 \
0 \
-4
\end{array}\right],\left[\begin{array}{l}
0 \
1 \
2 \
0 \
2
\end{array}\right]\right\}$

D)
$\left\{\left[\begin{array}{l}
1 \
0 \
0 \
0
\end{array}\right],\left[\begin{array}{l}
0 \
1 \
0 \
0
\end{array}\right],\left[\begin{array}{l}
0 \
0 \
1 \
0
\end{array}\right]\right\}$

Quizplus · Accepted Answer

The null space of a matrix consists of all vectors that satisfy $Ax = 0$. For the given matrix, the reduced row echelon form shows that the free variables are associated with the third and fifth columns. The basis vectors for the null space are constructed by setting one free variable to 1 and the rest to 0 in turn, and solving for the pivot variables, resulting in the vectors given in option B.

Find a Basis for the Null Space of the Matrix A=[10−40−4012020001100000]A = \left[ \begin{array} { r r r r r } 1 & 0 & - 4 & 0 & - 4 \\0 & 1 & 2 & 0 & 2 \\0 & 0 & 0 & 1 & 1 \\0 & 0 & 0 & 0 & 0\end{array} \right]A=​1000​0100​−4200​0010​−4210​​

Find a Basis for the Null Space of the Matrix $A = \left[ \begin{array} { r r r r r } 1 & 0 & - 4 & 0 & - 4 \\0 & 1 & 2 & 0 & 2 \\0 & 0 & 0 & 1 & 1 \\0 & 0 & 0 & 0 & 0\end{array} \right]$