display a matrix A and an echelon form of A. Find a basis for Col A and a basis for Nul A.

$$A=\begin{bmatrix}-3 & 9 & -2 & -7\\2 & -6 & 4 & 8\\3 & -9 & -2 & 2\end{bmatrix} = \begin{bmatrix}1 & -3 & 6 & 9\\0 & 0 & 4 & 5\\0 & 0 & 0 & 0\end{bmatrix}$$

display a

Adrian

Answered by Aaron 3 years ago

to get a basis for *Col A* use a linear combination of the column vectors of *A* for which there are pivot columns,

$$ \begin{bmatrix}-3 \\2 \\3 \end{bmatrix} , \begin{bmatrix}-2 \\4 \\-2 \end{bmatrix}$$

to get a basis for *Nul A* get the reduced echelon form for $Ax=0$ which corresponds to,

$$\begin{bmatrix}1 & -3 & 0 & 1.5 & 0\\0 & 0 & 1 & 1.25 & 0\\0 & 0 & 0 & 0 & 0\end{bmatrix}$$

where $x_2$ and $x_4$ are free variables,

$$x_1=3x_2-1.5x_4$$

$$x_3=-1.25x_4$$

write the solution in parametric vector form,

$$\begin{bmatrix}x_1 \\x_2 \\x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix}3x_2-1.5x+4 \\x_2 \\-1.25x_4 \\ x_4 \end{bmatrix} = x_2 \begin{bmatrix}3 \\1 \\0 \\0 \end{bmatrix} +x_4 \begin{bmatrix}-1.5 \\0 \\-1.25 \\1 \end{bmatrix}$$

so a basis for *Nul A* is

$$ \begin{bmatrix}3 \\1 \\0 \\0 \end{bmatrix} , \begin{bmatrix}-1.5 \\0 \\-1.25 \\1 \end{bmatrix}$$

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Asked: 3 years ago