Question

Chapter 7: Problem 4 1 4 -2 6 (1 point) Let A- 3 6 -24 12 36 | Find basis for the kernal and image of the linear transformation T defined by T 12 6 18 Z)-Ai SAMSUNG

Answer 1

Solution:

_{$\small Ker\left ( A \right )$} is the set of solution to $\small AX=0$ .

$\small \begin{bmatrix} 1 &4 & -2 & 6\\ 3& 12 & -6 &18 \\ -6 &-24 & 12 & -36 \end{bmatrix}\begin{bmatrix} a\\ b\\ c\\ d\\ \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0\\ \end{bmatrix}$

By _{$\small R_{2}-3R_{1},\: \: R_{3}+6R_{1}$}

$\small \begin{bmatrix} 1 &4 & -2 & 6\\ 0& 0 & 0 &0 \\ 0 &0 & 0 & 0 \end{bmatrix}\begin{bmatrix} a\\ b\\ c\\ d\\ \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0\\ \end{bmatrix}$

The rank of the matrix $\small A$ is  $\small 1$ which is less than the number of unknown _{$\small \left ( =4 \right )$} .

So the number of free variables is  $\small 4-1=3.$

$\small \therefore a+4b-2c+6d=0\Rightarrow a=-4b+2c-6d$

$\small \therefore X=\begin{bmatrix} a\\ b\\ c\\ d\\ \end{bmatrix}=\begin{bmatrix} -4b+2c-6d\\ b\\ c\\ d\\ \end{bmatrix}=b\begin{bmatrix} -4\\ 1\\ 0\\ 0\\ \end{bmatrix}+c\begin{bmatrix} 2\\ 0\\ 1\\ 0\\ \end{bmatrix}+d\begin{bmatrix} -6\\ 0\\ 0\\ 1\\ \end{bmatrix}$

$\small \therefore$Kernel basis:

$\small \left \{ \begin{bmatrix} -4\\ 1\\ 0\\ 0\\ \end{bmatrix},\: \begin{bmatrix} 2\\ 0\\ 1\\ 0\\ \end{bmatrix},\: \begin{bmatrix} -6\\ 0\\ 0\\ 1\\ \end{bmatrix} \right \}$

In the row reduced echelon form of $\small A$ , the first column form the pivot column.

So the first column of the original matrix form the basis for the image of $\small A$ .

Image basis:

$\small \left \{ \begin{bmatrix} 1\\ 3\\ -6\\ \end{bmatrix} \right \}$