Chapter 6.CR, Problem 1CR

To determine

(a)

To find:

The image of $v$.

$T:R^2\to R^2$, $T\left(v_1,v_2\right)=\left(v_1,v_1+2v_2\right)$, $v=\left(2,-3\right)$

Answer to Problem 1CR

Solution:

$w=\left(2,-4\right)$ is the image of $v$.

Explanation of Solution

Given:

The given function is,

$T\left(v_1,v_2\right)=\left(v_1,v_1+2v_2\right)$,

The points are,

$v=\left(2,-3\right)$, and $w=\left(4,12\right)$.

Approach:

If $v$ is in $\text{V}$ and $w$ is in $\text{W}$ such that $T\left(v\right)=w$, then $w$ is the image of $v$.

Calculation:

Given that,

$v=\left(2,-3\right)$

Therefore,

$\begin{array}{rll}T\left(v\right)& =& \left(2,2+2\left(-3\right)\right)\\ & =& \left(2,-4\right)\end{array}$

Therefore, the $w=\left(2,-4\right)$ is the image of $v$.

To determine

(b)

To find:

The preimage of $w$.

$T:R^2\to R^2$, $T\left(v_1,v_2\right)=\left(v_1,v_1+2v_2\right)$, $w=\left(4,12\right)$.

Answer to Problem 1CR

Solution:

The preimage of $w$ is $\left(4,4\right)$.

Explanation of Solution

Approach:

The set of vectors in $v$ is in $\text{V}$ such that $T\left(v\right)=w$, then the set $v$ is the preimage of $w$.

Calculation:

Here,

$\begin{array}{rll}T\left(v_1,v_2\right)& =& \left(v_1,v_1+2v_2\right)\\ \left(4,12\right)& =& \left(v_1,v_1+2v_2\right)\end{array}$

Write this system as a system of linear equations.

$\begin{array}{rll}4& =& v_1\\ 12& =& v_1+2v_2\end{array}$……$\left(1\right)$

Substitute $4$ for $v_1$ in equation $\left(1\right)$.

$\begin{array}{rll}12& =& 4+2v_2\\ 2v_2& =& 12-4\\ v_2& =& \frac{8}{2}\\ & =& 4\end{array}$

Therefore, the preimage of $w$ is $\left(4,4\right)$.