Question

Remaining Time: 45 minutes, 24 seconds. Question Completion Status: 2 4 6 7 9 10 20 216220 23 24 25 Dryopo Question 21 QUESTION 10A If the linear transformation T(v) = AV maps the basis vectors inputted as shown below, what is the matrix A? (O)-4-(0)-8-8-1 Input your answer in the box below without brackets. (Exactly as you would do it in a Zoom chat.) TTTT Paragraph Arial 3 (120) T- O SOM THES boc Pathop Moving to another question will save this response Type here to search o BI e MacBook Air

Answer 1

The linear transformation   is given by ,

$T\left ( \begin{bmatrix} a\\ 0 \\ 0 \end{bmatrix} \right )=\begin{bmatrix} 0\\ a \\ 0 \end{bmatrix}$ ,  
T b 0 b
and  $T\left ( \begin{bmatrix} 0\\ 0 \\ c \end{bmatrix} \right )=\begin{bmatrix} c\\ 0 \\ c \end{bmatrix}$

substituting  $a=1$ we get ,

$T\left ( \begin{bmatrix} 1\\ 0 \\ 0 \end{bmatrix} \right )=\begin{bmatrix} 0\\ 1 \\ 0 \end{bmatrix}$

substituting  $b=1$  we get ,

$T\left ( \begin{bmatrix} 0\\ 1 \\ 0 \end{bmatrix} \right )=\begin{bmatrix} 1\\ 0 \\ 1 \end{bmatrix}$

substituting  $c=1$  we get ,

$T\left ( \begin{bmatrix} 0\\ 0 \\ 1 \end{bmatrix} \right )=\begin{bmatrix} 1\\ 0 \\ 1 \end{bmatrix}$

So ,

$T\left ( \begin{bmatrix} 1\\ 0 \\ 0 \end{bmatrix} \right )=0.\begin{bmatrix} 1\\ 0 \\ 0 \end{bmatrix}+1.\begin{bmatrix} 0\\ 1 \\ 0 \end{bmatrix}+0.\begin{bmatrix} 0\\ 0 \\ 1 \end{bmatrix}$

$T\left ( \begin{bmatrix} 0\\ 1 \\ 0 \end{bmatrix} \right )=1.\begin{bmatrix} 1\\ 0 \\ 0 \end{bmatrix}+0.\begin{bmatrix} 0\\ 1 \\ 0 \end{bmatrix}+1.\begin{bmatrix} 0\\ 0 \\ 1 \end{bmatrix}$

$T\left ( \begin{bmatrix} 0\\ 0 \\ 1 \end{bmatrix} \right )=1.\begin{bmatrix} 1\\ 0 \\ 0 \end{bmatrix}+0.\begin{bmatrix} 0\\ 1 \\ 0 \end{bmatrix}+1.\begin{bmatrix} 0\\ 0 \\ 1 \end{bmatrix}$

Writ ting the coefficient column wise the required matrix   is ,

$A=\begin{bmatrix} 0 & 1 & 1\\ 1 & 0 & 0\\ 0 & 1 & 1 \end{bmatrix}$

Hence ,

Answer :  $A=\begin{bmatrix} 0 & 1 & 1\\ 1 & 0 & 0\\ 0 & 1 & 1 \end{bmatrix}$ .

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