 # Find a matrix transformation mapping \{(1,1,1),(0,1,0),(1,0,2)\} to \{(1,1,1),(0,1,0),(1,0,1)\} Connor Randall 2022-01-30 Answered
Find a matrix transformation mapping
$\left\{\left(1,1,1\right),\left(0,1,0\right),\left(1,0,2\right)\right\}$
to
$\left\{\left(1,1,1\right),\left(0,1,0\right),\left(1,0,1\right)\right\}$
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Step 1
We wish to find a $3×3$ matrix T such that $TA=B$ where
$\begin{array}{rlrl}A& =\left[\begin{array}{ccc}1& 0& 1\\ 1& 1& 0\\ 1& 0& 2\end{array}\right]& B& =\left[\begin{array}{ccc}1& 0& 1\\ 1& 1& 0\\ 1& 0& 1\end{array}\right]\end{array}$
Perhaps the quickest way to find T is to multiply the equation $TA=B$ on the right by ${A}^{-1}$ to obtain
$T=B{A}^{-1}$
###### Not exactly what you’re looking for? mihady54
Step 1
The columns of the matrix tell you where it sends the standard basis vectors. For instance if I am interested in the third column then I need to determine what the action of our linear operator is on the column vector ,
$\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right].$
This vector can be written as a linear combination of the vectors used to define the linear operator,
$\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 2\end{array}\right]-\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]-\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right].$
Multiplying both sides by our linear operator M we get,
$M\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]=M\left[\begin{array}{c}1\\ 0\\ 2\end{array}\right]-M\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]-M\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right].$
Note that we know what M does to the vectors on the right so we can just substitute those values in and add,
$M\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]-\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]-\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ -2\\ 0\end{array}\right].$
The resulting vector is the third column of our matrix,

A similar process will yield the other collumns.