Question #149667
Let T be a linear operator whose matrix with
respect to the basis {(1,1,1),(1,-1,0),(1,0,-1)} is the matrix [
4 0 0
0 1 0
0 0 1. ]
Obtain the matrix of T wrt the standard basis.
1
Expert's answer
2020-12-15T13:33:03-0500

[111110101];\begin{bmatrix} 1& 1& 1 \\ 1& -1& 0 \\ 1& 0& -1 \\ \end{bmatrix}; [400010001];\begin{bmatrix} 4& 0& 0 \\ 0& 1& 0 \\ 0& 0& 1 \\ \end{bmatrix};


[111110101]×T=[400010001];\begin{bmatrix} 1& 1& 1 \\ 1& -1& 0 \\ 1& 0& -1 \\ \end{bmatrix} \times T= \begin{bmatrix} 4& 0& 0 \\ 0& 1& 0 \\ 0& 0& 1 \\ \end{bmatrix};


T=[111110101]1[400010001];T=\begin{bmatrix} 1& 1& 1 \\ 1& -1& 0 \\ 1& 0& -1 \\ \end{bmatrix} ^{-1} \begin{bmatrix} 4& 0& 0 \\ 0& 1& 0 \\ 0& 0& 1 \\ \end{bmatrix};



Consider the matrix [111110101]\begin{bmatrix} 1& 1& 1 \\ 1& -1& 0 \\ 1& 0& -1 \\ \end{bmatrix} is A. We use Linear Row Reduction to find the Inverse Matrix of A:

A1=[1   11             100110              0101   01           001]1A^{-1}=\begin{bmatrix} 1& \ \ \ 1&1 \ \ \ \ \ \ \ \ \ \ \ \ \ 1&0&0 \\ 1& -1& 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0&1&0 \\ 1&\ \ \ 0& -1 \ \ \ \ \ \ \ \ \ \ \ 0&0&1 \\ \end{bmatrix} ^{-1} = {r1=r1+r2+r33r2=r1+r32×r23r3=r1+r22×r33}\begin{Bmatrix} r_1=\frac{r_1+r_2+r_3}{3} \\ r_2=\frac{r_1+r_3-2\times r_2}{3} \\ r_3=\frac{r_1+r_2-2\times r_3}{3} \end{Bmatrix} =


A1=[100          131313010          132313001          131323];A^{-1}=\begin{bmatrix} 1&0&0\ \ \ \ \ \ \ \ \ \ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ 0&1&0\ \ \ \ \ \ \ \ \ \ \frac{1}{3} & -\frac{2}{3} & \frac{1}{3} \\ 0&0&1\ \ \ \ \ \ \ \ \ \ \frac{1}{3} & \frac{1}{3} & -\frac{2}{3} \\ \end{bmatrix} ;


T=[131313132313131323][400010001];T=\begin{bmatrix} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & -\frac{2}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & -\frac{2}{3} \\ \end{bmatrix} \begin{bmatrix} 4& 0& 0 \\ 0& 1& 0 \\ 0& 0& 1 \\ \end{bmatrix};


T=[431313432313431323];T=\begin{bmatrix} \frac{4}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{4}{3} & -\frac{2}{3} & \frac{1}{3} \\ \frac{4}{3} & \frac{1}{3} & -\frac{2}{3} \\ \end{bmatrix} ; .


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