Answer to Question #158625 in Linear Algebra for dineth

Question

𝑇: ℝ3→ℝ3 defined by 𝑇(𝑥1,𝑥2,𝑥3)=(𝑥1+𝑥2,𝑥2+𝑥3,𝑥3+𝑥1).

Answer 1

$T=\begin{pmatrix} 1 & 1 & 0\\ 0 & 1 & 1\\ 1 & 0 & 1 \end{pmatrix}\\ det(T)=1\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}-1\begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix}+0\\ det(T)=2,\\ \text{then, T is invertible}\\ \text{Augmented matrix of T is,}\\ \begin{pmatrix} 1 & 1 & 0 & & 1 & 0 & 0\\ 0 & 1 & 1 & & 0 & 1 & 0\\ 1 & 0 & 1 & & 0 & 0 & 1 \end{pmatrix} \\-R_1+R_3 \implies R_3\\ \begin{pmatrix} 1 & 1 & 0 & & 1 & 0 & 0\\ 0 & 1 & 1 & & 0 & 1 & 0\\ 0 & -1 & 1 & & -1 & 0 & 1 \end{pmatrix} \\R_3+R_1 \implies R_1\\ \begin{pmatrix} 1 & 0 & 1 & & 0 & 0 & 1\\ 0 & 1 & 1 & & 0 & 1 & 0\\ 0 & -1 & 1 & & -1 & 0 & 1 \end{pmatrix} \\R_2+R_3 \implies R_3\\ \begin{pmatrix} 1 & 0 & 1 & & 0 & 0 & 1\\ 0 & 1 & 1 & & 0 & 1 & 0\\ 0 & 0 & 2 & & -1 & 1 & 1 \end{pmatrix} \\ \frac{1}{2}R_3\\ \begin{pmatrix} 1 & 0 & 1 & & 0 & 0 & 1\\ 0 & 1 & 1 & & 0 & 1 & 0\\ 0 & 0 & 1 & & -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{pmatrix} \\ -R_3+R_1 \implies R_1\\ -R_3+R_2 \implies R_2\\ \begin{pmatrix} 1 & 0 & 0 & & \frac{1}{2} & -\frac{1}{2} & \frac{3}{2} \\ 0 & 1 & 0 & & \frac{1}{2} & \frac{3}{2} & -\frac{1}{2}\\ 0 & 0 & 1 & & -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{pmatrix} \\ T^{-1}=\begin{pmatrix} \frac{1}{2} & -\frac{1}{2} & \frac{3}{2} \\ \frac{1}{2} & \frac{3}{2} & -\frac{1}{2}\\ -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{pmatrix} \\$