Question #158625

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


1
Expert's answer
2021-02-01T01:38:05-0500

T=(110011101)det(T)=1100110111+0det(T)=2,then, T is invertibleAugmented matrix of T is,(110100011010101001)R1+R3    R3(110100011010011101)R3+R1    R1(101001011010011101)R2+R3    R3(101001011010002111)12R3(101001011010001121212)R3+R1    R1R3+R2    R2(100121232010123212001121212)T1=(121232123212121212)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} \\


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