Question 3.8: Determine T^−1(x1,x2,x3) if T:R³ → R³ is the linear operator......
Determine \operatorname*{T^{-1}}(x_{1},x_{2},x_{3}) if T:R³ → R³ is the linear operator defined by the formula \operatorname*{T}(x_{1},x_{2},x_{3})=\bigl(3x_{1}+x_{2},-2x_{1}-4x_{2}+3x_{3},5x_{1}+4x_{2}-2x_{3}\bigr).
Let the standard matrix of the transformation
\operatorname*{T}(x_{1},x_{2},x_{3})=\bigl(3x_{1}+x_{2},-2x_{1}-4x_{2}+3x_{3},5x_{1}+4x_{2}-2x_{3}\bigr).
\\[1 em] \operatorname*{T}=\left[\ {\begin{array}{c c c}{3}&{1}&{0}\\ {-2}&{-4}&{3}\\{5}&{4}&{-2}\end{array}}\ \right]\\[1 em] \begin{array}{c}{\operatorname*{det}(T)=3(-4)-1(-4)-0(12)}\\[1 em] {=-12+4}\\[1 em] {=-8\neq0.}\end{array}
Therefore, T is one-one transformation.
Use the Gauss-Jordan method for finding T^{−1}
\pmb[\operatorname*T|\operatorname*I\pmb]=\left[\begin{array}{c c c |c c c}{{3}}&{{1}}&{{0}} & {{1}}&{{0}}&{{0}}\\ {{-2}}&{{-4}}&{{3}}&{{0}}&{{1}}&{{0}} \\ {{5}}&{{4}}&{{-2}}& {{0}}&{{0}}&{{1}}\end{array}\right]
\\[1 em] R_{1}\rightarrow \frac{1}{3}R_{1}
\\[1 em] \approx \left[\begin{array}{c c c |c c c}{{1}}&\frac{1}{3}&{{0}} & \frac{1}{3}&{{0}}&{{0}}\\[0.3 em] {{-2}}&{{-4}}&{{3}}&{{0}}&{{1}}&{{0}} \\[0.3 em] {{5}}&{{4}}&{{-2}}& {{0}}&{{0}}&{{1}} \end{array}\right]
\\[1 em] R_{2}\rightarrow R_{2}+2R_{1},\,R_{3}\rightarrow R_{3}-5R_{1}
\\[1 em] \approx \left[\begin{array}{c c c |c c c}{{1}}&\frac{1}{3}&{{0}} & \frac{1}{3}&{{0}}&{{0}}\\[0.5 em] {{0}}&\frac{-10}{3}&3&\frac{2}{3}&1&{{0}} \\[0.5 em] {{0}}&\frac{7}{3}& -2 & \frac{-5}{3}&{{0}}&{{1}}\end{array}\right]
\\[1 em] R_{2}\rightarrow- \frac{3}{10}R_{2}
\\[1 em] \approx \left[\begin{array}{c c c |c c c}{{1}}&\frac{1}{3}&{{0}} & \frac{1}{3}&{{0}}&{{0}}\\[0.5 em] {{0}}&1&-\frac{9}{10}&-\frac{1}{5}&-\frac{3}{10}&{{0}} \\[0.5 em] {{0}}&\frac{7}{3}& -2 & \frac{-5}{3}&{{0}}&{{1}}\end{array}\right]
\\[1 em] R_{1}\rightarrow R_{1}-\frac {1}{3}R_{2},\,R_{3}\rightarrow R_{3}-\frac {7}{3}R_{2}
\\[1 em] \approx \left[\begin{array}{c c c |c c c}{{1}}&0&\frac{3}{10} & \frac{2}{5}&\frac{1}{10}&{{0}}\\[0.5 em] {{0}}&1&\frac{-9}{10}&\frac{-1}{5}&\frac{-3}{10}&{{0}} \\[0.5 em] {{0}}&0& \frac{1}{10} & \frac{-6}{5}&\frac{7}{10}&{{1}}\end{array}\right]
\\[1 em] R_{3}\rightarrow 10R_{3}
\\[1 em] \approx \left[\begin{array}{c c c |c c c}{{1}}&0&\frac{3}{10} & \frac{2}{5}&\frac{1}{10}&{{0}}\\[0.5 em] {{0}}&1&\frac{-9}{10}&\frac{-1}{5}&\frac{-3}{10}&{{0}} \\[0.5 em] {{0}}&0& 1 & -12 & 7 & 10 \end{array}\right]
\\[1 em] R_{1}\rightarrow R_{1}-\frac {3}{10}R_{3},\,R_{2}\rightarrow R_{2}+\frac {9}{10}R_{3}
\\[1 em] \approx \left[\begin{array}{c c c |c c c}{{1}}&0&{{0}} & 4& -2 & -3 \\ 0 & 1 & 0 & -11 & 6 & 9 \\ 0 & 0 & 1 & -12 & 7 & 10 \end{array}\right]
\\[1 em] \pmb{\big[}\operatorname*{T^{-1}}\pmb{\big]}=\left[\begin{array}{c c c} {{4}}&{{-2}}&{{-3}}\\ {{-11}}&{{6}}&{{9}} \\ {{-12}}&{{7}}&{{10}}\end{array}\right].
Then, the inverse transformation is
\\[1 em] {T^{-1}}\left\lgroup\!\!{\left[\begin{array}{l}{\ \ x_{1}\ \ }\\ {\ \ x_{2}\ \ }\\ {\ \ x_{3}\ \ }\end{array}\right]}\!\!\right\rgroup=\left[\begin{array}{c c c} {{4}}&{{-2}}&{{-3}}\\ {{-11}}&{{6}}&{{9}} \\ {{-12}}&{{7}}&{{10}}\end{array}\right]\!\!\!\!\left[\ \ \begin{array}{l}{x_{1}}\\ {x_{2}}\\ {x_{3}}\end{array}\ \ \right]=\left[\begin{array}{c}{{4x_{1}-2x_{2}-3x_{3}}}\\ {{-11x_{1}+6x_{2}+9x_{3}}}\\ {{-12x_{1}+7x_{2}+10x_{3}}}\end{array}\right ]