Question 3.6: Let T:R³ → R³ be the linear operator defined by the formula ......
Let T:R³ → R³ be the linear operator defined by the formula T\left(x_{1},x_{2},x_{3}\right)=\left(x_{1}-x_{2}+x_{3},2x_{2}-x_{3},2x_{1}+3x_{2}\right). Determine whether T is one-one. If so, find T^{-1}(x_{1},x_{2},x_{3}).
The standard matrix of T is T={\left[\begin{array}{c c c}{1}&{-1}&{1}\\ {0}&{2}&{-1}\\ {2}&{3}&{0}\end{array}\right]}
\operatorname*{det}(A)=1(0+3)+1(0+2)+1(0-4)\\[0.7 em] =3+2-4 \\[0.7 em] =1 \neq 0 .
Therefore, T is one-one.
Using the Gauss-Jordan method to find T^{-1}
\pmb[T|I\pmb]=\left[\begin{array}{c c c |c c c}{{1}}&{{-1}}&{{1}} & {{1}}&{{0}}&{{0}}\\ {{0}}&{{2}}&{{-1}}&{{0}}&{{1}}&{{0}} \\ {{2}}&{{3}}&{{0}}& {{0}}&{{0}}&{{1}}\end{array}\right]
\\[1 em] R_{3}\rightarrow R_{3}-2R_{1}
\\[1 em] \approx \left[\begin{array}{c c c |c c c}{{1}}&{{-1}}&{{1}} & {{1}}&{{0}}&{{0}}\\ {{0}}&{{2}}&{{-1}}&{{0}}&{{1}}&{{0}} \\ {{0}}&{{5}}&{{-2}}& {{-2}}&{{0}}&{{1}}\end{array}\right]
\\[1 em] R_{2}\to{\frac{1}{2}}R_{2}
\\[1 em] \approx \left[\begin{array}{c c c |c c c}{{1}}&{{-1}}&{{1}} & {{1}}&{{0}}&{{0}}\\[0.5 em] {{0}}&{{1}}&{{-\frac{1}{2}}}&{{0}}&{{\frac{1}{2}}}&{{0}} \\[0.5 em] {{0}}&{{5}}&{{-2}}& {{-2}}&{{0}}&{{1}}\end{array}\right]
\\[1 em] R_{3}\rightarrow R_{3}-5R_{2},\,R_{1}\rightarrow R_{1}+R_{2}
\\[1 em] \approx \left[\begin{array}{c c c |c c c} {{1}}&{{0}}&{{\frac{1}{2}}} & {{1}}&{{\frac{1}{2}}}&{{0}}\\[0.5 em] {{0}}&{{1}}&{{-\frac{1}{2}}}&{{0}}&{{\frac{1}{2}}}&{{0}} \\[0.5 em] {{0}}&{{0}}&{{\frac{1}{2}}}& {{-2}}&{-{\frac{5}{2}}}&{{1}} \end{array}\right]
\\[1 em] R_{3}\rightarrow2R_{3}
\\[1 em] \approx \left[\begin{array}{c c c |c c c} &&& \\[-1 em] {{1}}&{{0}}&{{\frac{1}{2}}} & {{1}}&{{\frac{1}{2}}}&{{0}}\\[0.5 em] {{0}}&{{1}}&{{-\frac{1}{2}}}&{{0}}&{{\frac{1}{2}}}&{{0}} \\[0.5 em] {{0}}&{{0}}&{{1}}& {{-4}}&{{-5}}&{{2}} \end{array}\right]
\\[1 em] R_{2}\to R_{2}+{\frac{1}{2}}R_{3},\;R_{1}\to R_{1}-{\frac{1}{2}}R_3
\\[1 em] \pmb [\operatorname*{I}|\operatorname*{T^{-1}}\pmb ]=\left[\begin{array}{c c c |c c c}{{1}}&{{0}}&{{0}} & {{3}}&{{3}}&{{-1}}\\ {{0}}&{{1}}&{{0}}&{{-2}}&{{-2}}&{{1}} \\ {{0}}&{{0}}&{{1}}& {{-4}}&{{-5}}&{{2}}\end{array}\right]
\\[1 em] \operatorname*{T^{-1}}=\left[\begin{array}{c c c} {{3}}&{{3}}&{{-1}}\\ {{-2}}&{{-2}}&{{1}} \\ {{-4}}&{{-5}}&{{2}}\end{array}\right]
\\[1 em] \operatorname*{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} {{3}}&{{3}}&{{-1}}\\ {{-2}}&{{-2}}&{{1}} \\ {{-4}}&{{-5}}&{{2}}\end{array}\right]\!\!\left[\ \ \begin{array}{l}{x_{1}}\\ {x_{2}}\\ {x_{3}}\end{array}\ \ \right]=\left[\begin{array}{c}{{3x_{1}+3x_{2}-x_{3}}}\\ {{-2x_{1}-2x_{2}+x_{3}}}\\ {{-4x_{1}-5x_{2}+2x_{3}}}\end{array}\right ]
\\[1 em] \operatorname*{T^{-1}}\bigl(x_{1},x_{2},x_{3}\bigr)=\bigl(3x_{1}+3x_{2}-x_{3},-2x_{1}-2x_{2}+x_{3},-4x_{1}-5x_{2}+2x_{3}\bigr).