In Exercises 33 and 34, Tis a linear transformation from \({\mathbb{R}^2}\) into \({\mathbb{R}^2}\). Show that T is invertible and find a formula for \({T^{ - 1}}\).

33. \(T\left( {{x_1},{x_2}} \right) = \left( { - 5{x_1} + 9{x_2},4{x_1} - 7{x_2}} \right)\)

Write the transformation \(T\left( x \right)\) and \(x\) in the column vector of \(A\).

\[\begin{array}{c}T\left( x \right) = \left[ {\begin{array}{*{20}{c}}{ - 5{x_1} + 9{x_2}}\\{4{x_1} - 7{x_3}}\end{array}} \right]\\ = \left[ A \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 5}&9\\4&{ - 7}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right]\end{array}\]

Thus, the standard matrix of T is \(A = \left[ {\begin{array}{*{20}{c}}{ - 5}&9\\4&{ - 7}\end{array}} \right]\).

Theorem 4 states that \(A = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\). If \(ad - bc \ne 0\), then A is invertible.

\({A^{ - 1}} = \frac{1}{{ad - bc}}\left[ {\begin{array}{*{20}{c}}d&{ - b}\\{ - c}&a\end{array}} \right]\)

If \(ad - bc = 0\), then Aisnot invertible.

The linear transformation T is invertible since the determinant of the matrix is non-zero.

Let\(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be a linear transformation and Abe the standard matrix for T. Then, according toTheorem 9,Tis invertible if and only if Ais an invertible matrix. The linear transformation S,given by \[S\left( x \right) = {A^{ - 1}}{\mathop{\rm x}\nolimits} \], is a unique function satisfying the equations

1. \(S\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \)for all x in \({\mathbb{R}^n}\), and
2. \(T\left( {S\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \)for all x in \({\mathbb{R}^n}\).

According to theorem 9, transformation Tis invertible and the standard matrix of \({T^{ - 1}}\) is \({A^{ - 1}}\).

Use the formula for \(2 \times 2\) inverse.

\(\begin{array}{c}{A^{ - 1}} = \frac{1}{{35 - 36}}\left[ {\begin{array}{*{20}{c}}{ - 7}&{ - 9}\\{ - 4}&{ - 5}\end{array}} \right]\\ = \frac{1}{{ - 1}}\left[ {\begin{array}{*{20}{c}}{ - 7}&{ - 9}\\{ - 4}&{ - 5}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}7&9\\4&5\end{array}} \right]\end{array}\)

Therefore,

\[\begin{array}{c}{T^{ - 1}}\left( {{x_1},{x_2}} \right) = {A^{ - 1}}x\\ = \left[ {\begin{array}{*{20}{c}}7&9\\4&5\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right]\\ = \left( {7{x_1} + 9{x_2},4{x_1} + 5{x_2}} \right)\end{array}\]

Thus, the formula for \({T^{ - 1}}\) is \[{T^{ - 1}}\left( {{x_1},{x_2}} \right) = \left( {7{x_1} + 9{x_2},4{x_1} + 5{x_2}} \right)\].