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Q2.3-33Q

Expert-verifiedFound in: Page 93

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

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

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

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

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 *A *is **not 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 *A *be the standard matrix for *T*. Then, according to **Theorem 9, ***T *is invertible if and only if *A *is 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

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

** **

According to theorem 9, transformation *T *is 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 \[{T^{ - 1}}\left( {{x_1},{x_2}} \right) = \left( {7{x_1} + 9{x_2},4{x_1} + 5{x_2}} \right)\].

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