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### Linear Algebra and its Applications

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 $${T^{ - 1}}$$.34. $$T\left( {{x_1},{x_2}} \right) = \left( {6{x_1} - 8{x_2}, - 5{x_1} + 7{x_2}} \right)$$

The formula for $${T^{ - 1}}$$ is $${T^{ - 1}}\left( {{x_1},{x_2}} \right) = \left( {\frac{7}{2}{x_1} + 4{x_2},\frac{5}{2}{x_1} + 3{x_2}} \right)$$.

See the step by step solution

## Step 1: Determine the standard matrix T

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

\begin{aligned}{c}T\left( x \right) = \left( {\begin{aligned}{*{20}{c}}{6{x_1} - 8{x_2}}\\{ - 5{x_1} + 7{x_3}}\end{aligned}} \right)\\ = \left( A \right)\left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}6&{ - 8}\\{ - 5}&7\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\end{aligned}

Thus, the standard matrix of T is A = \left( {\begin{aligned}{*{20}{c}}6&{ - 8}\\{ - 5}&7\end{aligned}} \right).

## Step 2: Show that T  is invertible

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

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

If $$ad - bc = 0$$, then A is not invertible.

The linear transformation T is invertible since det$$A = 2 \ne 0$$.

## Step 3: Determine the formula for $${T^{ - 1}}$$

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

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 T is invertible and $${T^{ - 1}}\left( x \right) = Bx$$, where$$B = {A^{ - 1}}$$.

Use the formula for $$2 \times 2$$ inverse.

\begin{aligned}{c}{A^{ - 1}} = \frac{1}{{42 - 40}}\left( {\begin{aligned}{*{20}{c}}7&8\\5&6\end{aligned}} \right)\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}7&8\\5&6\end{aligned}} \right)\end{aligned}

Therefore,

\begin{aligned}{c}{T^{ - 1}}\left( {{x_1},{x_2}} \right) = {A^{ - 1}}x\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}7&8\\5&6\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\\ = \left( {\frac{7}{2}{x_1} + 4{x_2},\frac{5}{2}{x_1} + 3{x_2}} \right)\end{aligned}

Thus, the formula for $${T^{ - 1}}$$ is $${T^{ - 1}}\left( {{x_1},{x_2}} \right) = \left( {\frac{7}{2}{x_1} + 4{x_2},\frac{5}{2}{x_1} + 3{x_2}} \right)$$.