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Expert-verified Found in: Page 93 ### 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 $${T^{ - 1}}$$.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 ${T^{ - 1}}\left( {{x_1},{x_2}} \right) = \left( {7{x_1} + 9{x_2},4{x_1} + 5{x_2}} \right)$.

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 $$A$$.

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

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

## Step 2: Show that T  is invertible

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.

## 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 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 ${T^{ - 1}}\left( {{x_1},{x_2}} \right) = \left( {7{x_1} + 9{x_2},4{x_1} + 5{x_2}} \right)$. ### Want to see more solutions like these? 