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Q2.3-33Q

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Linear Algebra and its Applications
Found in: Page 93
Linear Algebra and its Applications

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

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Short Answer

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

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

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

Most popular questions for Math Textbooks

If A is an \(n \times n\) matrix and the transformation \({\bf{x}}| \to A{\bf{x}}\) is one-to-one, what else can you say about this transformation? Justify your answer.

Generalize the idea of Exercise 21(a) [not 21(b)] by constructing a \(5 \times 5\) matrix \(M = \left[ {\begin{array}{*{20}{c}}A&0\\C&D\end{array}} \right]\) such that \({M^2} = I\). Make C a nonzero \(2 \times 3\) matrix. Show that your construction works.

Use the inverse found in Exercise 1 to solve the system

\(\begin{aligned}{l}{\bf{8}}{{\bf{x}}_{\bf{1}}} + {\bf{6}}{{\bf{x}}_{\bf{2}}} = {\bf{2}}\\{\bf{5}}{{\bf{x}}_{\bf{1}}} + {\bf{4}}{{\bf{x}}_{\bf{2}}} = - {\bf{1}}\end{aligned}\)

Suppose the transfer function W(s) in Exercise 19 is invertible for some s. It can be showed that the inverse transfer function \(W{\left( s \right)^{ - {\bf{1}}}}\), which transforms outputs into inputs, is the Schur complement of \(A - BC - s{I_n}\) for the matrix below. Find the Sachur complement. See Exercise 15.

\(\left[ {\begin{array}{*{20}{c}}{A - BC - s{I_n}}&B\\{ - C}&{{I_m}}\end{array}} \right]\)

Suppose T and U are linear transformations from \({\mathbb{R}^n}\) to \({\mathbb{R}^n}\) such that \(T\left( {U{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\) . Is it true that \(U\left( {T{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\)? Why or why not?
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