Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Q2SE (page 93)

Find the matrix C whose inverse is \({C^{ - {\bf{1}}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{\bf{5}}\\{\bf{6}}&{\bf{7}}\end{aligned}} \right)\).

Short Answer

Expert verified

Matrix \(C = \left( {\begin{aligned}{*{20}{c}}{{{ - 7} \mathord{\left/

{\vphantom {{ - 7} 2}} \right.

\kern-\nulldelimiterspace} 2}}&{{5 \mathord{\left/

{\vphantom {5 2}} \right.

\kern-\nulldelimiterspace} 2}}\\3&{ - 2}\end{aligned}} \right)\).

Step by step solution

01

Use the inverse property

Note that C is invertible. Since \({C^{ - 1}}\) exists, therefore,

\({\left( {{C^{ - 1}}} \right)^{ - 1}} = C\)

02

Use the formula

\({\left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)^{ - 1}} = \frac{1}{{ad - bc}}\left( {\begin{aligned}{*{20}{c}}d&{ - b}\\{ - c}&a\end{aligned}} \right)\)when \(ad - bc \ne 0\).

03

Find matrix C

\(\begin{aligned}{c}C = {\left( {\begin{aligned}{*{20}{c}}4&5\\6&7\end{aligned}} \right)^{ - 1}}\\ = \frac{1}{{4(7) - 5\left( 6 \right)}}\left( {\begin{aligned}{*{20}{c}}7&{ - 5}\\{ - 6}&4\end{aligned}} \right)\\ = \frac{1}{{28 - 30}}\left( {\begin{aligned}{*{20}{c}}7&{ - 5}\\{ - 6}&4\end{aligned}} \right)\\ = - \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}7&{ - 5}\\{ - 6}&4\end{aligned}} \right)\\C = \left( {\begin{aligned}{*{20}{c}}{{{ - 7} \mathord{\left/

{\vphantom {{ - 7} 2}} \right.

\kern-\nulldelimiterspace} 2}}&{{5 \mathord{\left/

{\vphantom {5 2}} \right.

\kern-\nulldelimiterspace} 2}}\\3&{ - 2}\end{aligned}} \right)\end{aligned}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercises 1 and 2, compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let

\(A = \left( {\begin{aligned}{*{20}{c}}2&0&{ - 1}\\4&{ - 5}&2\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}7&{ - 5}&1\\1&{ - 4}&{ - 3}\end{aligned}} \right)\), \(C = \left( {\begin{aligned}{*{20}{c}}1&2\\{ - 2}&1\end{aligned}} \right)\), \(D = \left( {\begin{aligned}{*{20}{c}}3&5\\{ - 1}&4\end{aligned}} \right)\) and \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)

\( - 2A\), \(B - 2A\), \(AC\), \(CD\).

Prove Theorem 2(d). (Hint: The \(\left( {i,j} \right)\)- entry in \(\left( {rA} \right)B\) is \(\left( {r{a_{i1}}} \right){b_{1j}} + ... + \left( {r{a_{in}}} \right){b_{nj}}\).)

Suppose \({A_{{\bf{11}}}}\) is invertible. Find \(X\) and \(Y\) such that

\[\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{{A_{{\bf{12}}}}}\\{{A_{{\bf{21}}}}}&{{A_{{\bf{22}}}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\X&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{\bf{0}}\\{\bf{0}}&S\end{array}} \right]\left[ {\begin{array}{*{20}{c}}I&Y\\{\bf{0}}&I\end{array}} \right]\]

Where \(S = {A_{{\bf{22}}}} - {A_{21}}A_{{\bf{11}}}^{ - {\bf{1}}}{A_{{\bf{12}}}}\). The matrix \(S\) is called the Schur complement of \({A_{{\bf{11}}}}\). Likewise, if \({A_{{\bf{22}}}}\) is invertible, the matrix \({A_{{\bf{11}}}} - {A_{{\bf{12}}}}A_{{\bf{22}}}^{ - {\bf{1}}}{A_{{\bf{21}}}}\) is called the Schur complement of \({A_{{\bf{22}}}}\). Such expressions occur frequently in the theory of systems engineering, and elsewhere.

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{2}}\\{\bf{5}}&{{\bf{12}}}\end{aligned}} \right),{b_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{1}}}\\{\bf{3}}\end{aligned}} \right),{b_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{ - {\bf{5}}}\end{aligned}} \right),{b_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{\bf{6}}\end{aligned}} \right),\) and \({b_{\bf{4}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{5}}\end{aligned}} \right)\).

  1. Find \({A^{ - {\bf{1}}}}\), and use it to solve the four equations \(Ax = {b_{\bf{1}}},\)\(Ax = {b_2},\)\(Ax = {b_{\bf{3}}},\)\(Ax = {b_{\bf{4}}}\)\(\)
  2. The four equations in part (a) can be solved by the same set of row operations, since the coefficient matrix is the same in each case. Solve the four equations in part (a) by row reducing the augmented matrix \(\left( {\begin{aligned}{*{20}{c}}A&{{b_{\bf{1}}}}&{{b_{\bf{2}}}}&{{b_{\bf{3}}}}&{{b_{\bf{4}}}}\end{aligned}} \right)\).

Suppose block matrix \(A\) on the left side of (7) is invertible and \({A_{{\bf{11}}}}\) is invertible. Show that the Schur component \(S\) of \({A_{{\bf{11}}}}\) is invertible. [Hint: The outside factors on the right side of (7) are always invertible. Verify this.] When \(A\) and \({A_{{\bf{11}}}}\) are invertible, (7) leads to a formula for \({A^{ - {\bf{1}}}}\), using \({S^{ - {\bf{1}}}}\) \(A_{{\bf{11}}}^{ - {\bf{1}}}\), and the other entries in \(A\).
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free