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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

3. Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{5}}\\{ - {\bf{7}}}&{ - {\bf{5}}}\end{aligned}} \right)\).

The inverse of \(\left( {\begin{aligned}{*{20}{c}}8&5\\{ - 7}&{ - 5}\end{aligned}} \right)\) is \(\left( {\begin{aligned}{*{20}{c}}1&1\\{ - 1.4}&{ - 1.6}\end{aligned}} \right)\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Check if the matrix is invertible

\(\begin{aligned}{c}\det \left( {\left( {\begin{aligned}{*{20}{c}}8&5\\{ - 7}&{ - 5}\end{aligned}} \right)} \right) = 8\left( { - 5} \right) - 5\left( { - 7} \right)\\ = - 40 + 35\\\det \left( {\left( {\begin{aligned}{*{20}{c}}8&5\\{ - 7}&{ - 5}\end{aligned}} \right)} \right) = - 5 \ne 0\end{aligned}\)

This implies that \(\left( {\begin{aligned}{*{20}{c}}8&5\\{ - 7}&{ - 5}\end{aligned}} \right)\) is invertible.

Step 2: 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 \(ad - bc \ne 0\).

Step 3: Write the inverse matrix

\(\begin{aligned}{c}{\left( {\begin{aligned}{*{20}{c}}8&5\\{ - 7}&{ - 5}\end{aligned}} \right)^{ - 1}} = \frac{1}{{ - 5}}\left( {\begin{aligned}{*{20}{c}}{ - 5}&{ - 5}\\7&8\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&1\\{ - \frac{7}{5}}&{ - \frac{8}{5}}\end{aligned}} \right)\\{\left( {\begin{aligned}{*{20}{c}}8&5\\{ - 7}&{ - 5}\end{aligned}} \right)^{ - 1}} = \left( {\begin{aligned}{*{20}{c}}1&1\\{ - 1.4}&{ - 1.6}\end{aligned}} \right)\end{aligned}\)

Most popular questions for Math Textbooks

The inverse of \(\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\C&I&{\bf{0}}\\A&B&I\end{array}} \right]\) is \(\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\Z&I&{\bf{0}}\\X&Y&I\end{array}} \right]\). Find X, Y, and Z.

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5–8, find formulas for X, Y, and Z in terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint: Compute the product on the left, and set it equal to the right side.]

6. \[\left[ {\begin{array}{*{20}{c}}X&{\bf{0}}\\Y&Z\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&{\bf{0}}\\B&C\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\{\bf{0}}&I\end{array}} \right]\]

Suppose A is an \(n \times n\) matrix with the property that the equation \[A{\mathop{\rm x}\nolimits} = 0\] has at least one solution for each b in \({\mathbb{R}^n}\). Without using Theorem 5 or 8, explain why each equation Ax = b has in fact exactly one solution.

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 the Theorem 3(d) i.e., \({\left( {AB} \right)^T} = {B^T}{A^T}\).

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