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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

Suppose A, B, and C are invertible \(n \times n\) matrices. Show that ABC is also invertible by producing a matrix D such that \(\left( {ABC} \right)D = I\) and \(D\left( {ABC} \right) = I\).

It is proved that \(\left( {ABC} \right)D = I\) and \(D\left( {ABC} \right) = I\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Condition for an invertible matrix

Theorem 6 states that if A and B are \(n \times n\) invertible matrices, the inverse of AB is the product of the inverses of A and B in the reverse order. That is, \({\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}\).

Step 2: Show that \(\left( {ABC} \right)D = I\) and \(D\left( {ABC} \right) = I\)

Consider \(D = {C^{ - 1}}{B^{ - 1}}{A^{ - 1}}\). Then,

\(\begin{aligned}{c}\left( {ABC} \right){C^{ - 1}}{B^{ - 1}}{A^{ - 1}} = AB\left( {C{C^{ - 1}}} \right){B^{ - 1}}A\\ = ABI{B^{ - 1}}{A^{ - 1}}\\ = AI{A^{ - 1}}\\ = I\end{aligned}\)

And

\(\begin{aligned}{c}{C^{ - 1}}{B^{ - 1}}{A^{ - 1}}\left( {ABC} \right) = {C^{ - 1}}{B^{ - 1}}{A^{ - 1}}ABC\\ = {C^{ - 1}}{B^{ - 1}}IBC\\ = {C^{ - 1}}IC\\ = I\end{aligned}\)

Hence, it is proved that \(\left( {ABC} \right)D = I\) and \(D\left( {ABC} \right) = I\).

Most popular questions for Math Textbooks

Use partitioned matrices to prove by induction that for \(n = 2,3,...\), the \(n \times n\) matrices \(A\) shown below is invertible and \(B\) is its inverse.

\[A = \left[ {\begin{array}{*{20}{c}}1&0&0& \cdots &0\\1&1&0&{}&0\\1&1&1&{}&0\\ \vdots &{}&{}& \ddots &{}\\1&1&1& \ldots &1\end{array}} \right]\]

\[B = \left[ {\begin{array}{*{20}{c}}1&0&0& \cdots &0\\{ - 1}&1&0&{}&0\\0&{ - 1}&1&{}&0\\ \vdots &{}& \ddots & \ddots &{}\\0&{}& \ldots &{ - 1}&1\end{array}} \right]\]

For the induction step, assume A and B are \(\left( {k + 1} \right) \times \left( {k + 1} \right)\) matrices, and partition A and B in a form similar to that displayed in Exercises 23.

Suppose A is an \(n \times n\) matrix with the property that the equation \(Ax = 0\)has only the trivial solution. Without using the Invertible Matrix Theorem, explain directly why the equation \(Ax = b\) must have a solution for each b in \({\mathbb{R}^n}\).

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

Let \(A = \left( {\begin{aligned}{*{20}{c}}1&1&1\\1&2&3\\1&4&5\end{aligned}} \right)\), and \(D = \left( {\begin{aligned}{*{20}{c}}2&0&0\\0&3&0\\0&0&5\end{aligned}} \right)\). Compute \(AD\) and \(DA\). Explain how the columns or rows of A change when A is multiplied by D on the right or on the left. Find a \(3 \times 3\) matrix B, not the identity matrix or the zero matrix, such that \(AB = BA\).

In the rest of this exercise set and in those to follow, you should assume that each matrix expression is defined. That is, the sizes of the matrices (and vectors) involved match appropriately.

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