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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

Let A be an invertible \(n \times n\) matrix, and let \(B\) be an \(n \times p\) matrix. Explain why \({A^{ - 1}}B\) can be computed by row reduction: If\(\left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right) \sim ... \sim \left( {\begin{aligned}{*{20}{c}}I&X\end{aligned}} \right)\), then \(X = {A^{ - 1}}B\).

If A is larger than \(2 \times 2\), then row reduction of \(\left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right)\) is much faster than computing both \({A^{ - 1}}\) and \({A^{ - 1}}B\).

The solution \(\begin{aligned}{*{20}{c}}{{{\mathop{\rm u}\nolimits} _1}}&{...}&{{{\mathop{\rm u}\nolimits} _p}}\end{aligned}\) can be uniquely determined since A is an invertible matrix, and \(\left( {\begin{aligned}{*{20}{c}}A&{{b_1}}&{...}&{{b_p}}\end{aligned}} \right)\) must row reduced to \(\left( {\begin{aligned}{*{20}{c}}I&{{{\mathop{\rm u}\nolimits} _1}}&{...}&{{{\mathop{\rm u}\nolimits} _p}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}I&X\end{aligned}} \right)\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: The equation \(AX = B\) is equivalent to the p systems

It is known that \(A\) is a \(m \times n\) matrix. If B is a \(n \times p\) matrix with columns \({{\mathop{\rm b}\nolimits} _1},...,{{\mathop{\rm b}\nolimits} _p}\), the product AB is the \(m \times p\) matrix whose columns are \(A{{\mathop{\rm b}\nolimits} _1},...,A{{\mathop{\rm b}\nolimits} _p}\). That is, \(AB = A\left( {\begin{aligned}{*{20}{c}}{{b_1}}&{{b_2}}&{{b_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{A{b_1}}&{A{b_2}}&{A{b_3}}\end{aligned}} \right)\) .

Write the matrix \(B = \left( {\begin{aligned}{*{20}{c}}{{b_1}}&{...}&{{b_p}}\end{aligned}} \right)\) and \(X = \left( {\begin{aligned}{*{20}{c}}{{{\mathop{\rm u}\nolimits} _1}}&{...}&{{{\mathop{\rm u}\nolimits} _p}}\end{aligned}} \right)\). \(AX = \left( {\begin{aligned}{*{20}{c}}{A{{\mathop{\rm u}\nolimits} _1}}&{...}&{A{{\mathop{\rm u}\nolimits} _p}}\end{aligned}} \right)\) according to the matrix multiplication. Therefore, the equation \(AX = B\) is equivalent to the p system.

\(A{{\mathop{\rm u}\nolimits} _1} = {{\mathop{\rm b}\nolimits} _1},...,A{{\mathop{\rm u}\nolimits} _p} = {{\mathop{\rm b}\nolimits} _p}\)

Since A is the coefficient matrix in each system, these systems can be solved simultaneously by placing the augmented columns of each system next to A such that \(\left( {\begin{aligned}{*{20}{c}}A&{{b_1}}&{...}&{{b_p}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right)\).

Step 2: The row reduced matrix of \(\left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right)\)

If A is an invertible \(n \times n\) matrix, for each b in \({\mathbb{R}^n}\), the equation \(Ax = b\) has the unique solution \(x = {A^{ - 1}}b\).

The solution \(\begin{aligned}{*{20}{c}}{{{\mathop{\rm u}\nolimits} _1}}&{...}&{{{\mathop{\rm u}\nolimits} _p}}\end{aligned}\) can be uniquely determined since A is an invertible matrix, and \(\left( {\begin{aligned}{*{20}{c}}A&{{b_1}}&{...}&{{b_p}}\end{aligned}} \right)\)must row reduced to \(\left( {\begin{aligned}{*{20}{c}}I&{{{\mathop{\rm u}\nolimits} _1}}&{...}&{{{\mathop{\rm u}\nolimits} _p}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}I&X\end{aligned}} \right)\). Thus, \(X\) is a unique solution \({A^{ - 1}}B\) of the equation \(AX = B\).

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.

Suppose a linear transformation \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) has the property that \(T\left( {\mathop{\rm u}\nolimits} \right) = T\left( {\mathop{\rm v}\nolimits} \right)\) for some pair of distinct vectors u and v in \({\mathbb{R}^n}\). Can T map \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\)? Why or why not?

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.

Let \(S = \left( {\begin{aligned}{*{20}{c}}0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0\end{aligned}} \right)\). Compute \({S^k}\) for \(k = {\bf{2}},...,{\bf{6}}\).

[M] For block operations, it may be necessary to access or enter submatrices of a large matrix. Describe the functions or commands of your matrix program that accomplish the following tasks. Suppose A is a \(20 \times 30\) matrix.

  1. Display the submatrix of A from rows 15 to 20 and columns 5 to 10.
  2. Insert a \(5 \times 10\) matrix B into A, beginning at row 10 and column 20.
  3. Create a \(50 \times 50\) matrix of the form \(B = \left[ {\begin{array}{*{20}{c}}A&0\\0&{{A^T}}\end{array}} \right]\).

[Note: It may not be necessary to specify the zero blocks in B.]
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