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Q2.3-29Q

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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

If A is an \(n \times n\) matrix and the equation \(A{\bf{x}} = {\bf{b}}\) has more than one solution for some b, then the transformation \({\bf{x}}| \to A{\bf{x}}\) is not one-to-one. What else can you say about this transformation? Justify your answer.

The transformation is not invertible.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: State the invertible matrix theorem

The statements are identical according to the invertible matrix theorem, as shown below:

  1. The matrix is invertible.
  2. For some b, the matrix equation \[A{\bf{x}} = {\bf{b}}\] does not have a unique solution (more than one solution).
  3. The linear transformation \(x| \to Ax\) is one-to-one.
  4. The mapping of \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\) is equivalent to the linear transformation \(x| \to Ax\).

Step 2: Define the transformation

According to the invertible matrix theorem, if the matrix equation \[A{\bf{x}} = {\bf{b}}\] has more than one solution, the transformation is one-to-one.

From the given statement, the matrix equation has more than one solution for some b, but the linear transformation \(x| \to Ax\) is not one-to-one.

So, the given statement cannot be true.

Therefore, both the matrix and the transformation are not invertible.

Most popular questions for Math Textbooks

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.

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 \(A\).

[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.]

Suppose A is an \(m \times n\) matrix and there exist \(n \times m\) matrices C and D such that \(CA = {I_n}\) and \(AD = {I_m}\). Prove that \(m = n\) and \(C = D\). (Hint: Think about the product CAD.)

Suppose P is invertible and \(A = PB{P^{ - 1}}\). Solve for B in terms of A.
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