### Prove: Provided it exists, $(A^{-1})^{-1} = A$.To prove this mathematical statement, we'll work through the concept of the inverse of a matrix. 1. **Definition of the Inverse Matrix:** For a matrix $ A $, the inverse $ A^{-1} $ is defined such that: \[ A \cdot A^{-1} = A^{-1} \cdot A = I \] where $ I $ is the identity matrix.2. **Inverse of the Inverse Matrix:** Given $ A $ is an invertible matrix, its inverse $ A^{-1} $ also has an inverse, denoted as $ (A^{-1})^{-1} $.3. **Property of Matrix Inverses:** The inverse of an inverse matrix should satisfy the original property of inverses. That is: \[ A^{-1} \cdot (A^{-1})^{-1} = I \] and \[ (A^{-1})^{-1} \cdot A^{-1} = I \]4. **Proof:** By the definition of the inverse, multiplying $ A^{-1} $ by $ (A^{-1})^{-1} $ must yield the identity matrix: \[ A^{-1} \cdot (A^{-1})^{-1} = I \] To retrieve the original matrix $ A $ from $ (A^{-1})^{-1} $, notice that: \[ A \cdot A^{-1} = A^{-1} \cdot A = I \] implies that $ (A^{-1})^{-1} $ behaves in the same fashion as $ A $.5. **Conclusion:** Since $ (A^{-1})^{-1} $ satisfies the same property as $ A $, we can conclude: \[ (A^{-1})^{-1} = A \] Thus, we have proved the given statement. **Matrix Multiplication and Scalar Addition Proof****Problem Statement:**Prove the following matrix equation:If $ a, b \in \mathbb{R} $ and $ P $ is a $ m \times n $ matrix, then the following equality holds: \[ aP + bP = (a + b)P \]**Explanation:**Given scalars $ a $ and $ b $ in the set of real numbers $ \mathbb{R} $, and a matrix $ P $ of dimensions $ m \times n $:- The term $ aP $ refers to the scalar multiplication of matrix $ P $ by the scalar $ a $.- Similarly, $ bP $ is the scalar multiplication of matrix $ P $ by the scalar $ b $.- The expression $ aP + bP $ involves the addition of these two resultant matrices.**Proof:**To prove the given equation, consider the operations involved:1. **Scalar Multiplication:** When a matrix $ P $ is multiplied by a scalar $ c $ (where $ c $ is either $ a $ or $ b $): \[ cP = c \begin{pmatrix} p_{11} & p_{12} & \cdots & p_{1n} \\ p_{21} & p_{22} & \cdots & p_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ p_{m1} & p_{m2} & \cdots & p_{mn} \end{pmatrix} = \begin{pmatrix} cp_{11} & cp_{12} & \cdots & cp_{1n} \\ cp_{21} & cp_{22} & \cdots & cp_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ cp_{m1} & cp_{m2} & \cdots & cp_{mn} \end{pmatrix} \]2. **Matrix Addition:** Adding the resulting matrices $ aP $ and $ bP $: \[ aP + bP = \begin{pmatrix} ap_{11}

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Answered: Prove: Provided it exists, (A-1)¯ = A.

Math

Advanced Math

Prove: Provided it exists, (A-1)¯ = A.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Concept explainers

Limits

When it comes to calculus, limits are considered to be a very important topic of discussion. The main importance of limit lies in expressing the value of a function as it gets closer to a certain value. Also, limits can help you to understand other topics, such as continuity and differentiability better. In a broader sense, every calculus notion is a limit. Other branches of calculus have also been derived from limits.

Question

### Prove: Provided it exists, $(A^{-1})^{-1} = A$.

To prove this mathematical statement, we'll work through the concept of the inverse of a matrix.

1. **Definition of the Inverse Matrix:**
For a matrix $ A $, the inverse $ A^{-1} $ is defined such that:
\[
A \cdot A^{-1} = A^{-1} \cdot A = I
\]
where $ I $ is the identity matrix.

2. **Inverse of the Inverse Matrix:**
Given $ A $ is an invertible matrix, its inverse $ A^{-1} $ also has an inverse, denoted as $ (A^{-1})^{-1} $.

3. **Property of Matrix Inverses:**
The inverse of an inverse matrix should satisfy the original property of inverses. That is:
\[
A^{-1} \cdot (A^{-1})^{-1} = I
\]
and
\[
(A^{-1})^{-1} \cdot A^{-1} = I
\]

4. **Proof:**
By the definition of the inverse, multiplying $ A^{-1} $ by $ (A^{-1})^{-1} $ must yield the identity matrix:
\[
A^{-1} \cdot (A^{-1})^{-1} = I
\]
To retrieve the original matrix $ A $ from $ (A^{-1})^{-1} $, notice that:
\[
A \cdot A^{-1} = A^{-1} \cdot A = I
\]
implies that $ (A^{-1})^{-1} $ behaves in the same fashion as $ A $.

5. **Conclusion:**
Since $ (A^{-1})^{-1} $ satisfies the same property as $ A $, we can conclude:
\[
(A^{-1})^{-1} = A
\]
Thus, we have proved the given statement.

**Matrix Multiplication and Scalar Addition Proof**

**Problem Statement:**
Prove the following matrix equation:
If $ a, b \in \mathbb{R} $ and $ P $ is a $ m \times n $ matrix, then the following equality holds:

\[ aP + bP = (a + b)P \]

**Explanation:**
Given scalars $ a $ and $ b $ in the set of real numbers $ \mathbb{R} $, and a matrix $ P $ of dimensions $ m \times n $:
- The term $ aP $ refers to the scalar multiplication of matrix $ P $ by the scalar $ a $.
- Similarly, $ bP $ is the scalar multiplication of matrix $ P $ by the scalar $ b $.
- The expression $ aP + bP $ involves the addition of these two resultant matrices.

**Proof:**
To prove the given equation, consider the operations involved:

1. **Scalar Multiplication:**
When a matrix $ P $ is multiplied by a scalar $ c $ (where $ c $ is either $ a $ or $ b $):
\[ cP = c \begin{pmatrix}
p_{11} & p_{12} & \cdots & p_{1n} \\
p_{21} & p_{22} & \cdots & p_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
p_{m1} & p_{m2} & \cdots & p_{mn}
\end{pmatrix}
= \begin{pmatrix}
cp_{11} & cp_{12} & \cdots & cp_{1n} \\
cp_{21} & cp_{22} & \cdots & cp_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
cp_{m1} & cp_{m2} & \cdots & cp_{mn}
\end{pmatrix} \]

2. **Matrix Addition:**
Adding the resulting matrices $ aP $ and $ bP $:
\[ aP + bP = \begin{pmatrix}
ap_{11}

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