how that AC BC, even though A ± B. 0 1 1 0 4 3 В - 1 0 A = C = 0 1 4 3
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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![**Title: Matrix Multiplication Example**
**Objective:**
Show that \( AC = BC \), even though \( A \neq B \).
**Matrix Definitions:**
\[
A = \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}, \quad C = \begin{bmatrix} 4 & 3 \\ 4 & 3 \end{bmatrix}
\]
**Steps:**
1. **Matrix Multiplication \(AC\):**
\[
AC = \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 4 & 3 \\ 4 & 3 \end{bmatrix}
\]
The multiplication involves the following steps:
- Each element of the resulting matrix is obtained by multiplying and summing corresponding elements of rows from \(A\) with columns from \(C\).
2. **Matrix Multiplication \(BC\):**
\[
BC = \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 4 & 3 \\ 4 & 3 \end{bmatrix}
\]
Similarly, each element of the resulting matrix is obtained by multiplying and summing corresponding elements of rows from \(B\) with columns from \(C\).
**Explanation of Process:**
- The diagrams illustrate how each matrix multiplication involves row-by-column operations.
- Arrows indicate the direction of multiplication:
- Horizontal arrows correspond to the row operations.
- Vertical arrows correspond to the column operations.
By performing these calculations, you can show that although matrices \(A\) and \(B\) are different, their multiplications with the matrix \(C\) result in the same matrix. This demonstrates that \( AC = BC \) despite \( A \neq B \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9a7c37e1-d117-46e5-8cc2-cb36983cacd6%2Ffb0df24a-e02f-4cb3-be18-bd9c935c2793%2Fnqknfw_processed.png&w=3840&q=75)
Transcribed Image Text:**Title: Matrix Multiplication Example**
**Objective:**
Show that \( AC = BC \), even though \( A \neq B \).
**Matrix Definitions:**
\[
A = \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}, \quad C = \begin{bmatrix} 4 & 3 \\ 4 & 3 \end{bmatrix}
\]
**Steps:**
1. **Matrix Multiplication \(AC\):**
\[
AC = \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 4 & 3 \\ 4 & 3 \end{bmatrix}
\]
The multiplication involves the following steps:
- Each element of the resulting matrix is obtained by multiplying and summing corresponding elements of rows from \(A\) with columns from \(C\).
2. **Matrix Multiplication \(BC\):**
\[
BC = \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 4 & 3 \\ 4 & 3 \end{bmatrix}
\]
Similarly, each element of the resulting matrix is obtained by multiplying and summing corresponding elements of rows from \(B\) with columns from \(C\).
**Explanation of Process:**
- The diagrams illustrate how each matrix multiplication involves row-by-column operations.
- Arrows indicate the direction of multiplication:
- Horizontal arrows correspond to the row operations.
- Vertical arrows correspond to the column operations.
By performing these calculations, you can show that although matrices \(A\) and \(B\) are different, their multiplications with the matrix \(C\) result in the same matrix. This demonstrates that \( AC = BC \) despite \( A \neq B \).
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