Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Question
![**Matrix Multiplication: Find BA**
Consider two matrices A and B defined as follows:
\[
A = \begin{pmatrix}
7 & 0 \\
-1 & -3
\end{pmatrix}
\]
\[
B = \begin{pmatrix}
0 & -3 \\
-5 & -4
\end{pmatrix}
\]
To find the product \( BA \), use matrix multiplication rules where the element in the \(i\)-th row and \(j\)-th column of the resulting matrix is the dot product of the \(i\)-th row of matrix B and the \(j\)-th column of matrix A.
**Matrix A:**
\[
\begin{pmatrix}
7 & 0 \\
-1 & -3
\end{pmatrix}
\]
**Matrix B:**
\[
\begin{pmatrix}
0 & -3 \\
-5 & -4
\end{pmatrix}
\]
### Step-by-Step Calculation
1. **First row, first column**: \( (0 \cdot 7) + (-3 \cdot -1) = 0 + 3 = 3 \)
2. **First row, second column**: \( (0 \cdot 0) + (-3 \cdot -3) = 0 + 9 = 9 \)
3. **Second row, first column**: \( (-5 \cdot 7) + (-4 \cdot -1) = -35 + 4 = -31 \)
4. **Second row, second column**: \( (-5 \cdot 0) + (-4 \cdot -3) = 0 + 12 = 12 \)
**Resulting Matrix BA:**
\[
\begin{pmatrix}
3 & 9 \\
-31 & 12
\end{pmatrix}
\]
So,
\[
BA = \begin{pmatrix}
3 & 9 \\
-31 & 12
\end{pmatrix}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8b6c3136-3b78-4c3b-95a7-be26ea6fb45d%2F9b127814-41d2-42f4-931b-59d5142a3998%2Fmj6pu9k_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Matrix Multiplication: Find BA**
Consider two matrices A and B defined as follows:
\[
A = \begin{pmatrix}
7 & 0 \\
-1 & -3
\end{pmatrix}
\]
\[
B = \begin{pmatrix}
0 & -3 \\
-5 & -4
\end{pmatrix}
\]
To find the product \( BA \), use matrix multiplication rules where the element in the \(i\)-th row and \(j\)-th column of the resulting matrix is the dot product of the \(i\)-th row of matrix B and the \(j\)-th column of matrix A.
**Matrix A:**
\[
\begin{pmatrix}
7 & 0 \\
-1 & -3
\end{pmatrix}
\]
**Matrix B:**
\[
\begin{pmatrix}
0 & -3 \\
-5 & -4
\end{pmatrix}
\]
### Step-by-Step Calculation
1. **First row, first column**: \( (0 \cdot 7) + (-3 \cdot -1) = 0 + 3 = 3 \)
2. **First row, second column**: \( (0 \cdot 0) + (-3 \cdot -3) = 0 + 9 = 9 \)
3. **Second row, first column**: \( (-5 \cdot 7) + (-4 \cdot -1) = -35 + 4 = -31 \)
4. **Second row, second column**: \( (-5 \cdot 0) + (-4 \cdot -3) = 0 + 12 = 12 \)
**Resulting Matrix BA:**
\[
\begin{pmatrix}
3 & 9 \\
-31 & 12
\end{pmatrix}
\]
So,
\[
BA = \begin{pmatrix}
3 & 9 \\
-31 & 12
\end{pmatrix}
\]
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