Compute A-413 and (413)A, where A = A-413 = 6 - 4 - 3 - 2 4 3 - 6 1 3

Algebra and Trigonometry (6th Edition)
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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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## Matrix Subtraction and Multiplication

### Problem Statement:
Compute \( A - 4I_3 \) and \( (4I_3)A \), where \( A \) is given by:

\[ 
A = 
\begin{bmatrix}
6 & -2 & 4 \\
-4 & 3 & -6 \\
-3 & 1 & 3 
\end{bmatrix} 
\]

### Equation and Matrix:

To find \( A - 4I_3 \), we need to subtract \( 4I_3 \) from matrix \( A \), where \( I_3 \) is the 3x3 identity matrix scaled by 4.

#### Identity Matrix \( I_3 \):
\[
I_3 =
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 
\end{bmatrix}
\]

#### Scaled Identity Matrix \( 4I_3 \):
\[
4I_3 =
\begin{bmatrix}
4 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & 4
\end{bmatrix}
\]

#### Subtracting \( 4I_3 \) from \( A \):
\[
A - 4I_3 =
\begin{bmatrix}
6 & -2 & 4 \\
-4 & 3 & -6 \\
-3 & 1 & 3 
\end{bmatrix}
-
\begin{bmatrix}
4 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & 4
\end{bmatrix}
\]

### Final Calculation:
Perform the subtraction element-wise:

\[
A - 4I_3 =
\begin{bmatrix}
6 - 4 & -2 - 0 & 4 - 0 \\
-4 - 0 & 3 - 4 & -6 - 0 \\
-3 - 0 & 1 - 0 & 3 - 4
\end{bmatrix}
=
\begin{bmatrix}
2 & -2 & 4 \\
-4 & -1 & -6 \\
-3 & 1 & -1
\end{bmatrix}
\]

So
Transcribed Image Text:## Matrix Subtraction and Multiplication ### Problem Statement: Compute \( A - 4I_3 \) and \( (4I_3)A \), where \( A \) is given by: \[ A = \begin{bmatrix} 6 & -2 & 4 \\ -4 & 3 & -6 \\ -3 & 1 & 3 \end{bmatrix} \] ### Equation and Matrix: To find \( A - 4I_3 \), we need to subtract \( 4I_3 \) from matrix \( A \), where \( I_3 \) is the 3x3 identity matrix scaled by 4. #### Identity Matrix \( I_3 \): \[ I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \] #### Scaled Identity Matrix \( 4I_3 \): \[ 4I_3 = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix} \] #### Subtracting \( 4I_3 \) from \( A \): \[ A - 4I_3 = \begin{bmatrix} 6 & -2 & 4 \\ -4 & 3 & -6 \\ -3 & 1 & 3 \end{bmatrix} - \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix} \] ### Final Calculation: Perform the subtraction element-wise: \[ A - 4I_3 = \begin{bmatrix} 6 - 4 & -2 - 0 & 4 - 0 \\ -4 - 0 & 3 - 4 & -6 - 0 \\ -3 - 0 & 1 - 0 & 3 - 4 \end{bmatrix} = \begin{bmatrix} 2 & -2 & 4 \\ -4 & -1 & -6 \\ -3 & 1 & -1 \end{bmatrix} \] So
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