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Sagot :
Let's begin with the step-by-step calculation of the given tasks:
### 1. Calculation of [tex]\( A \cdot B \)[/tex]
Given matrices:
[tex]\[ A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}, \quad B = \begin{pmatrix} 1.2 & 1.4 & 3.1 \\ 2.2 & 1.1 & 5.6 \\ 3.7 & 4.2 & 6.7 \end{pmatrix} \][/tex]
Compute the product [tex]\( A \cdot B \)[/tex]:
[tex]\[ \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix} \cdot \begin{pmatrix} 1.2 & 1.4 & 3.1 \\ 2.2 & 1.1 & 5.6 \\ 3.7 & 4.2 & 6.7 \end{pmatrix} = \begin{pmatrix} c_{11} & c_{12} & c_{13} \\ ... & ... & ... \\ ... & ... & ... \end{pmatrix} \][/tex]
Using the true results:
[tex]\[ c_{11} = 16.7, \quad c_{12} = 16.2, \quad c_{13} = 34.4 \][/tex]
So we can fill in the appropriate checkmarks:
[tex]\[ c_{11} = 16.7 \checkmark, \quad c_{12} = 16.2 \checkmark, \quad c_{13} = 34.4 \checkmark \][/tex]
### 2. Final Calculation with vector and row vector
Given:
[tex]\[ 3 \begin{pmatrix} 3 \\ 19 \\ 12 \end{pmatrix} \begin{pmatrix} -1 & -3 & -5 \end{pmatrix} = \begin{pmatrix} d_{11} & d_{12} & d_{13} \\ d_{21} & d_{22} & d_{23} \\ -36 & -108 & -180 \end{pmatrix} \][/tex]
Computation for elements:
First, compute the product:
[tex]\[ \begin{pmatrix} 3 \\ 19 \\ 12 \end{pmatrix} \begin{pmatrix} -1 & -3 & -5 \end{pmatrix} = \begin{pmatrix} -3 & -9 & -15 \\ -19 & -57 & -95 \\ -12 & -36 & -60 \end{pmatrix} \][/tex]
Multiply by 3:
[tex]\[ 3 \cdot \begin{pmatrix} -3 & -9 & -15 \\ -19 & -57 & -95 \\ -12 & -36 & -60 \end{pmatrix} = \begin{pmatrix} d_{11} & d_{12} & d_{13} \\ d_{21} & d_{22} & d_{23} \\ -36 & -108 & -180 \end{pmatrix} \][/tex]
Thus,
[tex]\[ 3 \cdot \begin{pmatrix} -3 & -9 & -15 \\ -19 & -57 & -95 \\ -12 & -36 & -60 \end{pmatrix} = \begin{pmatrix} -9 & -27 & -45 \\ -57 & -171 & -285 \\ -36 & -108 & -180 \end{pmatrix} \][/tex]
So, fill in:
[tex]\[ d_{11} = -9 \checkmark, \quad d_{12} = -27 \checkmark, \quad d_{13} = -45 \checkmark \][/tex]
[tex]\[ d_{21} = -57 \checkmark, \quad d_{22} = -171 \checkmark, \quad d_{23} = -285 \checkmark \][/tex]
This completes the detailed step-by-step solution.
### 1. Calculation of [tex]\( A \cdot B \)[/tex]
Given matrices:
[tex]\[ A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}, \quad B = \begin{pmatrix} 1.2 & 1.4 & 3.1 \\ 2.2 & 1.1 & 5.6 \\ 3.7 & 4.2 & 6.7 \end{pmatrix} \][/tex]
Compute the product [tex]\( A \cdot B \)[/tex]:
[tex]\[ \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix} \cdot \begin{pmatrix} 1.2 & 1.4 & 3.1 \\ 2.2 & 1.1 & 5.6 \\ 3.7 & 4.2 & 6.7 \end{pmatrix} = \begin{pmatrix} c_{11} & c_{12} & c_{13} \\ ... & ... & ... \\ ... & ... & ... \end{pmatrix} \][/tex]
Using the true results:
[tex]\[ c_{11} = 16.7, \quad c_{12} = 16.2, \quad c_{13} = 34.4 \][/tex]
So we can fill in the appropriate checkmarks:
[tex]\[ c_{11} = 16.7 \checkmark, \quad c_{12} = 16.2 \checkmark, \quad c_{13} = 34.4 \checkmark \][/tex]
### 2. Final Calculation with vector and row vector
Given:
[tex]\[ 3 \begin{pmatrix} 3 \\ 19 \\ 12 \end{pmatrix} \begin{pmatrix} -1 & -3 & -5 \end{pmatrix} = \begin{pmatrix} d_{11} & d_{12} & d_{13} \\ d_{21} & d_{22} & d_{23} \\ -36 & -108 & -180 \end{pmatrix} \][/tex]
Computation for elements:
First, compute the product:
[tex]\[ \begin{pmatrix} 3 \\ 19 \\ 12 \end{pmatrix} \begin{pmatrix} -1 & -3 & -5 \end{pmatrix} = \begin{pmatrix} -3 & -9 & -15 \\ -19 & -57 & -95 \\ -12 & -36 & -60 \end{pmatrix} \][/tex]
Multiply by 3:
[tex]\[ 3 \cdot \begin{pmatrix} -3 & -9 & -15 \\ -19 & -57 & -95 \\ -12 & -36 & -60 \end{pmatrix} = \begin{pmatrix} d_{11} & d_{12} & d_{13} \\ d_{21} & d_{22} & d_{23} \\ -36 & -108 & -180 \end{pmatrix} \][/tex]
Thus,
[tex]\[ 3 \cdot \begin{pmatrix} -3 & -9 & -15 \\ -19 & -57 & -95 \\ -12 & -36 & -60 \end{pmatrix} = \begin{pmatrix} -9 & -27 & -45 \\ -57 & -171 & -285 \\ -36 & -108 & -180 \end{pmatrix} \][/tex]
So, fill in:
[tex]\[ d_{11} = -9 \checkmark, \quad d_{12} = -27 \checkmark, \quad d_{13} = -45 \checkmark \][/tex]
[tex]\[ d_{21} = -57 \checkmark, \quad d_{22} = -171 \checkmark, \quad d_{23} = -285 \checkmark \][/tex]
This completes the detailed step-by-step solution.
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