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2. Find the resale of the matrix:

a. [tex]\( A = \begin{pmatrix} 6 & 2 \\ 4 & 1 \end{pmatrix} \)[/tex]

b. [tex]\( M = \begin{pmatrix} 5 & -2 & 0 & 1 \\ -2 & -4 & 10 & 20 \end{pmatrix} \)[/tex]


Sagot :

It seems there might be some formatting issues in your question, but I understand the main task is to find the product of two matrices A and B. Let's break down the problem step-by-step.

### Matrices Given:
Matrix A:
[tex]\[ A = \begin{pmatrix} 6 & 2 \\ 4 & 1 \\ \end{pmatrix} \][/tex]

Matrix B:
[tex]\[ B = \begin{pmatrix} 5 & -2 & 0 & 1 \\ -2 & -4 & 10 & 20 \\ \end{pmatrix} \][/tex]

### Matrix Multiplication:
To find the product AB, we will multiply each element of the rows of A by each element of the columns of B and sum the results for each corresponding element in the resulting matrix.

#### Step-by-Step Calculation:

1. First row of A multiplied with columns of B:

- First column of B:
[tex]\( (6 5) + (2 -2) = 30 - 4 = 26 \)[/tex]

- Second column of B:
[tex]\( (6 -2) + (2 -4) = -12 - 8 = -20 \)[/tex]

- Third column of B:
[tex]\( (6 0) + (2 10) = 0 + 20 = 20 \)[/tex]

- Fourth column of B:
[tex]\( (6 1) + (2 20) = 6 + 40 = 46 \)[/tex]

Resulting row: [tex]\([26, -20, 20, 46]\)[/tex]

2. Second row of A multiplied with columns of B:

- First column of B:
[tex]\( (4 5) + (1 -2) = 20 - 2 = 18 \)[/tex]

- Second column of B:
[tex]\( (4 -2) + (1 -4) = -8 - 4 = -12 \)[/tex]

- Third column of B:
[tex]\( (4 0) + (1 10) = 0 + 10 = 10 \)[/tex]

- Fourth column of B:
[tex]\( (4 1) + (1 20) = 4 + 20 = 24 \)[/tex]

Resulting row: [tex]\([18, -12, 10, 24]\)[/tex]

### Final Result:
The product matrix AB is:
[tex]\[ AB = \begin{pmatrix} 26 & -20 & 20 & 46 \\ 18 & -12 & 10 & 24 \\ \end{pmatrix} \][/tex]

This is the desired result.
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