Sure, let's solve the given problem step by step.
### Problem
Given the matrices:
[tex]\[ A = \begin{pmatrix}
21 & 10 & 20 \\
5 & 13 & 19
\end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix}
14 & 10 & 3 \\
0 & 7 & 6
\end{pmatrix} \][/tex]
We are to find:
a) [tex]\( 5A - 3B \)[/tex]
b) [tex]\( 2A + 4B \)[/tex]
### Solution
#### Part (a): [tex]\( 5A - 3B \)[/tex]
1. Multiply each element in matrix [tex]\(A\)[/tex] by 5:
[tex]\[ 5A = 5 \begin{pmatrix}
21 & 10 & 20 \\
5 & 13 & 19
\end{pmatrix} = \begin{pmatrix}
5 \cdot 21 & 5 \cdot 10 & 5 \cdot 20 \\
5 \cdot 5 & 5 \cdot 13 & 5 \cdot 19
\end{pmatrix} = \begin{pmatrix}
105 & 50 & 100 \\
25 & 65 & 95
\end{pmatrix} \][/tex]
2. Multiply each element in matrix [tex]\(B\)[/tex] by 3:
[tex]\[ 3B = 3 \begin{pmatrix}
14 & 10 & 3 \\
0 & 7 & 6
\end{pmatrix} = \begin{pmatrix}
3 \cdot 14 & 3 \cdot 10 & 3 \cdot 3 \\
3 \cdot 0 & 3 \cdot 7 & 3 \cdot 6
\end{pmatrix} = \begin{pmatrix}
42 & 30 & 9 \\
0 & 21 & 18
\end{pmatrix} \][/tex]
3. Subtract the resultant matrix [tex]\( 3B \)[/tex] from [tex]\( 5A \)[/tex]:
[tex]\[ 5A - 3B = \begin{pmatrix}
105 & 50 & 100 \\
25 & 65 & 95
\end{pmatrix} - \begin{pmatrix}
42 & 30 & 9 \\
0 & 21 & 18
\end{pmatrix} = \begin{pmatrix}
105 - 42 & 50 - 30 & 100 - 9 \\
25 - 0 & 65 - 21 & 95 - 18
\end{pmatrix} = \begin{pmatrix}
63 & 20 & 91 \\
25 & 44 & 77
\end{pmatrix} \][/tex]
So, [tex]\( 5A - 3B = \begin{pmatrix}
63 & 20 & 91 \\
25 & 44 & 77
\end{pmatrix} \)[/tex].
#### Part (b): [tex]\( 2A + 4B \)[/tex]
1. Multiply each element in matrix [tex]\(A\)[/tex] by 2:
[tex]\[ 2A = 2 \begin{pmatrix}
21 & 10 & 20 \\
5 & 13 & 19
\end{pmatrix} = \begin{pmatrix}
2 \cdot 21 & 2 \cdot 10 & 2 \cdot 20 \\
2 \cdot 5 & 2 \cdot 13 & 2 \cdot 19
\end{pmatrix} = \begin{pmatrix}
42 & 20 & 40 \\
10 & 26 & 38
\end{pmatrix} \][/tex]
2. Multiply each element in matrix [tex]\(B\)[/tex] by 4:
[tex]\[ 4B = 4 \begin{pmatrix}
14 & 10 & 3 \\
0 & 7 & 6
\end{pmatrix} = \begin{pmatrix}
4 \cdot 14 & 4 \cdot 10 & 4 \cdot 3 \\
4 \cdot 0 & 4 \cdot 7 & 4 \cdot 6
\end{pmatrix} = \begin{pmatrix}
56 & 40 & 12 \\
0 & 28 & 24
\end{pmatrix} \][/tex]
3. Add the resultant matrix [tex]\( 4B \)[/tex] to [tex]\( 2A \)[/tex]:
[tex]\[ 2A + 4B = \begin{pmatrix}
42 & 20 & 40 \\
10 & 26 & 38
\end{pmatrix} + \begin{pmatrix}
56 & 40 & 12 \\
0 & 28 & 24
\end{pmatrix} = \begin{pmatrix}
42 + 56 & 20 + 40 & 40 + 12 \\
10 + 0 & 26 + 28 & 38 + 24
\end{pmatrix} = \begin{pmatrix}
98 & 60 & 52 \\
10 & 54 & 62
\end{pmatrix} \][/tex]
So, [tex]\( 2A + 4B = \begin{pmatrix}
98 & 60 & 52 \\
10 & 54 & 62
\end{pmatrix} \)[/tex].
### Summary
- [tex]\( 5A - 3B \)[/tex] results in: [tex]\(\begin{pmatrix} 63 & 20 & 91 \\ 25 & 44 & 77 \end{pmatrix}\)[/tex]
- [tex]\( 2A + 4B \)[/tex] results in: [tex]\(\begin{pmatrix} 98 & 60 & 52 \\ 10 & 54 & 62 \end{pmatrix}\)[/tex]
