Answered

3) Sejam:

[tex]\[ A = \left[\begin{array}{cc} 3 & 2 \\ 0 & -4 \end{array}\right], \quad B = \left[\begin{array}{cc} 0 & 10 \\ -4 & 6 \end{array}\right], \quad C = \left[\begin{array}{ccc} 5 & 0 & 10 \\ 15 & -20 & 15 \\ 0 & -5 & 0 \end{array}\right] \][/tex]

Calcule:

a) [tex]\( 3A = \left[\begin{array}{cc} 9 & 6 \\ 0 & -12 \end{array}\right] \)[/tex]

b) [tex]\( \frac{1}{2}B = \left[\begin{array}{cc} 0 & 5 \\ -2 & 3 \end{array}\right] \)[/tex]

c) [tex]\( -2B = \left[\begin{array}{cc} 0 & -20 \\ 8 & -12 \end{array}\right] \)[/tex]

d) [tex]\( \frac{1}{5}C = \left[\begin{array}{ccc} 1 & 0 & 2 \\ 3 & -4 & 3 \\ 0 & -1 & 0 \end{array}\right] \)[/tex]

e) [tex]\( 2A + 3B = \left[\begin{array}{cc} 6 & 4 \\ 0 & -8 \end{array}\right] + \left[\begin{array}{cc} 0 & 30 \\ -12 & 18 \end{array}\right] = \left[\begin{array}{cc} 6 & 34 \\ -12 & 10 \end{array}\right] \)[/tex]

f) [tex]\( B - 3A = \left[\begin{array}{cc} 0 & 10 \\ -4 & 6 \end{array}\right] - \left[\begin{array}{cc} 9 & 6 \\ 0 & -12 \end{array}\right] = \left[\begin{array}{cc} -9 & 4 \\ -4 & 18 \end{array}\right] \)[/tex]



Answer :

GinnyAnswer
Let's go through each part of the problem step by step.

### a) [tex]\( 3A \)[/tex]

Given matrix [tex]\( A \)[/tex]:
[tex]\[ A = \begin{pmatrix} 3 & 2 \\ 0 & -4 \end{pmatrix} \][/tex]

To find [tex]\( 3A \)[/tex], we multiply every element of [tex]\( A \)[/tex] by 3:
[tex]\[ 3A = 3 \cdot \begin{pmatrix} 3 & 2 \\ 0 & -4 \end{pmatrix} = \begin{pmatrix} 9 & 6 \\ 0 & -12 \end{pmatrix} \][/tex]

Here, [tex]\( b \)[/tex] is the element in the first row and second column of the resulting matrix. Thus, [tex]\( b = 6 \)[/tex].

So,
[tex]\[ 3A = \begin{pmatrix} 9 & 6 \\ 0 & -12 \end{pmatrix} \][/tex]

### b) [tex]\(\frac{1}{2} B\)[/tex]

Given matrix [tex]\( B \)[/tex]:
[tex]\[ B = \begin{pmatrix} 0 & 10 \\ -4 & 6 \end{pmatrix} \][/tex]

To calculate [tex]\( \frac{1}{2} B \)[/tex], we multiply each element of [tex]\( B \)[/tex] by [tex]\( \frac{1}{2} \)[/tex]:
[tex]\[ \frac{1}{2} B = \frac{1}{2} \cdot \begin{pmatrix} 0 & 10 \\ -4 & 6 \end{pmatrix} = \begin{pmatrix} 0 & 5 \\ -2 & 3 \end{pmatrix} \][/tex]

### c) [tex]\(-2B\)[/tex]

To calculate [tex]\( -2B \)[/tex], we multiply each element of [tex]\( B \)[/tex] by -2:
[tex]\[ -2B = -2 \cdot \begin{pmatrix} 0 & 10 \\ -4 & 6 \end{pmatrix} = \begin{pmatrix} 0 & -20 \\ 8 & -12 \end{pmatrix} \][/tex]

### e) [tex]\( 2A + 3B \)[/tex]

First, we calculate [tex]\( 2A \)[/tex]:
[tex]\[ 2A = 2 \cdot \begin{pmatrix} 3 & 2 \\ 0 & -4 \end{pmatrix} = \begin{pmatrix} 6 & 4 \\ 0 & -8 \end{pmatrix} \][/tex]

Next, we calculate [tex]\( 3B \)[/tex]:
[tex]\[ 3B = 3 \cdot \begin{pmatrix} 0 & 10 \\ -4 & 6 \end{pmatrix} = \begin{pmatrix} 0 & 30 \\ -12 & 18 \end{pmatrix} \][/tex]

Now, add [tex]\( 2A \)[/tex] and [tex]\( 3B \)[/tex]:
[tex]\[ 2A + 3B = \begin{pmatrix} 6 & 4 \\ 0 & -8 \end{pmatrix} + \begin{pmatrix} 0 & 30 \\ -12 & 18 \end{pmatrix} = \begin{pmatrix} 6 & 34 \\ -12 & 10 \end{pmatrix} \][/tex]

### d) [tex]\(\frac{1}{5} C\)[/tex]

Given matrix [tex]\( C \)[/tex]:
[tex]\[ C = \begin{pmatrix} 5 & 0 & 10 \\ 15 & -20 & 15 \\ 0 & -5 & 0 \end{pmatrix} \][/tex]

To calculate [tex]\( \frac{1}{5} C \)[/tex], we multiply each element of [tex]\( C \)[/tex] by [tex]\( \frac{1}{5} \)[/tex]:
[tex]\[ \frac{1}{5} C = \frac{1}{5} \cdot \begin{pmatrix} 5 & 0 & 10 \\ 15 & -20 & 15 \\ 0 & -5 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 2 \\ 3 & -4 & 3 \\ 0 & -1 & 0 \end{pmatrix} \][/tex]

### Given Matrices Sum Calculation

The given sum of matrices is:
[tex]\[ \begin{pmatrix} 6 & 4 \\ 0 & -8 \end{pmatrix} + \begin{pmatrix} 0 & 30 \\ -12 & 18 \end{pmatrix} = \begin{pmatrix} 6 & 34 \\ -12 & 10 \end{pmatrix} \][/tex]

This matches the result from part e.

### f) [tex]\( B - 3A \)[/tex]

Finally, to compute [tex]\( B - 3A \)[/tex]:
[tex]\[ 3A = \begin{pmatrix} 9 & 6 \\ 0 & -12 \end{pmatrix} \][/tex]

Now subtract [tex]\( 3A \)[/tex] from [tex]\( B \)[/tex]:
[tex]\[ B - 3A = \begin{pmatrix} 0 & 10 \\ -4 & 6 \end{pmatrix} - \begin{pmatrix} 9 & 6 \\ 0 & -12 \end{pmatrix} = \begin{pmatrix} -9 & 4 \\ -4 & 18 \end{pmatrix} \][/tex]

### Summary

- [tex]\( 3A = \begin{pmatrix} 9 & 6 \\ 0 & -12 \end{pmatrix} \)[/tex]
- [tex]\( b = 6 \)[/tex]
- [tex]\( \frac{1}{2} B = \begin{pmatrix} 0 & 5 \\ -2 & 3 \end{pmatrix} \)[/tex]
- [tex]\( -2B = \begin{pmatrix} 0 & -20 \\ 8 & -12 \end{pmatrix} \)[/tex]
- [tex]\( 2A + 3B = \begin{pmatrix} 6 & 34 \\ -12 & 10 \end{pmatrix} \)[/tex]
- [tex]\( \frac{1}{5} C = \begin{pmatrix} 1 & 0 & 2 \\ 3 & -4 & 3 \\ 0 & -1 & 0 \end{pmatrix} \)[/tex]
- Sum of given matrices: [tex]\( \begin{pmatrix} 6 & 34 \\ -12 & 10 \end{pmatrix} \)[/tex]
- [tex]\( B - 3A = \begin{pmatrix} -9 & 4 \\ -4 & 18 \end{pmatrix} \)[/tex]