Answered

1. If [tex]$A=\left[\begin{array}{ll}2 & 5 \\ 4 & 5\end{array}\right]$[/tex] and [tex]$B=\left[\begin{array}{ll}2 & 4 \\ 0 & 8\end{array}\right]$[/tex], then find [tex][tex]$A+B$[/tex][/tex] and [tex]$A-B$[/tex].

2. Let [tex]$A=\left[\begin{array}{ccc}3 & 1 & 2 \\ 0 & 8 & 7 \\ 9 & 11 & -4\end{array}\right]$[/tex] and [tex][tex]$B=\left[\begin{array}{ccc}7 & 1 & 9 \\ 3 & 6 & -1 \\ 4 & -6 & 12\end{array}\right]$[/tex][/tex], then find [tex]$5A + 5B$[/tex].



Answer :

GinnyAnswer
Let's solve the problem step-by-step.

### Problem 1

Given matrices:

[tex]\[ A = \begin{bmatrix} 2 & 5 \\ 4 & 5 \end{bmatrix}, \quad B = \begin{bmatrix} 2 & 4 \\ 0 & 8 \end{bmatrix} \][/tex]

#### To find [tex]\( A + B \)[/tex]:

We perform element-wise addition of the matrices:

[tex]\[ A + B = \begin{bmatrix} 2 & 5 \\ 4 & 5 \end{bmatrix} + \begin{bmatrix} 2 & 4 \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 2+2 & 5+4 \\ 4+0 & 5+8 \end{bmatrix} = \begin{bmatrix} 4 & 9 \\ 4 & 13 \end{bmatrix} \][/tex]

So, we have:

[tex]\[ A + B = \begin{bmatrix} 4 & 9 \\ 4 & 13 \end{bmatrix} \][/tex]

#### To find [tex]\( A - B \)[/tex]:

We perform element-wise subtraction of the matrices:

[tex]\[ A - B = \begin{bmatrix} 2 & 5 \\ 4 & 5 \end{bmatrix} - \begin{bmatrix} 2 & 4 \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 2-2 & 5-4 \\ 4-0 & 5-8 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 4 & -3 \end{bmatrix} \][/tex]

So, we have:

[tex]\[ A - B = \begin{bmatrix} 0 & 1 \\ 4 & -3 \end{bmatrix} \][/tex]

### Problem 2

Given matrices:

[tex]\[ A = \begin{bmatrix} 3 & 1 & 2 \\ 0 & 8 & 7 \\ 9 & 11 & -4 \end{bmatrix}, \quad B = \begin{bmatrix} 7 & 1 & 9 \\ 3 & 6 & -1 \\ 4 & -6 & 12 \end{bmatrix} \][/tex]

#### To find [tex]\( 5A + 5B \)[/tex]:

First, we calculate [tex]\( 5A \)[/tex] and [tex]\( 5B \)[/tex] separately and then add the results:

[tex]\[ 5A = 5 \begin{bmatrix} 3 & 1 & 2 \\ 0 & 8 & 7 \\ 9 & 11 & -4 \end{bmatrix} = \begin{bmatrix} 5 \cdot 3 & 5 \cdot 1 & 5 \cdot 2 \\ 5 \cdot 0 & 5 \cdot 8 & 5 \cdot 7 \\ 5 \cdot 9 & 5 \cdot 11 & 5 \cdot -4 \end{bmatrix} = \begin{bmatrix} 15 & 5 & 10 \\ 0 & 40 & 35 \\ 45 & 55 & -20 \end{bmatrix} \][/tex]

[tex]\[ 5B = 5 \begin{bmatrix} 7 & 1 & 9 \\ 3 & 6 & -1 \\ 4 & -6 & 12 \end{bmatrix} = \begin{bmatrix} 5 \cdot 7 & 5 \cdot 1 & 5 \cdot 9 \\ 5 \cdot 3 & 5 \cdot 6 & 5 \cdot -1 \\ 5 \cdot 4 & 5 \cdot -6 & 5 \cdot 12 \end{bmatrix} = \begin{bmatrix} 35 & 5 & 45 \\ 15 & 30 & -5 \\ 20 & -30 & 60 \end{bmatrix} \][/tex]

Now we add [tex]\( 5A \)[/tex] and [tex]\( 5B \)[/tex]:

[tex]\[ 5A + 5B = \begin{bmatrix} 15 & 5 & 10 \\ 0 & 40 & 35 \\ 45 & 55 & -20 \end{bmatrix} + \begin{bmatrix} 35 & 5 & 45 \\ 15 & 30 & -5 \\ 20 & -30 & 60 \end{bmatrix} = \begin{bmatrix} 15+35 & 5+5 & 10+45 \\ 0+15 & 40+30 & 35-5 \\ 45+20 & 55-30 & -20+60 \end{bmatrix} = \begin{bmatrix} 50 & 10 & 55 \\ 15 & 70 & 30 \\ 65 & 25 & 40 \end{bmatrix} \][/tex]

So, we have:

[tex]\[ 5A + 5B = \begin{bmatrix} 50 & 10 & 55 \\ 15 & 70 & 30 \\ 65 & 25 & 40 \end{bmatrix} \][/tex]

This completes the solutions for both problems.

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