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If [tex]$M=\left[\begin{array}{cc}2 & -1 \\ 4 & 6\end{array}\right]$[/tex] and [tex]$N=\left[\begin{array}{cc}-3 & 0 \\ 4 & 1\end{array}\right]$[/tex], find [tex]$M ^{\prime}+ N ^{\prime}$[/tex] and [tex]$( M + N )^{\prime}$[/tex].

Are [tex]$M ^{\prime}+ N ^{\prime}$[/tex] and [tex]$( M + N )^{\prime}$[/tex] equal?

Sagot :

GinnyAnsweridn
Sure, let's work through the problem step-by-step.

Given:
[tex]\[ M = \begin{pmatrix} 2 & -1 \\ 4 & 6 \end{pmatrix} \][/tex]
[tex]\[ N = \begin{pmatrix} -3 & 0 \\ 4 & 1 \end{pmatrix} \][/tex]

### Step 1: Calculate the transposes of [tex]\( M \)[/tex] and [tex]\( N \)[/tex]

1. Transpose of [tex]\( M \)[/tex]
[tex]\[ M' = \begin{pmatrix} 2 & 4 \\ -1 & 6 \end{pmatrix} \][/tex]

2. Transpose of [tex]\( N \)[/tex]
[tex]\[ N' = \begin{pmatrix} -3 & 4 \\ 0 & 1 \end{pmatrix} \][/tex]

### Step 2: Calculate the sum of the transposes [tex]\( M' + N' \)[/tex]
[tex]\[ M' + N' = \begin{pmatrix} 2 & 4 \\ -1 & 6 \end{pmatrix} + \begin{pmatrix} -3 & 4 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 + (-3) & 4 + 4 \\ -1 + 0 & 6 + 1 \end{pmatrix} = \begin{pmatrix} -1 & 8 \\ -1 & 7 \end{pmatrix} \][/tex]

### Step 3: Calculate [tex]\( M + N \)[/tex]
[tex]\[ M + N = \begin{pmatrix} 2 & -1 \\ 4 & 6 \end{pmatrix} + \begin{pmatrix} -3 & 0 \\ 4 & 1 \end{pmatrix} = \begin{pmatrix} 2 + (-3) & -1 + 0 \\ 4 + 4 & 6 + 1 \end{pmatrix} = \begin{pmatrix} -1 & -1 \\ 8 & 7 \end{pmatrix} \][/tex]

### Step 4: Calculate the transpose of [tex]\( M + N \)[/tex]
[tex]\[ (M + N)' = \left( \begin{pmatrix} -1 & -1 \\ 8 & 7 \end{pmatrix} \right)' = \begin{pmatrix} -1 & 8 \\ -1 & 7 \end{pmatrix} \][/tex]

### Step 5: Compare [tex]\( M' + N' \)[/tex] with [tex]\( (M + N)' \)[/tex]
We have:
[tex]\[ M' + N' = \begin{pmatrix} -1 & 8 \\ -1 & 7 \end{pmatrix} \][/tex]
[tex]\[ (M + N)' = \begin{pmatrix} -1 & 8 \\ -1 & 7 \end{pmatrix} \][/tex]

Since [tex]\( M' + N' = (M + N)' \)[/tex], they are equal.
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