To solve this problem, subtract the corresponding numbers in each matrix: \(\displaystyle ( 9-4=5, -2-5=-7, 4-(-6)=10, -5-(-3)=-2)\). Then, put those answers in the corresponding places in the answer matrix, so that you get: \(\displaystyle \begin{bmatrix} 5 &-7 \\ 10&-2 \end{bmatrix}\).

In order to add matrices, they have to be of the same dimension. In this case, they are both 2x2.

The formula to add matrices is as follows:

\(\displaystyle \begin{bmatrix} a& b\\ c& d \end{bmatrix}+\begin{bmatrix} e& f\\ g& h \end{bmatrix}=\begin{bmatrix} a+e& b+f\\ c+g& d+h \end{bmatrix}\)

Plugging in our values to solve we get:

\(\displaystyle \begin{bmatrix} 11& 9\\ 7& -5 \end{bmatrix}+\begin{bmatrix} 6& 9\\ 10& 13 \end{bmatrix}=\begin{bmatrix} 11+6& 9+9\\ 7+10& -5+13 \end{bmatrix}=\begin{bmatrix} 17& 18\\ 17& 8 \end{bmatrix}\)

The first step in solving this problem is to multiply the matrix by the scalar. The formula is as follows:

\(\displaystyle S\begin{bmatrix}a &b\\c &d\end{bmatrix}= \begin{bmatrix} Sa &Sb \\ Sc & Sd\end{bmatrix}\)

\(\displaystyle 3 \begin{bmatrix}2 &0 \\1 &4 \end{bmatrix} =\begin{bmatrix} 3(2)&3(0)\\3(1)& 3(4)\end{bmatrix}= \begin{bmatrix} 6 &0 \\3 &12\end{bmatrix}\)

In order to add matrices, they have to be of the same dimension.

In this case, they are both 2x2. So, we can then add the matrices together. The result is as follows:

\(\displaystyle \begin{bmatrix} 6 &0\\3 &12\end{bmatrix}+\begin{bmatrix} 2 & 7 \\5 &1 \end{bmatrix}=\begin{bmatrix} 6+2 &0+7\\3+5& 12+1\end{bmatrix}\)

\(\displaystyle =\begin{bmatrix} 8 &7\\8&13\end{bmatrix}\)

We can perform the addition since the matrices have the same sizes.

\(\displaystyle A+B=\begin{bmatrix} 1 & 2 & 3 &\cdots & n\\ 1 &2 &3 &\cdots &n \\ 1& 2 &3 &\cdots & n\\ \vdots&\vdots &\vdots &\vdots &\vdots \\ 1& 2 &3 &\cdots &n \end{bmatrix}+\)   \(\displaystyle \begin{bmatrix} n&n-1 &n-2 &\cdots &1 \\ n&n-1 &n-2 &\cdots &1\\\vdots & \vdots&\vdots &\vdots &\vdots \\ n&n-1 &n-2 &\cdots &1\\ n&n-1 &n-2 &\cdots &1 \end{bmatrix}\)

Looking at the first row of entries we get:

\(\displaystyle \begin{bmatrix} 1+n & 2+n-1=1+n & 3+n-2=1+n ...\end{bmatrix}\)

Note that any entry in the sum of \(\displaystyle A+B\) is equal to \(\displaystyle n+1\).

Thus the sum becomes:

\(\displaystyle \begin{bmatrix}n+1 &n+1 &n+1 &n+1 &n+1 \\ n+1 &n+1 &n+1 &n+1 &n+1 \\ n+1 &n+1 &n+1 &n+1 &n+1 \\ \vdots &\vdtos &\vdots &\vdots &\vdots \\ n+1 &n+1 &n+1 &n+1 &n+1 \end{bmatrix}\)

Note: For addition of matrices, we do it componentwise.

Note: Adding the first two columns of \(\displaystyle A\) and \(\displaystyle B\) we obtain \(\displaystyle 2\) for every row of the first column of the resulting matrix.

In all other case, we obtain 0 everywhere. This gives the matrix:

\(\displaystyle A+B=\begin{bmatrix} 2&0&\cdots &0 \\ 2&0&\cdots &0 \\ \vdots &\vdots &\vdots &\vdots \\ 2& 0&\cdots &0 \end{bmatrix}\)

Adding componentwise (adding entry by entry) we obtain zeros everywhere except for the last row where we get \(\displaystyle 2+2\) to obain \(\displaystyle 4\) in every component of the row.

This gives the matrix:

\(\displaystyle A+B=\begin{bmatrix} 0&0&\cdots &0 \\ 0&0&\cdots &0 \\ \vdots &\vdots &\vdots &\vdots \\ 4& 4&\cdots &4 \end{bmatrix}\)

Since A and B have the same size, we can perform addition.

The addition is performed componentwise.

The entry located at  (i,j) of matrix A  is added to the entry located at (i,j) of the matrix B.

In this case i=1 and j=1, 2 ,3 ,4,5

performing this operation we obtain:

\(\displaystyle \begin{bmatrix} 2 &4 & 6 &8 &10 \end{bmatrix}\)

Since A and B have the same size, we can perform addition.

The addition is performed componentwise.

The entry located at  (i,j) of matrix A  is added to the entry located at (i,j) of the matrix B.

Performing this operation we obtain:

\(\displaystyle \begin{bmatrix} 2 \\2 \\2 \\2 \\2 \end{bmatrix}\)

Since we are assuming that the two matrices have the same size, we can performthe matrices addition.

We know that when adding matrices, we add them componenwise. Let (i,j) be any entry of the addition matrix. We add the entry from A to the entry from B:

since the entries from A are the same and given by 1 and the entries from B are the same and given by \(\displaystyle (-1)^m\), we add these two to obtain :

\(\displaystyle 1+(-1)^m\) and we know that m is odd integer, hence \(\displaystyle (-1)^m=-1\).

Therefore the entry of the sum matrix is 0

Therefore our matrix is given by:

 \(\displaystyle \begin{bmatrix} 0 &0 &0 &0 \\ 0&0 &0 &0 \\ \vdots&\vdots&\vdots&\vdots\\ 0& 0 &0 &0 \end{bmatrix}\)

Matrix addition is very easy! All that you need to do is add each correlative member to each other. Think of it like this:

\(\displaystyle \begin{bmatrix} 14&22\\ -3 &4 \end{bmatrix}+\begin{bmatrix} -3&12\\ 5 &-10 \end{bmatrix}=\begin{bmatrix} 14-3&22+12\\ -3+5 &4-10 \end{bmatrix}\)

Now, just simplify:

\(\displaystyle \begin{bmatrix} 11&34\\ 2 &-6 \end{bmatrix}\)

There is your answer!