Even though the row matrix and the column matrix above have the same entries, they have a different “shape”, or “size”, so we must think of them has being different matrices. Let's give the definitions to make this precise.
Two matrices $A$ and $B$ have the same size if they have the same number of rows, and they have the same number of columns.
If two matrices do not have the same size, we say they have different sizes.
Two matrices $A$ and $B$ are said to be equal if both of the following conditions hold:
When $A$ and $B$ are equal matrices, we write $A=B$. Otherwise, we write $A\ne B$.
We want to define operations on matrices: some (useful) ways of taking two matrices and making a new matrix.
Before we begin, a remark about $1\times 1$ matrices. These are of the form $[a_{11}]$ where $a_{11}$ is just a number. The square brackets $[\,]$ don't really matter here; they just keep the inside of a matrix in one place. So really: a $1\times 1$ matrix is just a number. This means that special cases of the operations we define will be operations on ordinary numbers. You should check that in the special case when all the matrices involved are $1\times 1$ matrices, the operations become the ordinary operations on numbers, so we are generalising the familiar operations (addition, subtraction, multiplication and so on) from numbers to matrices.
If $A$ and $B$ are matrices of the same size, then $A+B$ is defined to be the matrix with the same size as $A$ and $B$ so that the $(i,j)$ entry of $A+B$ is $a_{ij}+b_{ij}$, for every $i,j$.
If $A$ and $B$ are matrices of different sizes, then $A+B$ is undefined.
\[ \begin{bmatrix}1&2&-2\\3&0&5\end{bmatrix}+\begin{bmatrix}-2&2&0\\1&1&1\end{bmatrix}=\begin{bmatrix}-1&4&-2\\4&1&6\end{bmatrix}.\]
\[ \begin{bmatrix}1&2&-2\\3&0&5\end{bmatrix}+\begin{bmatrix}-2&2\\1&1\end{bmatrix}\text{ is undefined.}\]
The $n\times m$ zero matrix is the $n\times m$ matrix so that every entry is $0$. We write this as $0_{n\times m}$. So \[ 0_{n\times m}=\begin{bmatrix} 0&0&\dots&0\\ 0&0&\dots&0\\ \vdots&\vdots&&\vdots\\ 0&0&\dots&0\end{bmatrix}\] where this matrix has $n$ rows and $m$ columns.
Show that if $A$ is any $n\times m$ matrix, then \[ 0_{n\times m}+A=A=A+0_{n\times m}.\] Remember that when checking that matrices are equal, you have to check that they have the same size, and that all the entries are the same.
If $A$ and $B$ are matrices of the same size, then $A-B$ is defined to be the matrix with the same size as $A$ and $B$ so that the $(i,j)$ entry of $A-B$ is $a_{ij}-b_{ij}$, for every $i,j$.
If $A$ and $B$ are matrices of different sizes, then $A-B$ is undefined.
\[ \begin{bmatrix}1&2&-2\\3&0&5\end{bmatrix}-\begin{bmatrix}-2&2&0\\1&1&1\end{bmatrix}=\begin{bmatrix}3&0&-2\\2&-1&4\end{bmatrix}.\]
\[ \begin{bmatrix}1&2&-2\\3&0&5\end{bmatrix}-\begin{bmatrix}-2&2\\1&1\end{bmatrix}\text{ is undefined.}\]
In linear algebra, a scalar is just a fancy name for a number (in this course: a real number). The reason is that numbers are often used for scaling things up or down; for example, the scalar $3$ is often used to scale things up by a factor of $3$ (by multiplying by $3$).
If $c$ is a real number and $A$ is an $n\times m$ matrix, then we define the matrix $cA$ to be the $n\times m$ matrix given by multiplying every entry of $A$ by $c$. In other words, the $(i,j)$ entry of $cA$ is $ca_{i,j}$.
If $A=\begin{bmatrix}1&0&-3\\3&-4&1\end{bmatrix}$, then $3A=\begin{bmatrix}3&0&-9\\9&-12&3\end{bmatrix}$. In other words, \[ 3\begin{bmatrix}1&0&-3\\3&-4&1\end{bmatrix}=\begin{bmatrix}3&0&-9\\9&-12&3\end{bmatrix}.\]
We write $-A$ as a shorthand for $-1A$; so the $(i,j)$ entry of $-A$ is $-a_{ij}$. For example, \[ -\begin{bmatrix}-1&0&3\\3&-4&1\end{bmatrix}=\begin{bmatrix}1&0&-3\\-3&4&-1\end{bmatrix}.\]
Prove that $A-B=A+(-B)$ for any matrices $A$ and $B$ of the same size.
If $a=\begin{bmatrix}a_1&a_2&\dots&a_n\end{bmatrix}$ is a $1\times n$ row vector and $b=\begin{bmatrix}b_1\\b_2\\\vdots\\b_n\end{bmatrix}$ is an $n\times 1$ column vector, then the row-column product, or simply the product of $a$ and $b$ is defined to be \[ ab=\begin{bmatrix}a_1&a_2&\dots&a_n\end{bmatrix}\begin{bmatrix}b_1\\b_2\\\vdots\\b_n\end{bmatrix}=a_1b_1+a_2b_2+\dots+a_nb_n.\]
If we want to emphasize that we are multiplying in this way, we sometimes write $a\cdot b$ instead of $ab$.
The product $ab$ of a $1\times m$ row vector $a$ with an $n\times 1$ column vector $b$ is undefined if $m\ne n$.
This generalises row-column multiplication. The idea is that you build a new matrix from all possible row-column products. The formal definition will appear later, but here's an example: \[ \def\r{\begin{bmatrix}1&0&5\end{bmatrix}}\def\rr{\begin{bmatrix}2&-1&3\end{bmatrix}}\begin{bmatrix}1&0&5\\2&-1&3\end{bmatrix}\begin{bmatrix} 1&2\\3&4\\5&6\end{bmatrix} \def\s{\begin{bmatrix}1\\3\\5\end{bmatrix}}\def\ss{\begin{bmatrix}2\\4\\6\end{bmatrix}} = \begin{bmatrix}{\r\s}&{\r\ss}\\{\rr\s}&{\rr\ss}\end{bmatrix}=\begin{bmatrix}26&32\\14&18\end{bmatrix}.\]