As discussed at the beginning of this page, each number system with an algebra has its own rules for addition and multiplication. Matrices have the operations called "addition" and "multiplication", but "multiplication by a number" is also a defined operation. The matrix system is set up in such a way that whenever one of these operations is applied, the result is still within the matrix system.
Addition is defined by adding corresponding elements of two or more matrices. In order to be "conformable to addition", the matrices must have the same order, that is, they must have the same number of rows and the same number of columns (Allen, 1961). The matrices $$A=\begin{bmatrix} 1 & 0 \\ 2 & 1\\ \end{bmatrix},$$ and $$ B=\begin{bmatrix} 4 & -1 & 3 \\ 1 & 2 & 5\\ \end{bmatrix},$$ would not be conformable to addition because they are not of the same order. Consider instead the matrices $$E= \begin{bmatrix} -3 & -1 & -3 \\ -6 & -2 & 9\\ \end{bmatrix}, F=\begin{bmatrix} 10 & 0 & -3 \\ 1 & 9 & -2\\ \end{bmatrix}.$$ Then $$E+F=\begin{bmatrix} -3+10 & -1+0 & -3+(-3) \\ -6+1 & -2+9 & 9+(-2)\\ \end{bmatrix} =\begin{bmatrix} 7 & -1 & -6 \\ -5 & 7 & 7\\ \end{bmatrix}.$$
Since it doesn't matter which order you add the corresponding elements of the matrices (i.e. \(-3+10=10+(-3)\)), the order of adding matrices also doesn't matter. This means that matrix addition is commutative, or that \(E+F=F+E\). Similarly, since the addition of elements of a matrix is associative (i.e. \((-3+10)+4=-3+(10+4)\)), the addition of matrices is also associative (\((E+F)+B=E+(F+B)\)).
In real-number addition, the additive identity is zero. This means that when \(0\) is added to any number, it doesn't change the number's value. In matrix addition, the additive identity is called the zero matrix. It is a matrix where each element is \(0\), and it can take on any order. Then any matrix added to the zero matrix is itself. For example, let \(G\) be the zero matrix. Then $$F+G=\begin{bmatrix} 10 & 0 & -3 \\ 1 & 9 & -2\\ \end{bmatrix} +\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0\\ \end{bmatrix} =\begin{bmatrix} 10+0 & 0+0 & -3+0 \\ 1+0 & 9+0 & -2+0\end{bmatrix} =\begin{bmatrix} 10 & 0 & -3 \\ 1 & 9 & -2\\ \end{bmatrix} =F.$$
Similar to additive inverses in the real number system, matrices have additive inverses. The additive inverse of \(A\) is represented as \(-A\) and is defined by \(A+(-A)=0\), where \(0\) is the zero matrix. The existence of additive inverses allows one to solve for variable matrices within an equation where there is a sum, such as \(B+X=E\). To find the unknown matrix \(X\), all that is necessary is adding the additive inverse of \(B\) to both sides of the equation. Then \(X=E+(-B)\). Let's try it out. $$\begin{bmatrix} 4 & -1 & 3 \\ 1 & 2 & 5\\ \end{bmatrix} + X=\begin{bmatrix} -3 & -1 & -3 \\ -6 & -2 & 9\\ \end{bmatrix} \Rightarrow X=\begin{bmatrix} -3 & -1 & -3 \\ -6 & -2 & 9\\ \end{bmatrix} - \begin{bmatrix} 4 & -1 & 3 \\ 1 & 2 & 5\\ \end{bmatrix}$$ $$=\begin{bmatrix} -3-4 & -1-(-1) & -3-3 \\ -6-1 & -2-2 & 9-5\\ \end{bmatrix}=\begin{bmatrix} -7 & 0 & -6 \\ -7 & -4 & 4\\ \end{bmatrix} \Rightarrow X=\begin{bmatrix} -7 & 0 & -6 \\ -7 & -4 & 4\\ \end{bmatrix} $$
This may seem simple enough but hang on tight to see what happens with multiplication.
Matrices have multiplication by a number as a defined operation. This is called scalar multiplication, where the number multiplying the matrix is called the scalar. To perform scalar multiplication, each element in the matrix is multiplied by the scalar. Consider the example $$4F=4\begin{bmatrix} 10 & 0 & -3 \\ 1 & 9 & -2\\ \end{bmatrix}=\begin{bmatrix} 4\times10 & 4\times0 & 4\times-3 \\ 4\times1 & 4\times9 & 4\times-2\\ \end{bmatrix}=\begin{bmatrix} 40 & 0 & -12 \\ 4 & 36 & -8\\ \end{bmatrix}.$$
There is, however, another type of multiplication: multiplying a matrix by another matrix.
In order for multiplication of matrices to be defined, the first matrix must have the same number of columns as the number of rows of the second matrix (Allen, 1961). For example, a matrix of order \(2\times2\) multiplied by a matrix of \(2\times3\) is defined while the other way around is not. To calculate a matrix product, each row of the first matrix is multiplied by the corresponding column of the second. To illustrate, consider the product of $$A=\begin{bmatrix} 1 & 0 \\ 2 & 1\\ \end{bmatrix}$$ and $$B=\begin{bmatrix} 4 & -1 & 3 \\ 1 & 2 & 5\\ \end{bmatrix}$$ below (and watch the video for a more in-depth explanation):$$AB=\begin{bmatrix} 1 & 0 \\ 2 & 1\\ \end{bmatrix}\begin{bmatrix} 4 & -1 & 3 \\ 1 & 2 & 5\\ \end{bmatrix}\\ =\begin{bmatrix} (1)(4)+(1)(1) & (1)(-1)+(0)(2) & (1)(3)+(0)(5)\\ (2)(4)+(1)(1) & (2)(-1)+(1)(2) & (2)(3)+(1)(5)\\ \end{bmatrix}\\ =\begin{bmatrix} 4 & -1 & 3\\ 9 & 0 & 11\\ \end{bmatrix}.$$
Consider now \(BA\). In this case, the number of columns in the first matrix (\(B\)), doesn't match the number of rows in the second, so \(BA\) is undefined. This shows that \(BA \neq AB\), so matrix multiplication is not commutative.
In multiplication with real numbers, there exists a number that can be multiplied to any number without changing that number's value. This number, \(1\), is called the multiplicative identity. In matrix multiplication, there exists a similar multiplicative identity called the identity matrix. The identity matrix is a square matrix of any order, but it's defining feature is that is has \(1\) for each element along its principle diagonal (from the top left corner to the bottom right) and \(0\) for every other element. More formally, the identity matrix, denoted \(I\), is of order \(n\times n\) such that \(a_{ij}=1\) whenever \(i=j\) and \(a_{ij}=0\) whenever \(i \neq j\). For example, let the identity matrix, \(I\), be of order \(2\) (remember that since it is a square matrix, the order only needs to be described with one number, which is both the number of rows and columns). Then $$I=\begin{bmatrix} 1 & 0 \\ 0 & 1\\ \end{bmatrix}$$ and $$IA=\begin{bmatrix} 1 & 0 \\ 0 & 1\\ \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 2 & 1\\ \end{bmatrix}=\begin{bmatrix} (1)(1)+(0)(2) & (1)(0)+(0)(1)\\ (0)(1)+(1)(2) & (0)(2)+(1)(1)\\ \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 2 & 1\\ \end{bmatrix}=A$$
Often in the real number system, multiplication and division are seen as distinct. However, any equation using division can be written in a different way using multiplication. For example \(6\div 2=3\) can equivalently be represented as \(3 \times 2=6\) or \(6 \times \frac{1}{2}= 3\). In matrix algebra, division is not performed, but rather multiplication of multiplicative inverses. Multiplicative inverses are defined by the following relationship: assuming that there exists an inverse of \(J\) (denoted \(J^{-1}\)), \(JJ^{-1}=I=J^{-1}J\). This means that it is possible to solve for an unknown matrix \(X\) in the expression \(JX=K\) if \(J\) has an inverse since \(J^{-1}JX=J^{-1}K\), as long as the number of columns of \(J^{-1}\) is equal to the number of rows in \(K\). Then \(IX=J^{-1}K\), and thus \(X=J^{-1}K\).
There is a real number associated with every square matrix with elements in the real numbers. This number is called the determinant, and when the determinant is not \(0\), the matrix has an inverse, but if the determinant is \(0\), the matrix does not have an inverse. That means that in the case \(AX=B\), if \(det(A)=0\), there is not a matrix \(X\) that satisfies the relationship, but if \(det(A) \neq 0\), there is an inverse of \(A\), and \(X\) has a solution. The determinant for a \(2 \times 2\) matrix is defined by \(ad-bc\) in a matrix set up as $$\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}.$$
The formulas for the determinants of higher order square matrices are more complicated. If you are interested in learning how to find the determinants of \( 3 \times 3\) matrices, follow this link.