Matrix Addition

Definition If A and B are matrices of the same size, their sum A+B is the matrix formed by adding corresponding entries. If A=[aij] and B=[bij] this takes the form $$A+b=\left[ {{a}_{ij}}+{{b}_{ij}} \right]$$

Addition is not defined for matrices of different sizes!!!

If A,B and C are any matrices of the same size, then

A+B=B+A                (commutative law)

A+(B+C)=(A+B)+C (associative law)

Example 1

Compute –A, A-B and A+B-C if

$$A=\left[ \begin{matrix} 3 & -1 & 0 \\ 2 & 4 & 6 \\ 3 & 8 & 5 \\ \end{matrix} \right],B=\left[ \begin{matrix} 2 & 5 & 8 \\ 3 & 5 & 7 \\ 4 & 3 & 8 \\ \end{matrix} \right],\text{ and }C=\left[ \begin{matrix} 0 & 9 & 8 \\ 5 & 4 & 3 \\ 6 & 0 & 0 \\ \end{matrix} \right].$$

Solution:

$$\begin{align} & -A=\left[ \begin{matrix} -3 & 1 & 0 \\ -2 & -4 & -6 \\ -3 & -8 & -5 \\ \end{matrix} \right], \\ & A-B=\left[ \begin{matrix} 3-2 & -1-5 & 0-8 \\ 2-3 & 4-5 & 6-7 \\ 3-4 & 8-3 & 5-8 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -6 & -8 \\ -1 & -1 & -1 \\ -1 & 5 & -3 \\ \end{matrix} \right], \\ & A+B-C=\left[ \begin{matrix} 3+2-0 & -1+5-9 & 0+8-8 \\ 2+3-5 & 4+5-4 & 6+7-3 \\ 3+4-6 & 8+3-0 & 5+8-0 \\ \end{matrix} \right]=\left[ \begin{matrix} 5 & -5 & 0 \\ 0 & 5 & 10 \\ 1 & 11 & 13 \\ \end{matrix} \right] \\ \end{align}$$

Example 2

Solve 
$$\left[ \begin{matrix} 3 & 2 \\ -1 & 1 \\ \end{matrix} \right]+X=\left[ \begin{matrix} 0 & 1 \\ 1 & 1  \\ \end{matrix} \right],$$

where X is a matrix.

Solution 1: If we take a look at the previous equation we can conclude that the solution X has to be a 2x2 matrix. So let’s write this matrix:

$$X=\left[ \begin{matrix} x & y \\ z & w \\ \end{matrix} \right]$$ ,

then we get: 
$$\begin{align} & \left[ \begin{matrix} 0 & 1 \\ 1 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 3 & 2 \\ -1 & 1 \\ \end{matrix} \right]+\left[ \begin{matrix} x & y \\ z & w \\ \end{matrix} \right], \\ & \left[ \begin{matrix} 0 & 1 \\ 1 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 3+x & 2+y \\ -1+z & 1+w \\ \end{matrix} \right], \\ \end{align}$$

The rule of matrix equality gives: 

$$\begin{align} & 3+x=0, \\ & 2+y=1, \\ & -1+z=1, \\ & 1+w=1. \\ \end{align}$$

Solving the system of linear equations we get the following values:

$$x=-3,y=-1,z=2,w=0.$$

If we substitute these values in matrix X we get the solution of the equation in matrix form.

$$X=\left[ \begin{matrix} -3 & -1 \\ 2 & 0 \\ \end{matrix} \right].$$

Solution 2: We solve a numerical equation a + x =b by subtracting the number a from both sides to obtain x=b-a. This principle also works for matrices. So to solve the previous equation applying simple subtraction we get:

$$\begin{align} & X=\left[ \begin{matrix} 0 & 1 \\ 1 & 1 \\ \end{matrix} \right]-\left[ \begin{matrix} 3 & 2 \\ -1 & 1 \\ \end{matrix} \right], \\ & X=\left[ \begin{matrix} -3 & -1 \\ 2 & 0 \\ \end{matrix} \right] \\ \end{align}$$

The two solutions in the previous example are really different ways of doing the same thing. However, the first obtains four numerical equations, one for each entry, and solves them to get the four entries of X. The second solution solves the single matrix equation directly via matrix subtraction, and manipulation of entries comes in only at the end. The matrices themselves are manipulated. This ability to work with matrices as entities lies at the heart of matrix algebra.