  One of the basic operations that can be performed on matrices is the addition operation. Just as we add two or more integers, two or more matrices can also be added in a similar fashion. This is known as Addition of Matrices. Let’s learn about it in more detail.

### Suggested Videos        Matrix addition is the operation of adding two or matrices by adding the corresponding entry of each matrix together. [source:cornell]

The most important rule to know is that when adding two or more matrices, first make sure the matrices have the same dimensions. In order words, you can add a 2 x 3 with a 2 x 3 or a 2 x 2 with a 2 x 2. However, you cannot add a 3 x 2 with a 2 x 3 or a 2 x 2 with a 3 x 3. For example, the addition of two given matrices with dimension 2 × 2, [sourcce:mathisfun]

### 1. Commutative Law

If A = [aij], B = [bij] are matrices of the same order, say m × n, then A + B = B + A.

### 2. Associative Law

For any three matrices A = [aij], B = [bij], C = [cij] of the same order, say m × n, (A + B) + C = A + (B + C).

### 3. Existence of Additive Identity

Let A = [aij] be an m × n matrix and O be an m × n zero matrix, then A + O = O + A = A. In other words, O is the additive identity for matrix addition.

### 4. Existence of Additive Inverse

Let A = [aij]m×n be any matrix, then we have another matrix as – A = [–aij]m×n such that A + (–A) = (–A) + A= O. So – A is the additive inverse of A or negative of A.  ## Solved Examples on Addition of Matrices

Question: Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3, and p × k respectively. The restriction on n, k and p so that PY + WY will be defined are:

1. k = 3, p = n
2. k is arbitrary, p = 2
3. p is arbitrary, k = 3
4. k = 2, p = 3

Solution: In this, the order of × k, Order of × 3, Order of × k. Thus, the order of Pp×k, when 3. And the order of W× kwhere Thus option (A) is correct.

Question: If the sum of the matrices [x, x, y], [y, y, z] and [z, 0, 0] is the matrix [10, 5, 5], then what is the value of y?

1. -5
2.  0
3.  5
4.  10

Solution: [x, x, y] + [y, y, z] + [z, 0, 0] = [10, 5, 5]. Therefore x + y + z = 10, x +y = 5, y + z = 5 replacing x +y = 5 in x + y + z = 10. We have, z = 5, Also y + z = 5, Therefore y = 5 – z = 0. Thus option B is correct.

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