Operations on Matrices> Addition and Subtraction of Matrices

Let \[A = [a_{ij}]\] and \[B = [b_{ij}]\] be two matrices of the same order \[m \times n\].

Their sum \[C = A + B\] is defined as the matrix \[[c_{ij}]\] of order \[m \times n\], where

Let \[A = [a_{ij}]\] and \[B = [b_{ij}]\] be matrices of the same order \[m \times n\].

Their difference \[D = A - B\] is defined as the matrix \[[d_{ij}]\] where

Equivalently,

Given \[\mathrm{A=} \begin{bmatrix} \sqrt{3} & 1 & -1 \\ 2 & 3 & 0 \end{bmatrix}\] and \[\mathbf{B}= \begin{bmatrix} 2 & \sqrt{5} & 1 \\ \\ -2 & 3 & \frac{1}{2} \end{bmatrix}\], find A + B

Solution:

Since A and B are of the same order, 2 × 3. Therefore, the addition of A and B is defined and is given by

\[\mathrm{A}+\mathrm{B}= \begin{bmatrix} 2+\sqrt{3} & 1+\sqrt{5} & 1-1 \\ 2-2 & 3+3 & 0+\frac{1}{2} \end{bmatrix}= \begin{bmatrix} 2+\sqrt{3} & 1+\sqrt{5} & 0 \\ 0 & 6 & \frac{1}{2} \end{bmatrix}\]

If \[\mathbf{A}= \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix}\] and \[\mathbf{B}= \begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix}\], then find 2A – B.

Solution:

We have

2A – B = \[2{ \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix}}-{ \begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix}}\]

\[= \begin{bmatrix} 2 & 4 & 6 \\ 4 & 6 & 2 \end{bmatrix}+ \begin{bmatrix} -3 & 1 & -3 \\ 1 & 0 & -2 \end{bmatrix}\]

\[= \begin{bmatrix} 2-3 & 4+1 & 6-3 \\ 4+1 & 6+0 & 2-2 \end{bmatrix}= \begin{bmatrix} -1 & 5 & 3 \\ 5 & 6 & 0 \end{bmatrix}\]

• Matrices must be of same order for addition and subtraction.

• \[A + B = [a_{ij} + b_{ij}]\].

• A - B = A + (-B).

• Addition is commutative: A + B = B + A.

• Addition is associative: (A + B) + C = A + (B + C).

• Zero matrix is additive identity: A + O = A.

• Negative of a matrix is additive inverse: \[A + (-A) = O\].

• If order differs \[\rightarrow\] operation not defined.