# Addition and subtraction of matrices

If

*A*= [*a*]_{ij}_{m}_{ Ã— n}and*B*= [*b*]_{ij}_{m}_{ Ã— n}are two matrices of the same order, their sum*A*+*B*is defined to be the matrix of order*m*Ã—*n*such that (*A*+*B*)*=*_{ij}*a*+_{ij}*b*for_{ij}*i*= 1, 2, ...,*m*and*j*= 1, 2, ...,*n*

*Notes:*- Only matrices of the same order can be added or subtracted.
- Addition of matrices is commutative as well as associative.
- Cancellation laws hold well in case of addition.
- That is ,
*A*+*B*=*A*+*C*â‡’*B*=*C*

# Scalar multiplication

The matrix obtained by multiplying every element of a matrix

*A*by a scalar*Î»*is called the scalar multiple of*A*by*Î»*and is denoted by*Î»**A*, i.e., if*A*= [*a*] then_{ij}*Î»A*= [*Î»a*]._{ij}# Multiplication of matrices

Two matrices

*A*and*B*are conformable for the product*AB*if the number of columns in*A*(pre-multiplier) is same as the number of rows in*B*(post-multiplier). Thus, if*A*= [*a*]_{ij}_{m}_{ Ã— n}and*B*= [*b*]_{ij}_{n }_{Ã— p}are two matrices of order*m*Ã—*n*and*n*Ã—*p*respectively, then their product*AB*is of order*m*Ã—*p*and is defined as(

*AB*)*=*_{ij}= [

*a*_{i}_{1}*a*_{i}_{2}...*a*]_{in}= (

*i*th row of*A*) (*j*th column of*B*) ...(1)*i*= 1, 2, ...,

*m*and

*j*= 1, 2, ...,

*p*

*Notes:*- Commutative law does not necessarily hold for matrices.
- If
*AB*=*BA*then matrices*A*and*B*are called commutative matrices. - If
*AB*= â€“*BA*then matrices*A*and*B*are called anti-commutative matrices. - Matrix multiplication is associative
*A*(*BC*) = (*AB*)*C.* - Matrix multiplication is distributive with respect to addition, i.e.,
*A*(*B*+*C*) =*AB*+*AC*. - The matrices possess divisors of zero, i.e.,if the product
*AB*=*O*, it is not necessary that at least one of the matrices should be zero matrix. For example, if*A*= and*B*= , then*AB*= while neither*A*nor*B*is the null matrix. - Cancellation law does not necessarily hold,i.e., if
*AB*=*AC*then in general*B*â‰*C*, even if*A*â‰*O*. - Matrix multiplication
*A*â‹…*A*is represented as*A*^{2}. Thus*A*=^{n}*A*â‹…*A*...*n*times. - If
*A*= diag.(*a*_{1},*a*_{2},*a*_{3}, ...,*a*) and_{n}*B*= diag.(*b*_{1},*b*_{2},*b*_{3}, ...,*b*), then_{n}*A*â‹…*B*= diag.(*a*_{1}*b*_{1},*a*_{2}*b*_{2}, ...,*a*). Thus_{n}b_{n}*A*= dign.(^{n}*a*_{1},^{n}*a*_{2},^{n}*a*_{3}, ...,^{n}*a*)_{n}^{n} - If
*A*and*B*are diagonal matrices of the same order then*AB*=*BA*or diagonal matrices are commutative. - If
*A*and*B*are commutative then*A*+*B*)^{2}= (*A*+*B*)(*A*+*B*)*A*^{2}+*AB*+*BA*+*B*^{2}g*A*^{2}+ 2*AB*+*B*^{2}*A*+*B*)^{3}=*A*^{3}+ 3*A*^{2}*B*+ 3*AB*^{2}+*B*^{3}.*A*+*B*)=^{n}^{n}C_{0}*A*+^{n}^{n}C_{1}*A*^{n}^{ â€“ 1}*B*+^{n}C_{2}*A*^{n}^{ â€“ 2}*B*^{2}+ ... +^{n}C_{n}B^{n}.*A*and*I*are always commutative. Hence, (*I*+*A*)=^{n}^{n}C_{0}+^{n}C_{1}*A*+^{n}C_{1}*A*+... +^{2}.^{n}C_{n}A^{n}