A<-matrix(c(1,3,1,0,1,0,1,1,2),nrow=3,byrow=TRUE)
A
## [,1] [,2] [,3]
## [1,] 1 3 1
## [2,] 0 1 0
## [3,] 1 1 2
B<-matrix(c(2,-5,-1,0,1,0,-1,2,1),nrow=3,byrow=TRUE)
B
## [,1] [,2] [,3]
## [1,] 2 -5 -1
## [2,] 0 1 0
## [3,] -1 2 1
C<-A%*%B
C
## [,1] [,2] [,3]
## [1,] 1 0 0
## [2,] 0 1 0
## [3,] 0 0 1
As we can see this results in an identity matrix
We know that A*AInverse will lead to an identity matrix Let us test if B is inverse of A
print(solve(A))
## [,1] [,2] [,3]
## [1,] 2 -5 -1
## [2,] 0 1 0
## [3,] -1 2 1
This is exactly the matrix B
Let us look at the step by step multiplication process We will name variables according to row and column numbers (for example C11=row1column1 of matrixC)
C11<-A[1,1]*B[1,1]+A[1,2]*B[2,1]+A[1,3]*B[3,1]
C21<-A[2,1]*B[1,1]+A[2,2]*B[2,1]+A[2,3]*B[3,1]
C31<-A[3,1]*B[1,1]+A[3,2]*B[2,1]+A[3,3]*B[3,1]
C12<-A[1,1]*B[1,2]+A[1,2]*B[2,2]+A[1,3]*B[3,2]
C22<-A[2,1]*B[1,2]+A[2,2]*B[2,2]+A[2,3]*B[3,2]
C32<-A[3,1]*B[1,2]+A[3,2]*B[2,2]+A[3,3]*B[3,2]
C13<-A[1,1]*B[1,3]+A[1,2]*B[2,3]+A[1,3]*B[3,3]
C23<-A[2,1]*B[1,3]+A[2,2]*B[2,3]+A[2,3]*B[3,3]
C33<-A[3,1]*B[1,3]+A[3,2]*B[2,3]+A[3,3]*B[3,3]
M<-matrix(c(C11,C21,C31,C12,C22,C32,C13,C23,C33), nrow=3, byrow=TRUE)
M
## [,1] [,2] [,3]
## [1,] 1 0 0
## [2,] 0 1 0
## [3,] 0 0 1
This gives thesame result as using R.