Fill a matrix using R *apply functions - r

Question

I have a matrix with that needs to be filled up with values. The first row of the matrix will have the same value and the subsequent row values would be generated using a function based on the first row value
I can do this using nested for loop like this. The outer loop goes over the column, sets the first row in that column to value. Then the inner loop fills up the rest of the rows in that column using the fn. The function itself takes the previous row value as its input.
fn <- function(value){ value + 1 }
myMatrix <- matrix(NA,5,3)
value <- 100
for(col in 1:ncol(myMatrix)){
myMatrix[1,col]<-value #First row value for all the columns should be the same
for(row in 2:nrow(myMatrix)){
#Rest of the row values generated using fn
myMatrix[row,col] <- fn(myMatrix[row-1,col])
}
}
myMatrix
I don't want to use a for loop and would like to specifically achieve this using one of R's vectorized *apply functions. I tried this but its not working.
fn <- function(value){ value + 1 }
myMatrix2 <- matrix(NA,5,3)
value <- 100
sapply(1:ncol(myMatrix2), function(col){
myMatrix2[1,col]<-value
sapply(2:nrow(myMatrix2),function(row){
fn(myMatrix2[row-1,col])
})
})
EDIT :
I was able to achieve it using sapply and the <<- assignment operator for filling up the matrix. But, is there a more cleaner/efficient way to do it using the *apply family ?
fn <- function(value){ value + 1 }
myMatrix2 <- matrix(NA,5,3)
value <- 100
myMatrix2[1,]<-value #first row of the matrix to have the same value
sapply(1:ncol(myMatrix2), function(col){
sapply(2:nrow(myMatrix2),function(row){
myMatrix2[row,col] <<- fn(myMatrix2[row-1,col])
})
})
myMatrix2