Usually, when how turns out to be difficult, it's because of losing sight of what.
Every R problem can be thought of with advantage as the interaction of three objects— an existing object, x , a desired object,y , and a function, f, that will return a value of y given x as an argument. In other words, school algebra— f(x) = y. Any of the objects can be composites.
In the code given in the OP, x is a matrix dim 3 8, m and y is a logical vector of length 8. The objective is to populate y according to the equality tests for m[1,][1] == m[2,][2] and m[1,][1] == m[3,][1], which could be generalized to every other row if desired. For this use case, colSums is convenient, but not necessary. An alternative is presented below. Understand the application of the arguments to the subset operator, [ — m[,] subsets everything, m[1,] returns all columns for row 1, m[,1] returns all rows for column 1, m[2:3,3:4]andm[c(1,3),c(5,7)]` specified rows and columns.
See the FAQ: How to do a minimal reproducible example reprex for beginners to better frame questions.
# final element changed from 0 to 1 to conform to desired output
dat <- c(0,0,0,0,1,1,1,1,0,0,1,1,0,0,1,1,0,1,0,1,0,1,1,1)
m <- matrix(dat, nrow = 3, ncol = 8)
a_function <- function(x) colSums(x) == 0 | colSums(x) == 3
a_function(m)
#> [1] TRUE FALSE FALSE FALSE FALSE FALSE FALSE TRUE
another_function <- function(x,y) pi * m[y,] > 0
another_function(m,1:3)
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
#> [1,] FALSE FALSE TRUE FALSE FALSE TRUE FALSE TRUE
#> [2,] FALSE TRUE TRUE TRUE FALSE FALSE TRUE TRUE
#> [3,] FALSE TRUE FALSE TRUE TRUE TRUE FALSE TRUE
a_third_function <- function(x,y,z) x[y+1,] == x[y+z,]
a_third_function(m,1,2)
#> [1] TRUE TRUE FALSE TRUE FALSE FALSE FALSE TRUE