# create a matrix
    mat <- matrix(1:50, nrow=10, ncol=5)
    mat
    
    mat2 <- matrix(1:50, nrow=10, ncol=5, byrow=T)
    mat2    

# matrix is a vector with attributes
    typeof(1:5)
    typeof(mat)
    mode(mat)
    class(mat)
    str(mat)
    
    attributes(1:5) # << no atts for a vector
    attributes(mat) # << dim att for a matrix
    
    attributes(mat)$dim <- NULL
    mat
    class(mat)

    vec <- 1:50
    attributes(vec)
    attributes(vec) <- list(dim=c(10,5))
    vec    
    
# matrix math
    rm(list=ls()) # << a funky-looking command that means "empty the global environment"
    x <- matrix(1:10, 5, 2)
    x
    
    y <- x
    x + y
    
    # vector recycling is still a thing
    x + 3    
    x + c(100, 200, 300)
    
    # element-wise versus matrix multiplication
    x*y
    t(x) %*% y
    crossprod(x, y)
    x %*% t(y)
    
    # matrix operations you would expect
    t(x)
    z <- crossprod(x, y); z
    diag(z)
    iden <- diag(10); iden
    
    mat <- matrix(c(1,2,3,0,1,4,5,6,0), 3, 3); mat
    solve(mat)
    solve(solve(mat))
    
    sqmat <- matrix(c(2,-1,0,-1,2,-1,0,-1,2), 3, 3); sqmat
    chol2inv(chol(sqmat))
    solve(chol2inv(chol(sqmat)))
    
# our first "apply" family function
    mat <- matrix(1:100, 10, 10); mat
    min(mat)
    sum(mat)
    rowSums(mat)
    colSums(mat)
    
    mat
    apply(mat, 1, max)
    apply(mat, 2, max)    
    
# subsetting and element-assignment
    mat <- matrix(1:100, 10, 10); mat
    mat[3,2]
    mat[1:3, 1:2]
    mat[4:5,4] <- c(1000, 2000); mat
    
    sub <- matrix(c(1,1,4,6, 4, 7), nrow=3, ncol=2, byrow=T); sub
    mat[sub]
    
    



## PRACTICE QUESTIONS

#1. Write your own code to extract the diagonal of an nxn matrix
    n <- 8
    mat <- matrix(1:(n^2), nrow=n, ncol=n); mat
   
    
     
#2. Write code to extract the first lower off-diagonal band of a matrix
#   i.e., for the matrix below, extract values (2, 8, 14, 20)
    n <- 5
    mat <- matrix(1:(n^2), nrow=n, ncol=n); mat
    
    
    
#3. Write code to generate a general matrix (i.e., of any dimension n × n) that follows this pattern
          [,1] [,2] [,3] [,4] [,5] [,6]
    [1,]    2    3    4    5    6    7
    [2,]    3    4    5    6    7    8
    [3,]    4    5    6    7    8    9
    [4,]    5    6    7    8    9   10
    [5,]    6    7    8    9   10   11
    [6,]    7    8    9   10   11   12
    
    
    
#4. use the "rnorm()" function to take 10,000 random draws from a normal distribution with 
#       mean 3, and standard deviation 5.
#   store these 10,000 draws in a 100x100 matrix and calculate the mean and stdev of each row.
#   then take the average of those 100 means and the average of those 100 stdev's





## CHALLENGE QUESTION

    # Demean each ROW of this matrix: subtract each row's mean from that row's values.
    set.seed(2026)
    mat <- matrix(rnorm(50, mean=10, sd=3), nrow=10, ncol=5)

    demeaned <- apply(mat, 1, function(r) r - mean(r))

    # No error. No warning. But every demeaned row should now average exactly zero:
    round(rowMeans(demeaned), 10)

    #   a) Those are not zeros. Compare dim(mat) with dim(demeaned) and explain what
    #      apply() handed back to you.
    #   b) round(colMeans(demeaned), 10) IS all zeros. Why does that confirm the
    #      diagnosis rather than contradict it?
    #   c) Fix it with a single extra function call, no loop.
    #   d) Run the column version, apply(mat, 2, function(cc) cc - mean(cc)), and
    #      check its shape. The same bug is present in the same place -- why does
    #      this one come out looking right? What would have to change about 'mat'
    #      for the row version to also come out looking right, and why is that the
    #      most dangerous case of all?
    #   e) Do the row version without apply(), using only rowMeans() and recycling.