used (Mb) gc trigger (Mb) limit (Mb) max used (Mb)
Ncells 610430 32.7 1386281 74.1 NA 715608 38.3
Vcells 1124044 8.6 8388608 64.0 49152 2010826 15.4cat("\f") # Clear the consolegraphics.off() # Clear plots. Can use par(mfrow=c(1,1)) Either look at the math camp notes, or on some simplified online blogs.
Thus, you should always check for variable values and duplicate rows in your data to be able to check if you can estimate your “beta” coefficients in your linear regression as you have to invert the your data matrix to find it. Most statistical software will automatically drop duplicate rows but will give you an error if the matrix is not invertible.
Will you be able to invert these 4 matrices ?
\[ \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ \end{bmatrix} \]
Yes, you can invert matrix A as its determinant is not 0. It is a non-singular matrix.
solve(A) # inverse of A [,1] [,2]
[1,] -2.0 1.0
[2,] 1.5 -0.5 [,1] [,2]
[1,] 1 1.110223e-16
[2,] 0 1.000000e+00 [,1] [,2]
[1,] 1.000000e+00 4.440892e-16
[2,] -5.551115e-17 1.000000e+00Combine rows under each other.
Combine columns after each other.
Loading required package: RConics [,1] [,2] [,3]
[1,] 1 0 1
[2,] 2 2 1
[3,] 0 0 3adj_R <- adjoint(R) # Compute the classical adjoint (also called adjugate) of a square matrix. The adjoint is the transpose of the cofactor matrix.
adj_R [,1] [,2] [,3]
[1,] 6 0 -2
[2,] -6 3 1
[3,] 0 0 2det(R)[1] 6R_inverse <- adj_R/det(R)
R_inverse [,1] [,2] [,3]
[1,] 1 0.0 -0.3333333
[2,] -1 0.5 0.1666667
[3,] 0 0.0 0.3333333solve(R) [,1] [,2] [,3]
[1,] 1 0.0 -0.3333333
[2,] -1 0.5 0.1666667
[3,] 0 0.0 0.3333333Row 1 and row 3 are the same !
\[ \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 1 & 2 & 3 \\ \end{bmatrix} \]
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 4 5 6
[3,] 1 2 3typeof(B)[1] "double"?det
det(B)[1] 0??inverse
# Use the solve() function to calculate the inverse.
# solve(B)
# Using tryCatch to catch and print the error message
result <- tryCatch({
solve(B)
}, error = function(e) {
cat("Error: ", conditionMessage(e), "\n")
})Error: Lapack routine dgesv: system is exactly singular: U[3,3] = 0 \[ \begin{bmatrix} 1 & 2 & 2 \\ -2 & 0.5 & 0 \\ -2 & 5 & 4 \\ \end{bmatrix} \]
You may not be able to visually eyeball that the third row is a linear combination of the first two rows, with an equal weight (of 2) on row 1 and row 2. But calculating the determinant identifies this (i.e. det(C) = 0) and thus you know you will not be able to calculate the inverse of the matrix C as it should not exist. solve(C) command gives us an error, as expected.
[,1] [,2] [,3]
[1,] 1 2.0 2
[2,] -2 0.5 0
[3,] -2 5.0 4det(C)[1] 0#solve(C)
# Using tryCatch to catch and print the error message
result <- tryCatch({
solve(C)
}, error = function(e) {
cat("Error: ", conditionMessage(e), "\n")
})Error: Lapack routine dgesv: system is exactly singular: U[3,3] = 0 You may not be able to visually eyeball that the third column is a linear combination of the first two columns, with a weight of -2 on column 1 and a weight of 3 on the second column. Seeing the pattern is harder here than matrix C above. But again calculating the determinant identifies this (i.e. det(D) = 0) and thus you know you will not be able to calculate the inverse of the matrix D as it should not exist. solve(D) command gives us an error, as expected.
\[ \begin{bmatrix} 1 & -6 & -20 \\ 2 & -1 & -7 \\ 3 & -3 & 3 \\ \end{bmatrix} \]
[,1] [,2] [,3]
[1,] 1 -6 -20
[2,] 2 -1 -7
[3,] 3 3 3det(D)[1] 1.049161e-14#solve(D)
# Using tryCatch to catch and print the error message
result <- tryCatch({
solve(D)
}, error = function(e) {
cat("Error: ", conditionMessage(e), "\n")
})Error: system is computationally singular: reciprocal condition number = 2.77556e-18 Determinant of E is not defined if you have NA , non-numerical or missing values.
Do not replace NA values with 0. Think about how to impute missing values.
The appendix below might be of particular interest.