6.1 8A.
Gauss.Elimination = function(A){
# Input A: augmented matrix. Make sure that he diagonal
# elements of the coefficient matrix are non-zero!
A0 = A
n = dim(A0)[1] # number of rows
for(i in 1:(n-1)){ # iterator fo pivot element A0[i,i]
# i = 1, 2, ..., n-1
#~~~~~~~~~~~~~~~~~~~~~ Make the function robust ~~~~~~~~~~~~~~~~~~~~~~~~~#
if(A0[i,i] == 0){
non.0 = which(A0[,i] != 0)
#cat("\n\n i =",i, ", non.0 =", non.0,". (i+1):n =", (i+1):n, ".")
id = intersect(non.0, (i+1):n)[1] # which() equiv to find() in MATLAB
if(!is.na(id)){
tempi = A0[i,] # put the i-th row in a place-holder
A0[i,] = A0[id,] # row swapping
A0[id,] = tempi
}
}
#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#
if(A0[i,i] != 0){ # A0[i,i] is in the denominator of recursive equation
for(j in (i+1):n){ # to make the augmented matrix in the row echelon form
#cat("\n\n j =",j,".")
A0[j,] = A0[j,] - (A0[j,i]/A0[i,i])*A0[i,]
}
}
}
A0
}##
A1 = matrix(c(1/2,1/4,-1/8,0,1/3,-1/6,1/9,1,1/7,1/7,1/10,2), ncol =4, byrow = TRUE)
Gauss.Elimination(A = A1)## [,1] [,2] [,3] [,4]
## [1,] 0.5 0.2500000 -0.1250000 0.000000
## [2,] 0.0 -0.3333333 0.1944444 1.000000
## [3,] 0.0 0.0000000 0.1773810 2.214286Now solve for x1, x2, x3 manually and x3 = 12.48 x2 = 4.28 x1 = .9799
8B.
##
A1 = matrix(c(2.71,1,1032,12,4.12,-1,500,11.49,3.33,2,-200,41), ncol =4, byrow = TRUE)
Gauss.Elimination(A = A1)## [,1] [,2] [,3] [,4]
## [1,] 2.71 1.000000 1032.000 12.000000
## [2,] 0.00 -2.520295 -1068.945 -6.753542
## [3,] 0.00 0.000000 -1795.204 24.188009Now solve manually for x1, x2, x3 and x3 = -.013 x2 = 8.395 x1 = 6.462
8C.
##
A1 = matrix(c(pi,(2)^(1/2),-1,1,0,exp(1),-1,1,2,1,1,1,-(3)^(1/2),1,2,-1,-1,1,-(5)^(1/2),3), ncol =5, byrow = TRUE)
Gauss.Elimination(A = A1)## [,1] [,2] [,3] [,4] [,5]
## [1,] 3.141593 1.414214 -1.0000000 1.0000000 0.000000
## [2,] 0.000000 -2.223657 1.8652560 1.1347440 1.000000
## [3,] 0.000000 0.000000 -0.9525206 0.9622774 2.247269
## [4,] 0.000000 0.000000 0.0000000 -1.9756173 3.272882x4 = -1.657 x3 = -4.033 x2 = -4.678 x1 = 1.349
8D.
##
A1 = matrix(c(1,1,-1,1,-1,2,2,2,1,-1,1,4,3,1,-3,-2,3,8,4,1,-1,4,-5,16,16,-1,1,-1,-1,32), ncol =6, byrow = TRUE)
Gauss.Elimination(A = A1)## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 1 1 -1 1.0 -1.0000000 2.0000
## [2,] 0 -2 0 -5.0 6.0000000 2.0000
## [3,] 0 0 3 -3.0 3.0000000 0.0000
## [4,] 0 0 0 10.5 -13.0000000 5.0000
## [5,] 0 0 0 0.0 -0.3809524 -37.2381x5 = 97.75 x4 = 121.5 x3 = 23.75 x2 = -11.5 x1 = 13.5
####
Gauss.Jordan = function(A){
# A is the augmented matrix
A1 = Gauss.Elimination(A) # perform Gauss elimination
n = length(A1[,1])
id.0 = rep(1,n) # indicator of zero A[i,i]
for (i in 1:n){
if(A1[i,i] == 0) id.0[i] = 0
}
if(sum(id.0) == n){ # A[i,i] != 0
k = n
A2 = A1[k:1, c(k:1,(n+1))] # re-arrange rows and columns
A3 = Gauss.Elimination(A2) # perform Gauss elimination
A4 = A3[k:1, c(k:1,(n+1))] # backward re-arrangement
} else{ # if exists A[i,i] =0
k = which(id.0==0)[1]-1 # select rows and columns for re-arrangement
A2 = A1[c(k:1,(k+1):n), c(k:1,(k+1):(n+1))]
A3 = Gauss.Elimination(A2)
A4 = A3[c(k:1,(k+1):n), c(k:1,(k+1):(n+1))]
}
A4
}A1 = matrix(c(pi, (2)^(1/2), -1, 1, 0,exp(1),-1,1,2,1,1,1,-(3)^(1/2),1,2,-1,-1,1,-(5)^(1/2),3), ncol = 5, byrow = TRUE)
Gauss.Jordan(A = A1)## [,1] [,2] [,3] [,4] [,5]
## [1,] 3.141593 0.000000 0.0000000 0.000000 4.239418
## [2,] 0.000000 -2.223657 0.0000000 0.000000 10.402239
## [3,] 0.000000 0.000000 -0.9525206 0.000000 3.841414
## [4,] 0.000000 0.000000 0.0000000 -1.975617 3.272882x4 = -1.66 x3 = -4.03 x2 = -4.68 x1 = 1.35
6.3 1C.
# matrix multiplication in R - example
a = matrix(c(2,3,0,0,-1,2,0,2,-3), ncol=3, nrow=3)# matrix multiplication in R - example
b = matrix(c(2,5,1), ncol=1, nrow=3)a%*%b## [,1]
## [1,] 4
## [2,] 3
## [3,] 73C.
# matrix multiplication in R - example
a = matrix(c(2,4,5,-3,3,2,1,0,-4), ncol=3, nrow=3)# matrix multiplication in R - example
b = matrix(c(0,1,2,1,0,3,-2,-1,-2), ncol=3, nrow=3)a%*%b## [,1] [,2] [,3]
## [1,] -1 5 -3
## [2,] 3 4 -11
## [3,] -6 -7 -45D.
M.inv = function(A){
# Input A: Must be a square matrix
n = dim(A)[1] # number of rows
A0 = cbind(A, diag(rep(1,n)))
##
for(i in 1:(n-1)){ # iterator fo pivot element A0[i,i], i < n.
#~~~~~~~~~~~~~~~~~~~~~ Make the function robust ~~~~~~~~~~~~~~~~~~~~~~~~~#
if(A0[i,i] == 0){
non.0 = which(A0[,i] != 0)
#cat("\n\n i =",i, ", non.0 =", non.0,". (i+1):n =", (i+1):n, ".")
id = intersect(non.0, (i+1):n)[1] # which() equiv to find() in MATLAB
if(!is.na(id)){
tempi = A0[i,] # put the i-th row in a place-holder
A0[i,] = A0[id,] # row swapping
A0[id,] = tempi
}
}
#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#
if(A0[i,i] != 0){ # A0[i,i] is in the denominator of recursive equation
for(j in (i+1):n){ # to make the augmented matrix in the row echelon form
#cat("\n\n j =",j,".")
A0[j,] = A0[j,] - (A0[j,i]/A0[i,i])*A0[i,]
}
}
}
A1=A0
n = length(A1[,1])
id.0 = rep(1,n) # indicator of zero A[i,i]
for (i in 1:n){
if(A1[i,i] == 0) id.0[i] = 0
}
if(sum(id.0) == n){ # A[i,i] != 0
for (i in 1:n){
A1[i,] = A1[i,] /A1[i,i] # Divide each row by A1[i,i]
} # so that new A1[i,i] = 1.
k = n
A2 = A1[k:1, c(k:1,(n+1):(2*n))] # re-arrange rows and columns
#_______________________________________________________________##
for(i in 1:(n-1)){
if(A2[i,i] != 0){ # A0[i,i] is in the denominator of recursive equation
for(j in (i+1):n){ # to make the augmented matrix in the row echelon form
A2[j,] = A2[j,] - (A2[j,i]/A2[i,i])*A2[i,]
}
}
}
#--------------------------------------------------------------##
A4 = A2[k:1, c(k:1,(n+1):(2*n))] # backward re-arrangement
round(A4[, (n+1):(2*n)],7)
} else if(dim(A)[1] != dim(A)[2]){
cat("\n\nA square matrix is needed!")
}else {# if exists A[i,i] =0
cat("The matrix is singular")
}
}A=matrix(c(4,0,0,0,6,7,0,0,9,11,1,0,5,4,1,1), ncol = 4, byrow = TRUE)
M.inv(A)## [,1] [,2] [,3] [,4]
## [1,] 0.2500000 0.0000000 0 0
## [2,] -0.2142857 0.1428571 0 0
## [3,] 0.1071429 -1.5714286 1 0
## [4,] -0.5000000 1.0000000 -1 17A.
Gauss.Elimination = function(A){
# Input A: augmented matrix. Make sure that he diagonal
# elements of the coefficient matrix are non-zero!
A0 = A
n = dim(A0)[1] # number of rows
for(i in 1:(n-1)){ # iterator fo pivot element A0[i,i]
# i = 1, 2, ..., n-1
#~~~~~~~~~~~~~~~~~~~~~ Make the function robust ~~~~~~~~~~~~~~~~~~~~~~~~~#
if(A0[i,i] == 0){
non.0 = which(A0[,i] != 0)
#cat("\n\n i =",i, ", non.0 =", non.0,". (i+1):n =", (i+1):n, ".")
id = intersect(non.0, (i+1):n)[1] # which() equiv to find() in MATLAB
if(!is.na(id)){
tempi = A0[i,] # put the i-th row in a place-holder
A0[i,] = A0[id,] # row swapping
A0[id,] = tempi
}
}
#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#
if(A0[i,i] != 0){ # A0[i,i] is in the denominator of recursive equation
for(j in (i+1):n){ # to make the augmented matrix in the row echelon form
#cat("\n\n j =",j,".")
A0[j,] = A0[j,] - (A0[j,i]/A0[i,i])*A0[i,]
}
}
}
A0
}A1 = matrix(c(1,-1,2,-1,6,1,0,-1,1,4,2,1,3,-4,-2,0,-1,1,-1,5), ncol =5, byrow = TRUE)
Gauss.Elimination(A = A1)## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 -1 2 -1 6
## [2,] 0 1 -3 2 -2
## [3,] 0 0 8 -8 -8
## [4,] 0 0 0 -1 1x4 = -1 x3 = -2 x2 = -6 x1 = 3
A1 = matrix(c(1,-1,2,-1,1,1,0,-1,1,1,2,1,3,-4,2,0,-1,1,-1,-1), ncol =5, byrow = TRUE)
Gauss.Elimination(A = A1)## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 -1 2 -1 1
## [2,] 0 1 -3 2 0
## [3,] 0 0 8 -8 0
## [4,] 0 0 0 -1 -1x4 = 1 x3 = 1 x2 = 1 x1 = 1
7B.
A=matrix(c(1,-1,2,-1,1,0,-1,1,2,1,3,-4,0,-1,1,-1), ncol = 4, byrow = TRUE)
M.inv(A)## [,1] [,2] [,3] [,4]
## [1,] 0.125 0.625 0.125 0
## [2,] 0.125 -0.375 0.125 -1
## [3,] 0.875 -0.625 -0.125 -1
## [4,] 0.750 -0.250 -0.250 -1# matrix multiplication in R - example
a = matrix(c(1,1,2,-1), ncol=4, nrow=4)# matrix multiplication in R - example
b = matrix(c((1/8),(1/8),(7/8),(3/4),(5/8),(-3/8),(-5/8),(-1/4),(1/8),(1/8),(-1/8),(-1/4),0,-1,-1,-1), ncol=4, nrow=4)b%*%a## [,1] [,2] [,3] [,4]
## [1,] 1 1 1 1
## [2,] 1 1 1 1
## [3,] 1 1 1 1
## [4,] 1 1 1 1We get x1 = 1 x2 = 1 x3 = 1 x4 = 1
# matrix multiplication in R - example
a = matrix(c(6,4,-2,5), ncol=4, nrow=4)# matrix multiplication in R - example
b = matrix(c((1/8),(1/8),(7/8),(3/4),(5/8),(-3/8),(-5/8),(-1/4),(1/8),(1/8),(-1/8),(-1/4),0,-1,-1,-1), ncol=4, nrow=4)b%*%a## [,1] [,2] [,3] [,4]
## [1,] 3 3 3 3
## [2,] -6 -6 -6 -6
## [3,] -2 -2 -2 -2
## [4,] -1 -1 -1 -1Here we get x1 = 3 x2 = -6 x3 = -2 x4 = -1
7C. The second method requires more steps.
DET = function(A){
# Input A: Must be a square matrix
A0 = A
n = dim(A0)[1] # number of rows
for(i in 1:(n-1)){ # iterator fo pivot element A0[i,i]
# i = 1, 2, ..., n-1
#~~~~~~~~~~~~~~~~~~~~~ Make the function robust ~~~~~~~~~~~~~~~~~~~~~~~~~#
if(A0[i,i] == 0){
non.0 = which(A0[,i] != 0)
#cat("\n\n i =",i, ", non.0 =", non.0,". (i+1):n =", (i+1):n, ".")
id = intersect(non.0, (i+1):n)[1] # which() equiv to find() in MATLAB
if(!is.na(id)){
tempi = A0[i,] # put the i-th row in a place-holder
A0[i,] = A0[id,] # row swapping
A0[id,] = tempi
}
}
#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#
if(A0[i,i] != 0){ # A0[i,i] is in the denominator of recursive equation
for(j in (i+1):n){ # to make the augmented matrix in the row echelon form
#cat("\n\n j =",j,".")
A0[j,] = A0[j,] - (A0[j,i]/A0[i,i])*A0[i,]
}
}
}
DET = prod(diag(A0)) # prod() - vectorized multiplication, diag() extracts a[i,i]
if(dim(A)[1] != dim(A)[2]) {
cat("\n\nNeed a square matrix!")
} else{
list(Triangular.Metrix = round(A0,4), Determinant = round(DET,4))
}
}A = matrix(c(1,1,-1,1,1,2,-4,-2,2,1,1,5,-1,0,-2,-4), ncol = 4, byrow = TRUE)
DET(A)## $Triangular.Metrix
## [,1] [,2] [,3] [,4]
## [1,] 1 1 -1 1
## [2,] 0 1 -3 -3
## [3,] 0 0 0 0
## [4,] 0 0 0 0
##
## $Determinant
## [1] 06.5 1A
Gauss.Elimination = function(A){
# Input A: an n-by-m matrix
n = dim(A)[1] # number of rows
A0 = cbind(A, diag(rep(1,n))) # Extended augmented matrix
#P = diag(rep(1,n)) # identity matrix for permutation
for(i in 1:(n-1)){ # iterator fo pivot element A0[i,i]
# i = 1, 2, ..., n-1
#~~~~~~~~~~~~~~~~~~~~~ Make the function robust ~~~~~~~~~~~~~~~~~~~~~~~~~#
if(A0[i,i] == 0){
non.0 = which(A0[,i] != 0) # which() equiv to find() in MATLAB
#cat("\n\n i =",i, ", non.0 =", non.0,". (i+1):n =", (i+1):n, ".")
id = intersect(non.0, (i+1):n)[1] # vector of common row ID in the
# lower triangular part of the matrix.
if(!is.na(id)){ # if the common vector is not empty, do row-swapping!
tempi = A0[i,] # put the i-th row in a place-holder
A0[i,] = A0[id,] # row swapping
A0[id,] = tempi
### # updating permutation matrix
#tempP = P[i,]
#P[i,] = P[id,]
#P[id,] = tempP
}
}
#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#
if(A0[i,i] != 0){ # A0[i,i] is in the denominator of recursive equation
for(j in (i+1):n){ # to make the augmented matrix in the row echelon form
#cat("\n\n j =",j,".")
A0[j,] = A0[j,] - (A0[j,i]/A0[i,i])*A0[i,]
}
}
}
A0
}##
A1 = matrix(c(1,0,0,2,2,1,0,-1,-1,0,1,1), ncol =4, byrow = TRUE)
Gauss.Elimination(A = A1)## [,1] [,2] [,3] [,4] [,5] [,6] [,7]
## [1,] 1 0 0 2 1 0 0
## [2,] 0 1 0 -5 -2 1 0
## [3,] 0 0 1 3 1 0 1y1 = 2 y2 = -5 y3 = 3
Now we do
##
A1 = matrix(c(2,3,-1,2,0,-2,1,-5,0,0,3,3), ncol =4, byrow = TRUE)
Gauss.Elimination(A = A1)## [,1] [,2] [,3] [,4] [,5] [,6] [,7]
## [1,] 2 3 -1 2 1 0 0
## [2,] 0 -2 1 -5 0 1 0
## [3,] 0 0 3 3 0 0 1the solution of the system becomes x1=-3,x2=3,x3=1
7B.
##
A1 = matrix(c(1.012,-2.132,3.104,1.984,-2.132,4.096,-7.013,-5.049,3.104,-7.013,.014,-3.895), ncol =4, byrow = TRUE)
Gauss.Elimination(A = A1)## [,1] [,2] [,3] [,4] [,5] [,6] [,7]
## [1,] 1.012 -2.1320000 3.1040000 1.9840000 1.000000 0.000000 0
## [2,] 0.000 -0.3955257 -0.4737431 -0.8692688 2.106719 1.000000 0
## [3,] 0.000 0.0000000 -8.9391408 -8.9391408 -5.590528 -1.197756 1x3 = 1 x2 = 1 x1 = 1
7D.
##
A1 = matrix(c(2.1756,4.0231,-2.1732,5.1967,17.102,-4.0231,6,0,1.1973,-6.1593,-1,-5.2107,1.1111,0,3.0004,6.0235,7,0,-4.1561,0), ncol =5, byrow = TRUE)
Gauss.Elimination(A = A1)## [,1] [,2] [,3] [,4] [,5] [,6] [,7]
## [1,] 2.1756 4.023100e+00 -2.1732000 5.196700 17.10200 1.0000000 0.0000000
## [2,] 0.0000 1.343948e+01 -4.0186619 10.806991 25.46556 1.8491910 1.0000000
## [3,] 0.0000 4.440892e-16 -0.8929524 5.091694 17.23072 0.9221666 0.2501219
## [4,] 0.0000 2.376891e-15 0.0000000 12.036128 52.71598 2.7364815 1.6466670
## [,8] [,9]
## [1,] 0.000000 0
## [2,] 0.000000 0
## [3,] 1.000000 0
## [4,] 5.352283 1x4= 4.3798 x3= 5.6774 x2= 0.0707 x1= 2.9398
9A.
LU = function(A){
# Input A: an n-by-m matrix
n = dim(A)[1] # number of rows
A0 = cbind(A, diag(rep(1,n))) # Extended augmented matrix
P = diag(rep(1,n)) # identity matrix for permutation
for(i in 1:(n-1)){ # iterator fo pivot element A0[i,i]
# i = 1, 2, ..., n-1
#~~~~~~~~~~~~~~~~~~~~~ Make the function robust ~~~~~~~~~~~~~~~~~~~~~~~~~#
if(A0[i,i] == 0){
non.0 = which(A0[,i] != 0) # which() equiv to find() in MATLAB
#cat("\n\n i =",i, ", non.0 =", non.0,". (i+1):n =", (i+1):n, ".")
id = intersect(non.0, (i+1):n)[1] # vector of common row ID in the
# lower triangular part of the matrix.
if(!is.na(id)){ # if the common vector is not empty, do row-swapping!
tempi = A0[i,] # put the i-th row in a place-holder
A0[i,] = A0[id,] # row swapping
A0[id,] = tempi
### # updating permutation matrix
tempP = P[i,]
P[i,] = P[id,]
P[id,] = tempP
}
}
#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#
if(A0[i,i] != 0){ # A0[i,i] is in the denominator of recursive equation
for(j in (i+1):n){ # to make the augmented matrix in the row echelon form
#cat("\n\n j =",j,".")
A0[j,] = A0[j,] - (A0[j,i]/A0[i,i])*A0[i,]
}
}
}
PA = P%*%A
U = A0[,1:n]
L = solve(A0[, (n+1):(2*n)])
A = A
list(A = A, P = P, PA = PA, L = L, U = U)
}A=matrix(c(0,2,3,1,1,-1,0,-1,1), ncol = 3, byrow = TRUE)
P = matrix(c(0,1,0,1,0,0,0,0,1), ncol = 3, byrow = TRUE)
LU(A)## $A
## [,1] [,2] [,3]
## [1,] 0 2 3
## [2,] 1 1 -1
## [3,] 0 -1 1
##
## $P
## [,1] [,2] [,3]
## [1,] 0 1 0
## [2,] 1 0 0
## [3,] 0 0 1
##
## $PA
## [,1] [,2] [,3]
## [1,] 1 1 -1
## [2,] 0 2 3
## [3,] 0 -1 1
##
## $L
## [,1] [,2] [,3]
## [1,] 0 1.0 0
## [2,] 1 0.0 0
## [3,] 0 -0.5 1
##
## $U
## [,1] [,2] [,3]
## [1,] 1 1 -1.0
## [2,] 0 2 3.0
## [3,] 0 0 2.5P^(t)LU = [0,1,0,1,0,0,0,0,1][1,0,0,0,1,0,0,-0.5,1][1,1,-1,0,2,3,0,0,2.5] or \(P*\)L*$U from above
9D.
A=matrix(c(1,-2,3,0,1,-2,3,1,1,-2,2,-2,2,1,3,-1), ncol = 4, byrow = TRUE)
P = matrix(c(1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1), ncol = 4, byrow = TRUE)
LU(A)## $A
## [,1] [,2] [,3] [,4]
## [1,] 1 -2 3 0
## [2,] 1 -2 3 1
## [3,] 1 -2 2 -2
## [4,] 2 1 3 -1
##
## $P
## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 0 0 1
## [3,] 0 0 1 0
## [4,] 0 1 0 0
##
## $PA
## [,1] [,2] [,3] [,4]
## [1,] 1 -2 3 0
## [2,] 2 1 3 -1
## [3,] 1 -2 2 -2
## [4,] 1 -2 3 1
##
## $L
## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 1 0 0 1
## [3,] 1 0 1 0
## [4,] 2 1 0 0
##
## $U
## [,1] [,2] [,3] [,4]
## [1,] 1 -2 3 0
## [2,] 0 5 -3 -1
## [3,] 0 0 -1 -2
## [4,] 0 0 0 1P^(t)LU = [1,0,0,0,0,0,0,1,0,0,1,0,0,1,0,0][1,0,0,0,2,1,0,0,1,0,1,0,1,0,0,1][1,-2,3,0,0,5,-3,-1,0,0,-1,-2,0,0,0,1]or \(P*\)L*$U from above
8.1 12.
x <- c(.04,.041,.055,.056,.062,.071,.071,.078,.082,.09,.092,.1,.105,.12,.123,.13,.14)
y <- c(26.5,28.1,25.2,26,24,25,26.4,27.2,25.6,25,26.8,24.8,27,25,27.3,26.9,26.2)
plot(x, y)
abline(lm(y ~ x))
summary(lm(y ~ x))$coefficients## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 25.921755 0.8434062 30.7346032 5.816098e-15
## x 1.600393 9.3007288 0.1720718 8.656812e-01line of best fit is y = 1.6x + 25.92