# Define variables
x <- "x"
y <- "y"
z <- "z"
# Equations
equation1 <- paste(x, "+", y, "-", z, "=", -5)
equation2 <- paste(x, "-", y, "-", z, "=", -3)
equation3 <- paste(x, "+", y, "-", z, "=", 0)
# Display equations
equation1## [1] "x + y - z = -5"## [1] "x - y - z = -3"## [1] "x + y - z = 0"# Coefficient matrix A
A <- matrix(c(1, 1, -1, # coefficients for x in equations 1, 2, and 3
1, -1, -1, # coefficients for y in equations 1, 2, and 3
1, 1, -1), # coefficients for z in equations 1, 2, and 3
nrow = 3, byrow = TRUE)
# Right-hand side vector B
B <- c(-5, -3, 0)
# Display A and B
print("Coefficient matrix A:")## [1] "Coefficient matrix A:"## [,1] [,2] [,3]
## [1,] 1 1 -1
## [2,] 1 -1 -1
## [3,] 1 1 -1## [1] "Right-hand side vector B:"## [1] -5 -3 0# Solve the system of equations
# X <- solve(A, B)
# Display the solution vector X
# print("Solution vector X:")the coefficient matrix A is singular, meaning it doesn’t have an inverse. In simpler terms, the system of equations is dependent or inconsistent.
You can see that the third row is the same as the first row. This implies that the third equation is redundant and does not provide additional information. In such a case, the system is dependent,
Since the system of equations is dependent. I will remove the Redundant Equation in the third equation and use the least squares method to find a solution
# Coefficient matrix A
A <- matrix(c(1, 1, -1, # coefficients for x in equations 1 and 2
1, -1, -1), # coefficients for y in equations 1 and 2
nrow = 2, byrow = TRUE)
# Right-hand side vector B
B <- c(-5, -3)
# Create a data frame for the linear regression
data <- data.frame(A)
colnames(data) <- c("x", "y", "z")
data$B <- B
# Perform linear regression
model <- lm(B ~ x + y + z, data = data)
# Display coefficients (solution vector X)
coefficients(model)## (Intercept) x y z
## -4 NA -1 NAIt appears that the linear regression model has encountered some issues.