For this problem first its necessary to create a systems of Matrices from the equations:
\[\mathbf{} \left[\begin{array} {rrr} 1 & 2 & -3 \\ 2 & 1 & -3 \\ -1 & 1 & 0 \end{array}\right] * \left[\begin{array} {r} x\\ y\\ z \end{array}\right] = \left[\begin{array} {r} 5\\ 13\\ -8 \end{array}\right] \]
From what we know of multiplaction properties, to continue solving this system it is necessary to take the inverse and multiple it to both sides, this would give the following:
\[ A * X = B \\ X = A^{-1} * B \]
THe inverse for a 3x3 Matrix.
For this problem, we can use the solve equation in R to solve these system of equations.
x <- matrix(c(1,2-3,2,1,-3,-1,1,0), nrow=3,ncol=3,byrow=T)## Warning in matrix(c(1, 2 - 3, 2, 1, -3, -1, 1, 0), nrow = 3, ncol = 3,
## byrow = T): data length [8] is not a sub-multiple or multiple of the number
## of rows [3]b <- c(5,13,-8)
x## [,1] [,2] [,3]
## [1,] 1 -1 2
## [2,] 1 -3 -1
## [3,] 1 0 1b## [1] 5 13 -8solve(x,b)## [1] -11.6 -9.4 3.6\[\mathbf{} \left[\begin{array} {rrr} 4 & -3 \\ -3 & 5 \\ 0 & 1 \end{array}\right] * \left[\begin{array} {r} 1 & 4 \\ 3 & -2 \end{array}\right] \]
For this problem we must take the dot product of all the rows in the first matrix and multple it by each column of the second matrix which looks like this:
\[ (4,-3) \cdot (1,3) = 4 * 1 + (-3) * 3 = -5 \] \[ (4,-3) \cdot (4,-2) = 4 * 4 + (-3) * -2 = 22 \] \[ (-3,5) \cdot (1,3) = -3 * 1 + 5 * 3 = 12\] \[ (-3,5) \cdot (4,-2) = -3 * 4 + 5 * -2 = -22 \] \[ (0,1) \cdot (1,3) = 0 * 1 + 1 * 3 = 3 \] \[ (0,1) \cdot (4,-2) = 0 * 4 + 1 * -2 = -2 \]
\[ \mathbf{} \left[\begin{array} {rrr} 4 & -3 \\ -3 & 5 \\ 0 & 1 \end{array}\right] * \left[\begin{array} {rr} 1 & 4 \\ 3 & -2 \end{array}\right] = \left[\begin{array} {rr} -5 & 22\\ 12 & -22\\ 3 & -2 \end{array}\right] \]
x <- matrix(c(4,-3,-3,5,0,1), nrow=3,ncol=2,byrow=T)
y <- matrix(c(1,4,3,-2), nrow=2,ncol=2,byrow=T)
x %*% y## [,1] [,2]
## [1,] -5 22
## [2,] 12 -22
## [3,] 3 -2