1. Using matrix operations, describe the solutions for the following family of equations:
  x + 2y - 3z = 5
  2x + y - 3z = 13
  -x + y = -8

\(\textbf{} \begin{bmatrix} 1 & 2 & -3 \\ 2 & 1 & -3 \\ -1 & 1 & 0 \end{bmatrix}\,\,\) \(\textbf{} \begin{bmatrix} x \\ y \\ z \end{bmatrix}\,\,\) = \(\textbf{} \begin{bmatrix} 5 \\ 13 \\ -8 \end{bmatrix}\,\)

\[\mathbf{(R2 < R2 - R1)} = \left[\begin{array} {rrr} 1 & 2 & -3 & | & 5 \\ 1 & -1 & 0 & | & 8 \\ -1 & 1 & 0 & | & -8 \end{array}\right] \]

\[\mathbf{(R3 < R3 + R2)} = \left[\begin{array} {rrr} 1 & 2 & -3 & | & 5 \\ 1 & -1 & 0 & | & 8 \\ 0 & 0 & 0 & | & 0 \end{array}\right] \]

\[\mathbf{(R2 < R2 - R1)} = \left[\begin{array} {rrr} 1 & 2 & -3 & | & 5\\ 0 & -3 & 3 & | & 3\\ 0 & 0 & 0 & | & 0 \end{array}\right] \]

\[\mathbf{(R2 < R2/(-3))} = \left[\begin{array} {rrr} 1 & 2 & -3 & | & 5\\ 0 & 1 & -1 & | & -1\\ 0 & 0 & 0 & | & 0 \end{array}\right] \]

\[\mathbf{(R1 < R1 - R2*2)} = \left[\begin{array} {rrr} 1 & 0 & -1 & | & 7\\ 0 & 1 & -1 & | & -1\\ 0 & 0 & 0 &| & 0 \end{array}\right] \]

x - z = 7 == x = z + 7

y - z = -1 == y = z - 1

Solution (x = z + 7, y = z - 1)

2. Provide a solution for #1, using R functions of your choice.
M_data <- c(1,2,-1,2,1,1,-3,-3,0,5,13,-8)
M <- matrix(M_data, nrow = 3, ncol = 4)
M
##      [,1] [,2] [,3] [,4]
## [1,]    1    2   -3    5
## [2,]    2    1   -3   13
## [3,]   -1    1    0   -8
Matrix <- editrules::echelon(M)
Matrix
##      [,1] [,2] [,3] [,4]
## [1,]    1    0   -1    7
## [2,]    0    1   -1   -1
## [3,]    0    0    0    0
Results_1 <- paste(Matrix[1,1], "y + ", Matrix[1,2], "x + ", Matrix[1,3], "z = ", Matrix[1,4])
Results_2 <- paste(Matrix[2,1], "y + ", Matrix[2,2], "x + ", Matrix[2,3], "z = ", Matrix[2,4])
Results_3 <- paste(Matrix[3,1], "y + ", Matrix[3,2], "x + ", Matrix[3,3], "z = ", Matrix[3,4])
Results_1
## [1] "1 y +  0 x +  -1 z =  7"
Results_2
## [1] "0 y +  1 x +  -1 z =  -1"
Results_3
## [1] "0 y +  0 x +  0 z =  0"
3. Solve for AB by hand:

\[\mathbf{A} = \left[\begin{array} {rrr} 4&-3\\ -3&5\\ 0&1 \end{array}\right] \]

\[\mathbf{B} = \left[\begin{array} {rrr} 1&4\\ 3&-2 \end{array}\right] \]

\[\mathbf{X} = \left[\begin{array} {rrr} (4*1 + -3*3)& (4*4 + -3*-2) \\ (-3*1 + 5*3) & (-3*4 + 5*-2) \\ (0*1 + 1*3) & (0*4 + 1 * -2) \end{array}\right] \]

\[\mathbf{X} = \left[\begin{array} {rrr} -5 & 22 \\ 12 & -22 \\ 3 & -2 \end{array}\right] \]

4. Solve AB from #3 using R functions of your choice.
A_data <- c(4,-3,0,-3,5,1)
A <- matrix(A_data, nrow = 3, ncol = 2)
A
##      [,1] [,2]
## [1,]    4   -3
## [2,]   -3    5
## [3,]    0    1
B_data <- c(1,3,4,-2)
B <- matrix(B_data, nrow = 2, ncol = 2)
B
##      [,1] [,2]
## [1,]    1    4
## [2,]    3   -2
A %*% B
##      [,1] [,2]
## [1,]   -5   22
## [2,]   12  -22
## [3,]    3   -2