Link to Problem: HERE

Let \(A\) be the coefficient matrix of the system of equations below. Is \(A\) nonsingular or singular?

\[ \begin{aligned} -x_1 + 5x_2 = -8 \\ -2x_1 + 5x_2 + 5x_3 + 2x_4 = 9 \\ -3x_1 + -x_2 + 3x_3 + x_4 = 3 \\ 7x_1 + 6x_2 + 5x_3 + x_4 = 30 \end{aligned} \]

Begin by transforming the system of equations into a matrix. Be sure to place \(0\) terms for missing coefficients.

\[\mathbf{A} = \left[\begin{array} {rrr} -1 & 5 & 0 & 0 \\ -2 & 5 & 5 & 2 \\ -3 & -1 & 3 & 1 \\ 7 & 6 & 5 & 1 \end{array}\right] \]

By definition, a matrix \(A\) is nonsingular if it row reduces to the identify matrix. This can be done by hand or with the rref function from the pracma library

A <- matrix(c(1, 5, 0, 0, -2, 5, 5, 2, -3, -1, 3, 1, 7, 6, 5, 1), 4, 4, byrow=TRUE)
pracma::rref(A)
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    0    0
## [2,]    0    1    0    0
## [3,]    0    0    1    0
## [4,]    0    0    0    1

As per Theorem NMRRI (Nonsingular Matrices Row Reduce to the Identity Matrix) the original matrix \(A\) is nonsingular.

Explain what you could infer about the solution set for the system based only on what you have learned about \(A\) being singular or nonsingular.

As per Theorme NME1: Nonsingular Matrix Equivalences, Round 1 there are a number of equivalent attributes:

  1. \(A\) is nonsingular
  2. \(A\) row-reduces to the identity matrix (proven above)
  3. The null space of \(A\) contains only the zero vector, \(N(A) = {0}\)
  4. The linear system \(LS(A, b)\) has a unique solution for every possible choice of \(b\).

In the case of the specific system given the equation, the unique solution is

b <- matrix(c(-8, 9, 3, 30))
solve(A, b)
##           [,1]
## [1,]  4.826087
## [2,] -2.565217
## [3,] -1.652174
## [4,] 19.869565

\[ \mathbf{X} = \left[\begin{array} {rrr} 4.826087 \\ -2.565217 \\ -1.652174 \\ 19.869565 \end{array}\right] \]