Inefficient Way of Solving SLEs

{R1=x+2y+3z=6R2=2x−3y+2z=14R3=3x+y−z=2Subtract R1 from R2 and R3{R1=x+2y+3z=6R2−2R1=−7y−4z=2R3−3R1=−5y−10z=−20=−5(y+2z=4)Swap row 2 and row 3 for simplicity, and divide out the -5 from the new row 2.{R1=x+2y+3z=6R2=y+2z=4R3=−7y−4z=2From here, you can isolate the y and x variables and solve for them.{R1=x+2y+3z=6R2=y+2z=4R3+7R2=10z=30→z=3Use back substitution to solve for other variables.R2=y+2(3)=4→y=−2R1=x+2(−2)+3(3)=6→x=1Thus, the answer is: x=1,y=−2,z=3\begin{cases} R_1 = x + 2y + 3z = 6 \\ R_2 = 2x - 3y + 2z = 14 \\ R_3 = 3x + y - z = 2 \end{cases} \\ \text{Subtract } R_1 \text{ from } R_2 \text{ and } R_3 \\ \begin{cases} R_1 = x + 2y + 3z = 6 \\ R_2 - 2R_1 = -7y - 4z = 2 \\ R_3 - 3R_1 = -5y - 10z = -20 = -5(y + 2z = 4) \end{cases} \\ \text{Swap row 2 and row 3 for simplicity, and divide out the -5 from the new row 2.} \\ \begin{cases} R_1 = x + 2y + 3z = 6 \\ R_2 = y + 2z = 4 \\ R_3 = -7y - 4z = 2 \end{cases} \\ \text{From here, you can isolate the y and x variables and solve for them.} \\ \begin{cases} R_1 = x + 2y + 3z = 6 \\ R_2 = y + 2z = 4 \\ R_3 + 7R_2 = 10z = 30 \rightarrow z = 3 \end{cases} \\ \text{Use back substitution to solve for other variables.} \\ R_2 = y + 2(3) = 4 \rightarrow y = -2 \\ R_1 = x + 2(-2) + 3(3) = 6 \rightarrow x = 1 \\ \text{Thus, the answer is: } x = 1, y = -2, z = 3

Solving SLEs with matrices

First, convert the system to an augmented matrix, augmented meaning that there is a line separating variables from constants.{x+2y+z=0x−y+3z=−23x−z=4→(12101−13−230−14)Use the same process of removing one coefficient at a time.R2−R1,R3−3R1(12100−32−20−6−44)R3−2R1(12100−32−200−88)Convert back into system of equations and perform back substitution{x+2y+z=0−3y+2z=−2−8z=8→z=−1−3y+2(−1)=−2→y=0x+2(0)+(−1)=0→x=1Thus, the answer is x=1,y=0,z=−1\text{First, convert the system to an augmented matrix, augmented meaning that there is a line separating variables from constants.} \\ \begin{cases} x + 2y + z = 0 \\ x - y + 3z = -2 \\ 3x - z = 4 \end{cases} \rightarrow \left( \begin{array}{ccc|c} 1 & 2 & 1 & 0 \\ 1 & -1 & 3 & -2 \\ 3 & 0 & -1 & 4 \end{array} \right) \\ \text{Use the same process of removing one coefficient at a time.}\\ R_2 - R_1, R_3 - 3R_1\\ \left( \begin{array}{ccc|c} 1 & 2 & 1 & 0 \\ 0 & -3 & 2 & -2 \\ 0 & -6 & -4 & 4 \end{array} \right) \\ \\ R_3 - 2R_1 \\ \left( \begin{array}{ccc|c} 1 & 2 & 1 & 0 \\ 0 & -3 & 2 & -2 \\ 0 & 0 & -8 & 8 \end{array} \right) \\ \text{Convert back into system of equations and perform back substitution}\\ \begin{cases} x + 2y + z = 0 \\ -3y + 2z = -2 \\ -8z = 8 \rightarrow z = -1 \end{cases}\\ -3y + 2(-1) = -2 \rightarrow y = 0\\ x + 2(0) + (-1) = 0 \rightarrow x = 1\\ \text{Thus, the answer is } x = 1, y = 0, z = -1

In this method, we use Elementary Row Operations which include the following:

  1. Exchanging two rows $$ \begin{bmatrix} 2 & 3 & 2
    1 & 0 & 4
    1 & 1 & 1 \end{bmatrix} \rightarrow

\begin{bmatrix} 2 & 3 & 2
1 & 1 & 1
1 & 0 & 4 \end{bmatrix} $$

  1. Multiplying or dividing a row by a non-zero constant $$ \begin{bmatrix} 2 & 3 & 2
    1 & 0 & 4
    5 & 5 & 5 \end{bmatrix} \rightarrow

\begin{bmatrix} 2 & 3 & 2
1 & 0 & 4
1 & 1 & 1 \end{bmatrix} $$

  1. Add or subtract a multiple of one row from another $$ R_1 - 2R_2 \begin{bmatrix} 2 & 3 & 2
    1 & 0 & 4
    1 & 1 & 1 \end{bmatrix} \rightarrow

\begin{bmatrix} 0 & 3 & -6
1 & 0 & 4
1 & 1 & 1 \end{bmatrix} $$

All are used to solve systems of linear equations, and importantly, all operations are reversible.

Pivots

In reducing a matrix that represents a system, our goal is to have as many pivots as possible and get the matrix down to row echelon form. A matrix in row echelon form looks like the following:

(12100−32−200−88)\left( \begin{array}{ccc|c} 1 & 2 & 1 & 0 \\ 0 & -3 & 2 & -2 \\ 0 & 0 & -8 & 8 \end{array} \right) \\

As shown above, row echoleon forms have a pivot in every row. A pivot is a value where either everything before (left of) it is a zero and everything below it is a zero (so 1, -3, and -8). It is also effective to reduce down the matrix to a form such as this:

(100x010y001z)\left( \begin{array}{ccc|c} 1 & 0 & 0 & x \\ 0 & 1 & 0 & y \\ 0 & 0 & 1 & z \end{array} \right) \\

This is reduced row echelon form and is easy to reach when you have the row echelon form.

If the row echelon form has zeroes at the bottom but a nonzero constant such as this:

(014−5135−2000−10)\left( \begin{array}{ccc|c} 0 & 1 & 4 & -5 \\ 1 & 3 & 5 & -2 \\ 0 & 0 & 0 & -10 \end{array} \right) \\

There are no solutions.

If the bottom instead looks something like this:

(014−5135−20000)\left( \begin{array}{ccc|c} 0 & 1 & 4 & -5 \\ 1 & 3 & 5 & -2 \\ 0 & 0 & 0 & 0 \end{array} \right) \\

There are infinitely many solutions.