Homogeneous System of Linear Equations
2026-02-28 08:24 Diff
https://brightchamps.com/en-us/math/algebra/homogeneous-system-of-linear-equations

In solving a system of linear equations, first find the values of the variables that make all the equations true at the same time. For example, solve the system:

\(2x + y = 8\)

\(3x -2y = -1\)

Step 1: Write it as augmented matrix


\( \begin{bmatrix} 2 & 1 & | & 8 \\ 3 & -2 & | & -1 \end{bmatrix} \)


Step 2: Replace the first element of the matrix with 1

To do that, we divide the first row by 2:

\( R_1 = \tfrac{1}{2} R_1 \left[\begin{array}{cc|c} 1 & 0.5 & 4 \\ 3 & -2 & -1 \end{array}\right] \)

Step 3: Eliminate the 3 below the leading 1
Using row operation:

\(R2 = R2 - 3 × R1\)
 

\(R2 = 3,-2,-1 - 3 - 1, 0.5, 4 = 3-3,-2-1.5,-1-12  = 0,-3.5,-13 \)


So now, the new matrix is:
\(\left[\begin{array}{cc|c} 1 & 0.5 & 4 \\ 3 & -2 & -1 \end{array}\right]\)
 

Step 4: Make the second row’s middle into 1
Divide row 2 by -3.5

\(R2 = 1-3.5  \)
 

\( R_2 = \frac{(0,\;1,\;-13)}{-3.5} = (0,\;1,\;3.714\ldots) \)

Or as fraction: \(133.5 = 13035 = 267\)


\( \left[\begin{array}{cc|c} 1 & 0.5 & 4 \\ 0 & 1 & 267 \end{array}\right] \)

Step 5: make 0.5 into 0 

\(R1= R1 -0.5 × R2\)


\( \left[\begin{array}{cc|c} 1 & 0 & 157 \\ 0 & 1 & 267 \end{array}\right] \)


\(x = 157\)  and \(y = 267\)