# Define system of linear equations. When a system of equation is consistent? Determine if the system -2x1 – 3x2 + 4x3 = 5, x2 – 2x3 = 4, x1 + 3x2 – x3 = 2 is consistent.

System of linear equations is a collection of one or more linear equations involving the same variables.

For example:

3x + 2y – z = 1

2x – 2y + 4z = -2

-x + y – z = 0

If system of linear equation has at least one solution then system of equation is consistent.

Soluton

Given Equations are

-2x1 – 3x2 + 4x3 = 5

x2 – 2x3 = 4

x1 + 3x2 – x3 = 2

The given equations can be written in the form

Ax = b

Where

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

x = $$\begin{bmatrix}x_1\\ x_2\\ x_3\end{bmatrix}$$

b = $$\begin{bmatrix}5\\ 4\\ 2\end{bmatrix}$$

Its augmented form is [A:b]

~ [A:b]

~ $$\begin{bmatrix}-2 & -3 & 4 & : & 5\\ 0 & 1 & -2 & : & 4\\ 1 & 3 & -1 & : & 2 \end{bmatrix}$$

R3 ↔ R1

~ $$\begin{bmatrix}1 & 3 & -1 & : & 2\\ 0 & 1 & -2 & : & 4\\ -2 & -3 & 4 & : & 5 \end{bmatrix}$$

R3 → R3 + 2R1

~ $$\begin{bmatrix}1 & 3 & -1 & : & 2\\ 0 & 1 & -2 & : & 4\\ 0 & 3 & 2 & : & 9 \end{bmatrix}$$

R1 → R1 – 3R2 ,  R3 → R3 – 3R2

~ $$\begin{bmatrix}1 & 0 & 5 & : & -10\\ 0 & 1 & -2 & : & 4\\ 0 & 0 & 8 & : & -3 \end{bmatrix}$$

R3 → R3 / 8

~ $$\begin{bmatrix}1 & 0 & 5 & : & -10\\ 0 & 1 & -2 & : & 4\\ 0 & 0 & 1 & : & \frac{-3}{8} \end{bmatrix}$$

R2 → R2 + 2R3 ,  R1 → R1 – 5R3

~ $$\begin{bmatrix}1 & 0 & 0 & : & \frac{-65}{8} \\ 0 & 1 & 0 & : & \frac{13}{4}\\ 0 & 0 & 1 & : & \frac{-3}{8} \end{bmatrix}$$

From the above, We can say that

x = $$\begin{bmatrix}x_1\\ x_2\\ x_3\end{bmatrix}$$ = $$\begin{bmatrix}\frac{-65}{8}\\ \frac{13}{4}\\ \frac{-3}{8}\end{bmatrix}$$