Question

Triangular Systems

Use back-substitution to solve the triangular system

$\left\{\begin{array}{l}x-3y+z=0\\ y-z=3\\ z=-2\end{array}\right.$

Short answer

Hence we get the solution of the triangular system as

$\begin{array}{l}x=5\\ y=1\\ z=-2\end{array}$

Step-by-step solution

Step 1. Given information

We are given the set of linear equations that is-

$\left\{\begin{array}{l}x-3y+z=0\\ y-z=3\\ z=-2\end{array}\right.$

Step 2. Concept used

The process of solving a linear system of equations that has been transformed into the row-echelon form or reduced row-echelon form is known as the back-substituting method.

We have to solve the given triangular system by using the back substitution method.

Step 3. Calculation

Let us name all the equations of systems of equations.

$\begin{array}{l}x-3y+z=\mathrm{0...}\left(1\right)\\ y-z=\mathrm{3...}\left(2\right)\\ z=-\mathrm{2...}\left(3\right)\end{array}$

Now we are substituting the equation $\left(3\right)$in equation $\left(2\right)$, we get

$\begin{array}{l}y-(-2)=3\\ y+2=3\\ y=3-2\\ y=1\end{array}$

Now by substituting the value of y and z in equation $\left(1\right)$, we get

$\begin{array}{l}x-3\left(1\right)+(-2)=0\\ x-3-2=0\\ x-5=0\\ x=5\end{array}$

So, we solve the triangular system.