# Gaussian-Jordan Elimination

## Gaussian-Jordan Elimination

Definition

Consider the $m\times n$ system of linear equations:
\begin{align*}
a_{1 1} x_1+a_{1 2}x_2+\cdots+a_{1 n}x_n& =b_1 \\
a_{2 1} x_1+a_{2 2}x_2+\cdots+a_{2 n}x_n& =b_2 \\
a_{3 1} x_1+a_{3 2}x_2+\cdots+a_{3 n}x_n& =b_3 \\
&\vdots \\
a_{m 1} x_1+a_{m 2}x_2+\cdots+a_{m n}x_n& =b_m \\
\end{align*}

1. The coefficient matrix of the system is
$\begin{bmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\ a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m 1} & a_{m 2} & \cdots & a_{m n} \end{bmatrix}$
2. The augmented matrix of the system is
$\left[\begin{array}{rrrr|r} a_{1 1} & a_{1 2} & \cdots & a_{1 n} & b_1 \\ a_{2 1} & a_{2 2} & \cdots & a_{2 n} & b_2\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ a_{m 1} & a_{m 2} & \cdots & a_{m n} & b_m \end{array}\right]$
3. [Gauss-Jordan Elimination] For a given system of linear equations, we can find a solution as follows.
This procedure is called Gauss-Jordan elimination.

1. Write the augmented matrix of the system of linear equations.
2. Use elementaray row operations to reduce the augmented matrix into (reduced) row echelon form.
3. Write the system of linear equation corresponding to the matrix in row echelon form.
4. Solve the system using back substitution.

=solution

### Problems

1. Solve the following system by transforming the augmented matrix to reduced echelon form. Indicate the elementary row operations you performed.
\begin{align*}
x_1+x_2-x_5&=1\\
x_2+2x_3+x_4+3x_5&=1\\
x_1-x_3+x_4+x_5&=0
\end{align*}

2. Solve the following system of linear equations using Gaussian elimination.
\begin{align*}
x+2y+3z &=4 \\
5x+6y+7z &=8\\
9x+10y+11z &=12
\end{align*}

3. Solve the following system of linear equations using Gauss-Jordan elimination.
\begin{align*}
6x+8y+6z+3w &=-3 \\
6x-8y+6z-3w &=3\\
8y \,\,\,\,\,\,\,\,\,\,\,- 6w &=6
\end{align*}

4. Solve the following system of linear equations by transforming its augmented matrix to reduced echelon form (Gauss-Jordan elimination). Find the vector form for the general solution.
\begin{align*}
x_1-x_3-3x_5&=1\\
3x_1+x_2-x_3+x_4-9x_5&=3\\
x_1-x_3+x_4-2x_5&=1.
\end{align*}

5. The given matrix is the augmented matrix for a system of linear equations. Give the vector form for the general solution.
$\left[\begin{array}{rrrrr|r} 1 & 0 & -1 & 0 &-2 & 0 \\ 0 & 1 & 2 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 \\ \end{array} \right].$

6. Solve the following system of linear equations and give the vector form for the general solution.
\begin{align*}
x_1 -x_3 -2x_5&=1 \\
x_2+3x_3-x_5 &=2 \\
2x_1 -2x_3 +x_4 -3x_5 &= 0
\end{align*}
(The Ohio State University)

7. Find a cubic polynomial $p(x)=a+bx+cx^2+dx^3$ such that $p(1)=1, p'(1)=5, p(-1)=3$, and $p'(-1)=1$.