# Quiz 1. Gauss-Jordan Elimination / Homogeneous System. Math 2568 Spring 2017. ## Problem 262

(a) Solve the following system by transforming the augmented matrix to reduced echelon form (Gauss-Jordan elimination). 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*}

(b) Determine all possibilities for the solution set of a homogeneous system of $2$ equations in $2$ unknowns that has a solution $x_1=1, x_2=5$. Add to solve later

## Solution.

### (a) Solve the following system by Gauss-Jordan elimination

The augmented matrix of the system is
$\left[\begin{array}{rrrrr|r} 1 & 1 & 0 & 0 &-1 & 1 \\ 0 & 1 & 2 & 1 & 3 & 1 \\ 1 & 0 & -1 & 1 & 1 & 0 \\ \end{array} \right].$ We apply elementary row operations as follows.
\begin{align*}
\left[\begin{array}{rrrrr|r}
1 & 1 & 0 & 0 &-1 & 1 \\
0 & 1 & 2 & 1 & 3 & 1 \\
1 & 0 & -1 & 1 & 1 & 0 \\
\end{array} \right] \xrightarrow{R_3-R_1}
\left[\begin{array}{rrrrr|r}
1 & 1 & 0 & 0 &-1 & 1 \\
0 & 1 & 2 & 1 & 3 & 1 \\
0 & -1 & -1 & 1 & 2 & -1 \\
\end{array} \right] \xrightarrow{\substack{R_1-R_2\\ R_3+R_2}}\\[6pt] \left[\begin{array}{rrrrr|r}
1 & 0 & -2 & -1 &-4 & 0 \\
0 & 1 & 2 & 1 & 3 & 1 \\
0 & 0 & 1 & 2 & 5 & 0 \\
\end{array} \right] \xrightarrow{\substack{R_1+2R_3\\ R_2-2R_3}}
\left[\begin{array}{rrrrr|r}
1 & 0 & 0 & 3 & 6 & 0 \\
0 & 1 & 0 & -3 & -7 & 1 \\
0 & 0 & 1 & 2 & 5 & 0 \\
\end{array} \right].
\end{align*}

The last matrix is in reduced row echelon form.
The unknowns $x_1, x_2, x_3$ correspond to the leading $1$’s. Thus they are dependent variables and $x_4$ and $x_5$ are independent (free) variables.
From the last matrix, we see that the solution set is given by
\begin{align*}
x_1&=-3x_4-6x_5\\
x_2&=3x_4+7x_5+1\\
x_3&=-2x_4-5x_5
\end{align*}
for any numbers $x_4, x_5$.

### (b) Determine all possibilities for the solution set of a homogeneous system

A homogeneous system is always consistent because it has the zero solution $x_1=0, x_2=0$.
So, we already know the possibilities for the solution set are either a unique solution or infinitely many solutions.

Note that the homogeneous system we are inspecting has a solution $x_1=1, x_2=5$, which is different from the zero solution $x_1=0, x_2=0$. Thus, the homogeneous system has at least two solutions.
Thus, the only possibility is that the homogeneous system has infinitely many solutions.

## Comment.

These are Quiz 1 problems for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017.

### List of Quiz Problems of Linear Algebra (Math 2568) at OSU in Spring 2017

There were 13 weekly quizzes. Here is the list of links to the quiz problems and solutions. Add to solve later

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