# Solve the System of Linear Equations and Give the Vector Form for the General Solution

## Problem 296

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, linear algebra midterm exam problem)

## Solution.

We solve the system by Gauss-Jordan elimination.
The augmented matrix of the system is given by

$\left[\begin{array}{rrrrr|r} 1 & 0 & -1 & 0 &-2 & 1 \\ 0 & 1 & 3 & 0 & -1 & 2 \\ 2 & 0 & -2 & 1 & -3 & 0 \\ \end{array}\right].$ We apply the elementary row operations as follows.
\begin{align*}
\left[\begin{array}{rrrrr|r}
1 & 0 & -1 & 0 &-2 & 1 \\
0 & 1 & 3 & 0 & -1 & 2 \\
2 & 0 & -2 & 1 & -3 & 0 \\
\end{array}\right] \xrightarrow{R_3-2R_1}
\left[\begin{array}{rrrrr|r}
1 & 0 & -1 & 0 &-2 & 1 \\
0 & 1 & 3 & 0 & -1 & 2 \\
0 & 0 & 0 & 1 & 1 & -2 \\
\end{array}\right].
\end{align*}
Then the last matrix is in reduced row echelon form.
The variables $x_1, x_2, x_4$ correspond to the leading $1$’s of the last matrix, hence they are dependent variables and the rest $x_3, x_5$ are free variables.
From the last matrix we obtain the general solution
\begin{align*}
x_1&=x_3+2x_5+1\\
x_2&=-3x_3+x_5+2\\
x_4&=-x_5-2.
\end{align*}
The vector form for the general solution is obtained by substituting these into the vector $\mathbf{x}$.
We have
\begin{align*}
\mathbf{x}&=\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5
\end{bmatrix}=\begin{bmatrix}
x_3+2x_5+1 \\
-3x_3+x_5+2 \\
x_3 \\
-x_5-2 \\
x_5
\end{bmatrix}\10pt] &=x_3\begin{bmatrix} 1 \\ -3 \\ 1 \\ 0 \\ 0 \end{bmatrix}+x_5\begin{bmatrix} 2 \\ 1 \\ 0 \\ -1 \\ 1 \end{bmatrix} +\begin{bmatrix} 1 \\ 2 \\ 0 \\ -2 \\ 0 \end{bmatrix}. \end{align*} Therefore, the vector form for the general solution is given by \[\mathbf{x}=x_3\begin{bmatrix} 1 \\ -3 \\ 1 \\ 0 \\ 0 \end{bmatrix}+x_5\begin{bmatrix} 2 \\ 1 \\ 0 \\ -1 \\ 1 \end{bmatrix} +\begin{bmatrix} 1 \\ 2 \\ 0 \\ -2 \\ 0 \end{bmatrix}, where $x_3, x_5$ are free variables.

## Comment.

This is one of the midterm exam 1 problems of linear algebra (Math 2568) at the Ohio State University.

## Midterm 1 problems and solutions

The following list is the problems and solutions/proofs of midterm exam 1 of linear algebra at the Ohio State University in Spring 2017.

1. Problem 1 and its solution: Possibilities for the solution set of a system of linear equations
2. Problem 2 and its solution (The current page): The vector form of the general solution of a system
3. Problem 3 and its solution: Matrix operations (transpose and inverse matrices)
4. Problem 4 and its solution: Linear combination
5. Problem 5 and its solution: Inverse matrix
6. Problem 6 and its solution: Nonsingular matrix satisfying a relation
7. Problem 7 and its solution: Solve a system by the inverse matrix
8. Problem 8 and its solution:A proof problem about nonsingular matrix

### 8 Responses

1. 02/13/2017

[…] Problem 2 and its solution: The vector form of the general solution of a system […]

2. 02/13/2017

[…] Problem 2 and its solution: The vector form of the general solution of a system […]

3. 02/13/2017

[…] Problem 2 and its solution: The vector form of the general solution of a system […]

4. 02/13/2017

[…] Problem 2 and its solution: The vector form of the general solution of a system […]

5. 02/14/2017

[…] Problem 2 and its solution: The vector form of the general solution of a system […]

6. 02/14/2017

[…] Problem 2 and its solution: The vector form of the general solution of a system […]

7. 07/19/2017

[…] For a similar question, check out the post ↴ Solve the System of Linear Equations and Give the Vector Form for the General Solution. […]

8. 07/27/2017

[…] Problem 2 and its solution: The vector form of the general solution of a system […]

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