# 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)

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## 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.

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

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