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

Ohio State University exam problems and solutions in mathematics

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.

  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

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    […] 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. […]

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