Find the vector form solution $\mathbf{x}$ of the equation $A\mathbf{x}=\mathbf{0}$, where $A=\begin{bmatrix}
1 & 1 & 1 & 1 &2 \\
1 & 2 & 4 & 0 & 5 \\
3 & 2 & 0 & 5 & 2 \\
\end{bmatrix}$. Also, find two linearly independent vectors $\mathbf{x}$ satisfying $A\mathbf{x}=\mathbf{0}$.

Solution.

Find the vector form solution $\mathbf{x}$ of the equation $A\mathbf{x}=\mathbf{0}$

We reduce the augmented matrix as follows:
\begin{align*}
[A\mid \mathbf{0}]= \left[\begin{array}{rrrrr|r}
1 & 1 & 1 & 1 &2 & 0 \\
1 & 2 & 4 & 0 & 5 & 0 \\
3 & 2 & 0 & 5 & 2 & 0 \\
\end{array} \right] \xrightarrow[R_3-3R_1]{R_2-R_1}
\left[\begin{array}{rrrrr|r}
1 & 1 & 1 & 1 &2 & 0 \\
0 & 1 & 3 & -1 & 3 & 0 \\
0 & -1 & -3 & 2 & -4 & 0 \\
\end{array} \right] \\[6pt] \xrightarrow[R_3+R_2]{R_1-R_2}
\left[\begin{array}{rrrrr|r}
1 & 0 & -2 & 2 &-1 & 0 \\
0 & 1 & 3 & -1 & 3 & 0 \\
0 & 0 & 0 & 1 & -1 & 0 \\
\end{array} \right] \xrightarrow[R_2+R_3]{R_1-2R_3}
\left[\begin{array}{rrrrr|r}
1 & 0 & -2 & 0 &1 & 0 \\
0 & 1 & 3 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 & -1 & 0 \\
\end{array} \right].
\end{align*}
Hence, the solution of the system is
\begin{align*}
x_1&=2x_3-x_5\\
x_2&=-3x_3-2x_5\\
x_4&=x_5,
\end{align*}
where $x_3, x_5$ are free variables.

The vector form of the general solution is
\begin{align*}
\mathbf{x}&=\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5
\end{bmatrix}=\begin{bmatrix}
2x_3-x_5 \\
-3x_3-2x_5 \\
x_3 \\
x_5 \\
x_5
\end{bmatrix}\\[6pt] &=x_3\begin{bmatrix}
2 \\
-3 \\
1 \\
0 \\
0
\end{bmatrix}+x_5\begin{bmatrix}
-1 \\
-2 \\
0 \\
1 \\
1
\end{bmatrix}.
\end{align*}

Find two linearly independent vectors $\mathbf{x}$ satisfying $A\mathbf{x}=\mathbf{0}$

For example, setting $x_3=1, x_5=0$, we see that $\begin{bmatrix}
2 \\
-3 \\
1 \\
0 \\
0
\end{bmatrix}$ is a solution. Similarly, setting $x_3=0, x_5=1$, we see that $\begin{bmatrix}
-1 \\
-2 \\
0 \\
1 \\
1
\end{bmatrix}$ is another solution.
It is straightforward to check that these two vectors are linearly independent.

Common Mistake

This is a midterm exam problem of Lienar Algebra at the Ohio State University.

For the second part, some students chose the zero vector.
But note that the zero vector and another nonzero vector are always linearly dependent.

The post Find the Vector Form Solution to the Matrix Equation $A\mathbf{x}=\mathbf{0}$ first appeared on Problems in Mathematics.

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

The post Solve the System of Linear Equations and Give the Vector Form for the General Solution first appeared on Problems in Mathematics.

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

(b) Let
\[A=\begin{bmatrix}
1 & 2 & 3 \\
4 &5 &6
\end{bmatrix}, B=\begin{bmatrix}
1 & 0 & 1 \\
0 &1 &0
\end{bmatrix}, C=\begin{bmatrix}
1 & 2\\
0& 6
\end{bmatrix}, \mathbf{v}=\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix}.\] Then compute and simplify the following expression.
\[\mathbf{v}^{\trans}\left( A^{\trans}-(A-B)^{\trans}\right)C.\]

Solution.

(a) Give the vector form for the general solution

The system of linear equation represented by the augmented matrix is
\begin{align*}
x_1-x_3-2x_5&=0\\
x_2+2x_3-x_5&=0\\
x_4+x_5=0.
\end{align*}
Solving the preceding system, we obtain
\begin{align*}
x_1&=x_3+2x_5\\
x_2&=-2x_3+x_5\\
x_4&=-x_5.
\end{align*}
Here $x_2, x_5$ are free variables and the rest are dependent variables.

Hence the general solution $\mathbf{x}$ can be expressed as
\begin{align*}
\mathbf{x}&=\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5
\end{bmatrix}=\begin{bmatrix}
x_3+2x_5 \\
-2x_3+x_5 \\
x_3 \\
-x_5 \\
x_5
\end{bmatrix}\\[10pt] &=\begin{bmatrix}
x_3 \\
-2x_3 \\
x_3 \\
0 \\
0
\end{bmatrix}+\begin{bmatrix}
2x_5 \\
x_5 \\
0 \\
-x_5 \\
x_5
\end{bmatrix}=x_3\begin{bmatrix}
1 \\
-2 \\
1 \\
0 \\
0
\end{bmatrix}+x_5\begin{bmatrix}
2 \\
1 \\
0 \\
-1\\
1
\end{bmatrix}.
\end{align*}
Therefore, the vector form for the general solution is
\[\mathbf{x}=x_3\begin{bmatrix}
1 \\
-2 \\
1 \\
0 \\
0
\end{bmatrix}+x_5\begin{bmatrix}
2 \\
1 \\
0 \\
-1\\
1
\end{bmatrix}.\]

(b) Matrix product and transpose

First of all, note that by the property of the transpose we have
\[(A-B)^{\trans}=A^{\trans}-B^{\trans}.\] Hence we can simply the middle part:
\begin{align*}
A^{\trans}-(A-B)^{\trans}&= A^{\trans}-(A^{\trans}-B^{\trans})\\
&=A^{\trans}-A^{\trans}+B^{\trans}=B^{\trans}.
\end{align*}
Thus the expression becomes
\[\mathbf{v}^{\trans}\left( A^{\trans}-(A-B)^{\trans}\right)C=\mathbf{v}^{\trans}B^{\trans}C.\] The transposes of $\mathbf{v}$ and $B$ are
\[\mathbf{v}^{\trans}=\begin{bmatrix}
0 & 1 & 0 \\
\end{bmatrix} \text{ and } B^{\trans}=\begin{bmatrix}
1 & 0 \\
0 & 1 \\
1 &0
\end{bmatrix}.\] Thus we have
\begin{align*}
&\mathbf{v}^{\trans}\left( A^{\trans}-(A-B)^{\trans}\right)C=\mathbf{v}^{\trans}B^{\trans}C\\
&=\begin{bmatrix}
0 & 1 & 0 \\
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\
0 & 1 \\
1 &0
\end{bmatrix}
\begin{bmatrix}
1 & 2\\
0& 6
\end{bmatrix}\\
&=\begin{bmatrix}
0 & 1
\end{bmatrix}\begin{bmatrix}
1 & 2\\
0& 6
\end{bmatrix}
=\begin{bmatrix}
0 & 6
\end{bmatrix}.
\end{align*}

In conclusion, we have obtained
\[\mathbf{v}^{\trans}\left( A^{\trans}-(A-B)^{\trans}\right)C=\begin{bmatrix}
0 & 6
\end{bmatrix}.\]

Comment.

These are Quiz 2 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.

• Quiz 1. Gauss-Jordan elimination / homogeneous system.
• Quiz 2. The vector form for the general solution / Transpose matrices.
• Quiz 3. Condition that vectors are linearly dependent/ orthogonal vectors are linearly independent
• Quiz 4. Inverse matrix/ Nonsingular matrix satisfying a relation
• Quiz 5. Example and non-example of subspaces in 3-dimensional space
• Quiz 6. Determine vectors in null space, range / Find a basis of null space
• Quiz 7. Find a basis of the range, rank, and nullity of a matrix
• Quiz 8. Determine subsets are subspaces: functions taking integer values / set of skew-symmetric matrices
• Quiz 9. Find a basis of the subspace spanned by four matrices
• Quiz 10. Find orthogonal basis / Find value of linear transformation
• Quiz 11. Find eigenvalues and eigenvectors/ Properties of determinants
• Quiz 12. Find eigenvalues and their algebraic and geometric multiplicities
• Quiz 13 (Part 1). Diagonalize a matrix.
• Quiz 13 (Part 2). Find eigenvalues and eigenvectors of a special matrix

The post Quiz 2. The Vector Form For the General Solution / Transpose Matrices. Math 2568 Spring 2017. first appeared on Problems in Mathematics.

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*}

Solution.

The augmented matrix of the given system is
\begin{align*}
\left[\begin{array}{rrrrr|r}
1 & 0 & -1 & 0 &-3 & 1 \\
3 & 1 & -1 & 1 & -9 & 3 \\
1 & 0 & -1 & 1 & -2 & 1 \\
\end{array} \right].
\end{align*}
We apply the elementary row operations as follows.
We have
\begin{align*}
\left[\begin{array}{rrrrr|r}
1 & 0 & -1 & 0 &-3 & 1 \\
3 & 1 & -1 & 1 & -9 & 3 \\
1 & 0 & -1 & 1 & -2 & 1 \\
\end{array} \right] \xrightarrow{\substack{R_2-3R_1\\R_3-R_1}}
\left[\begin{array}{rrrrr|r}
1 & 0 & -1 & 0 &-3 & 1 \\
0 & 1 & 2 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 1 & 0 \\
\end{array} \right]\\[10pt] \xrightarrow{R_2-R_3}
\left[\begin{array}{rrrrr|r}
1 & 0 & -1 & 0 &-3 & 1 \\
0 & 1 & 2 & 0 & -1 & 0 \\
0 & 0 & 0 & 1 & 1 & 0 \\
\end{array} \right].
\end{align*}
The last matrix is in reduced row echelon form.
From this reduction, we see that the general solution is
\begin{align*}
x_1&=x_3+3x_5+1\\
x_2&=-2x_3+x_5\\
x_4&=-x_5.
\end{align*}
Here $x_3, x_5$ are free (independent) variables and $x_1, x_2, x_4$ are dependent variables.

To find the vector form for the general solution, we substitute these equations into the vector $\mathbf{x}$ as follows.
We have
\begin{align*}
\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5
\end{bmatrix}&=
\begin{bmatrix}
x_3+3x_5+1 \\
-2x_3+x_5 \\
x_3 \\
-x_5 \\
x_5
\end{bmatrix}\\[10pt] &=
\begin{bmatrix}
x_3 \\
-2x_3 \\
x_3 \\
0 \\
0
\end{bmatrix}
+
\begin{bmatrix}
3x_5 \\
x_5 \\
0 \\
-x_5 \\
x_5
\end{bmatrix}
+
\begin{bmatrix}
1 \\
0 \\
0 \\
0 \\
0
\end{bmatrix}\\[10pt] &=x_3\begin{bmatrix}
1 \\
-2 \\
1 \\
0 \\
0
\end{bmatrix}+x_5 \begin{bmatrix}
3 \\
1 \\
0 \\
-1 \\
1
\end{bmatrix}+\begin{bmatrix}
1 \\
0 \\
0 \\
0 \\
0
\end{bmatrix}.
\end{align*}
Therefore the vector form for the general solution is given by
\[\mathbf{x}=x_3\begin{bmatrix}
1 \\
-2 \\
1 \\
0 \\
0
\end{bmatrix}+x_5 \begin{bmatrix}
3 \\
1 \\
0 \\
-1 \\
1
\end{bmatrix}+\begin{bmatrix}
1 \\
0 \\
0 \\
0 \\
0
\end{bmatrix},\] where $x_3, x_5$ are free variables.

Related Question.

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

The post Vector Form for the General Solution of a System of Linear Equations first appeared on Problems in Mathematics.