(a) Find a function
\[g(\theta) = a \cos(\theta) + b \cos(2 \theta) + c \cos(3 \theta)\] such that $g(0) = g(\pi/2) = g(\pi) = 0$, where $a, b, c$ are constants.

(b) Find real numbers $a, b, c$ such that the function
\[g(\theta) = a \cos(\theta) + b \cos(2 \theta) + c \cos(3 \theta)\] satisfies $g(0) = 3$, $g(\pi/2) = 1$, and $g(\pi) = -5$.

Solution.

(a) Condition: $g(0) = g(\pi/2) = g(\pi) = 0$

Each condition required on $g$ can be turned into an equation involving the constants $a, b, c$. In particular, we have the system of linear equations
\begin{align*}
g(0) &= a + b + c = 0 \\[6pt] g \left( \frac{\pi}{2} \right) &= -b = 0 \\[6pt] g(\pi) &= -a + b – c = 0.
\end{align*}

To solve this system, we will use Gauss-Jordan elimination to reduce its augmented matrix.
\begin{align*}
\left[\begin{array}{rrr|r} 1 & 1 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ -1 & 1 & -1 & 0 \end{array} \right] \xrightarrow[R_3 + R_2]{R_1 + R_2} \left[\begin{array}{rrr|r} 1 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ -1 & 0 & -1 & 0 \end{array} \right] \xrightarrow[R_3 + R_1]{(-1) R_2} \left[\begin{array}{rrr|r} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right].
\end{align*}

The solution can now be read off: $a+c = 0$ and $b=0$.
Thus a general solution is of the form $g(\theta) = a \cos(\theta) – a \cos ( 3 \theta) $, where $a$ is any real number.

(b) Condition: $g(0) = 3$, $g(\pi/2) = 1$, and $g(\pi) = -5$

Each condition required on $g$ can be turned into an equation involving the constants $a, b, c$. In particular, we have

\begin{align*}
g(0) &= a + b + c = 3\\
g(\pi/2) &= -b = 1\\
g(\pi) &= -a + b – c = -5.
\end{align*}

To solve this system, we will use Gauss-Jordan elimination to reduce its augmented matrix.
\begin{align*}
\left[\begin{array}{rrr|r} 1 & 1 & 1 & 3 \\ 0 & -1 & 0 & 1 \\ -1 & 1 & -1 & -5 \end{array} \right] \xrightarrow[R_3 + R_2]{R_1 + R_2} \left[\begin{array}{rrr|r} 1 & 0 & 1 & 4 \\ 0 & -1 & 0 & 1 \\ -1 & 0 & -1 & -4 \end{array} \right] \xrightarrow[R_3 + R_1]{(-1) R_2} \left[\begin{array}{rrr|r} 1 & 0 & 1 & 4 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 \end{array} \right] \end{align*}

The solution can now be read off: $a+c = 4$ and $b=-1$. Thus a general solution is of the form
$$ g(\theta) = a \cos(\theta) – \cos(2 \theta) + (4 – a) \cos(3 \theta) , $$
where $a$ is any real number.

The post Determine Trigonometric Functions with Given Conditions first appeared on Problems in Mathematics.

Let
\[\mathbf{v}_1=\begin{bmatrix}
1 \\
5 \\
-1
\end{bmatrix}, \mathbf{v}_2=\begin{bmatrix}
1 \\
4 \\
3
\end{bmatrix}, \mathbf{v}_3=\begin{bmatrix}
1 \\
2 \\
1
\end{bmatrix}, \mathbf{b}=\begin{bmatrix}
2 \\
13 \\
6
\end{bmatrix}.\] Express the vector $\mathbf{b}$ as a linear combination of the vector $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$.

(The Ohio State University, Linear Algebra Midterm Exam Problem)
 

Solution.

We want to find scalars $x_1, x_2, x_3$ such that
\[\mathbf{b}=x_1\mathbf{v}_1+x_2\mathbf{v}_2+x_3\mathbf{v}_3.\] This is equivalent to solving the matrix equation $A\mathbf{x}=\mathbf{b}$, where
\[A=[\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3]=\begin{bmatrix}
1 & 1 & 1 \\
5 &4 &2 \\
-1 & 3 & 1
\end{bmatrix} \text{ and } \mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}.\]

We solve this by Gauss-Jordan elimination.
The augmented matrix is
\[ \left[\begin{array}{rrr|r}
1 & 1 & 1 & 2 \\
5 &4 & 2 & 13 \\
-1 & 3 & 1 & 6
\end{array} \right].\] We apply elementary row operations as follows.
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 1 & 1 & 2 \\
5 &4 & 2 & 13 \\
-1 & 3 & 1 & 6
\end{array} \right] \xrightarrow{\substack{R_2-5R_1\\R_3+R_1}}
\left[\begin{array}{rrr|r}
1 & 1 & 1 & 2 \\
0 &-1 & -3 & 3 \\
0 & 4 & 2 & 8
\end{array} \right]\\[10pt] \xrightarrow{\substack{-R_2\\ \frac{1}{2}R_3}}
\left[\begin{array}{rrr|r}
1 & 1 & 1 & 2 \\
0 &1 & 3 & -3 \\
0 & 2 & 1 & 4
\end{array} \right] \xrightarrow{\substack{R_1-R_2\\ R_3-2R_2}}
\left[\begin{array}{rrr|r}
1 & 0 & -2 & 5 \\
0 &1 & 3 & -3 \\
0 & 0 & -5 & 10
\end{array} \right]\\[10pt] \xrightarrow{-\frac{1}{2}R_3}
\left[\begin{array}{rrr|r}
1 & 0 & -2 & 5 \\
0 &1 & 3 & -3 \\
0 & 0 & 1 & -2
\end{array} \right] \xrightarrow{\substack{R_1+2R_3 \\ R_2-3R_3}}
\left[\begin{array}{rrr|r}
1 & 0 & 0 & 1 \\
0 &1 & 0 & 3 \\
0 & 0 & 1 & -2
\end{array} \right].
\end{align*}
The last matrix is in reduced row echelon form.
Therefore, the solution is
\[x_1=1, x_2=3, x_3=-2.\] Thus the linear combination we are looking for is
\[\mathbf{b}=\mathbf{v}_1+3\mathbf{v}_2-2\mathbf{v}_3.\]

Comment.

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

Many of students started their solutions by writing the augmented matrix.
But it is good to start with the linear combination (because that’s what we need to find) and proceed to the augmented matrix to find the coefficients.

Double check

Once you obtained the linear combination
\[\mathbf{b}=\mathbf{v}_1+3\mathbf{v}_2-2\mathbf{v}_3,\] you should check whether this equality really holds.
Since vectors $\mathbf{b}, \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are given explicitly, checking this won’t take time.
If the equality doesn’t hold, then you must have made a computational error.

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 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 (The current page): 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 Express a Vector as a Linear Combination of Given Three Vectors 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.

Determine whether there exists a nonsingular matrix $A$ if
\[A^2=AB+2A,\] where $B$ is the following matrix.
If such a nonsingular matrix $A$ exists, find the inverse matrix $A^{-1}$.

(a) \[B=\begin{bmatrix}
-1 & 1 & -1 \\
0 &-1 &0 \\
1 & 2 & -2
\end{bmatrix}\]

(b) \[B=\begin{bmatrix}
-1 & 1 & -1 \\
0 &-1 &0 \\
2 & 1 & -4
\end{bmatrix}.\]

Solution.

Suppose that a nonsingular matrix $A$ satisfying $A^2=AB+2A$ exists.
Then $A$ is invertible since $A$ is nonsingular, and thus the inverse $A^{-1}$ exists.

Multiplying by $A^{-1}$ on the left, we have
\begin{align*}
A=&A^{-1}A^2=A^{-1}(AB+2A)\\
&=A^{-1}AB+2A^{-1}A\\
&=B+2I,
\end{align*}
where $I$ is the $3\times 3$ identity matrix.
Therefore, if such a nonsingular matrix exists, it must be
\[A=B+2I. \tag{*}\]

(a) The first case

Let us consider the case
\[B=\begin{bmatrix}
-1 & 1 & -1 \\
0 &-1 &0 \\
1 & 2 & -2
\end{bmatrix}.\] In this case, we have from (*)
\[A=B+2I=\begin{bmatrix}
1 & 1 & -1 \\
0 &1 &0 \\
1 & 2 & 0
\end{bmatrix}.\]

We still need to check that this matrix is in fact a nonsingular matrix.
To check the non-singularity and to find the inverse matrix at once, we consider the augmented matrix $[A\mid I]$ and apply elementary row operations.
We have
\begin{align*}
&[A\mid I] =
\left[\begin{array}{rrr|rrr}
1 & 1 & -1 & 1 &0 & 0 \\
0 & 1 & 0 & 0 & 1 & 0 \\
1 & 2 & 0 & 0 & 0 & 1 \\
\end{array} \right]\\[10pt] & \xrightarrow{R_3-R_1}
\left[\begin{array}{rrr|rrr}
1 & 1 & -1 & 1 &0 & 0 \\
0 & 1 & 0 & 0 & 1 & 0 \\
0 & 1 & 1 & -1 & 0 & 1 \\
\end{array} \right] \xrightarrow{\substack{R_1-R_2 \\ R_3-R_2}}
\left[\begin{array}{rrr|rrr}
1 & 0 & -1 & 1 & -1& 0 \\
0 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & -1 & -1 & 1 \\
\end{array} \right] \\[10pt] &\xrightarrow{R_1+R_3}
\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & 0 & -2& 1 \\
0 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & -1 & -1 & 1 \\
\end{array} \right].
\end{align*}

The left part of the last matrix is the identity matrix, and thus the matrix $A$ is invertible and the inverse matrix is the right half:
\[A^{-1}=\begin{bmatrix}
0 & -2 & 1 \\
0 &1 &0 \\
-1 & -1 & 1
\end{bmatrix}.\]

(b) The second case

Next let us consider the case
\[B=\begin{bmatrix}
-1 & 1 & -1 \\
0 &-1 &0 \\
2 & 1 & -4
\end{bmatrix}.\] By (*), if a nonsingular matrix $A$ exists, it must be
\[A=B+2I=\begin{bmatrix}
1 & 1 & -1 \\
0 &1 &0 \\
2 & 1 & -2
\end{bmatrix}.\]

We need to determine whether this matrix is actually nonsingular.
In fact, we prove that this matrix is singular.
That is, we show that $A\mathbf{x}=\mathbf{0}$ has a nonzero solution.
Consider the augmented matrix $[A\mid \mathbf{0}]$. By Gauss-Jordan elimination, we have
\begin{align*}
&[A\mid \mathbf{0}] = \left[\begin{array}{rrr|r}
1 & 1 & -1 & 0 \\
0 &1 & 0 & 0 \\
2 & 1 & -2 & 0
\end{array} \right] \xrightarrow{R_3-2R_1}\\[10pt] & \left[\begin{array}{rrr|r}
1 & 1 & -1 & 0 \\
0 &1 & 0 & 0 \\
0 & -1 & 0 & 0
\end{array} \right] \xrightarrow{\substack{R_1-R_2\\ R_3+R_2}}
\left[\begin{array}{rrr|r}
1 & 0& -1 & 0 \\
0 &1 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array} \right].
\end{align*}

The last matrix is in reduced row echelon form and it has a zero row. From this, we see that $x_3$ must be a free variable, and the matrix $A$ is singular.
(The general solution is $x_1=x_3, x_2=0$. Thus for example, $x_1=1, x_2=0, x_3=1$ is a nonzero solution of $A\mathbf{x}=\mathbf{0}$.)

Thus, we conclude that there is no nonsingular matrix $A$ satisfying $A^2=AB+2A$ in this case.

The post Find a Nonsingular Matrix Satisfying Some Relation first appeared on Problems in Mathematics.

Determine whether the following set of vectors is linearly independent or linearly dependent. If the set is linearly dependent, express one vector in the set as a linear combination of the others.
\[\left\{\, \begin{bmatrix}
1 \\
0 \\
-1 \\
0
\end{bmatrix}, \begin{bmatrix}
1 \\
2 \\
3 \\
4
\end{bmatrix}, \begin{bmatrix}
-1 \\
-2 \\
0 \\
1
\end{bmatrix},
\begin{bmatrix}
-2 \\
-2 \\
7 \\
11
\end{bmatrix}\, \right\}.\]

Solution.

Consider the linear combination
\[x_1\begin{bmatrix}
1 \\
0 \\
-1 \\
0
\end{bmatrix}+x_2 \begin{bmatrix}
1 \\
2 \\
3 \\
4
\end{bmatrix}+x_3\begin{bmatrix}
-1 \\
-2 \\
0 \\
1
\end{bmatrix}+x_4
\begin{bmatrix}
-2 \\
-2 \\
7 \\
11
\end{bmatrix}=\mathbf{0} \tag{*}\] with variables $x_1, x_2, x_3, x_4$.
We determine whether there is $(x_1, x_2, x_3, x_4)\neq (0,0,0,0)$ satisfying the linear combination (*).

The linear combination (*) is written as the matrix equation
\[\begin{bmatrix}
1 & 1 & -1 & -2 \\
0 &2 & -2 & -2 \\
-1 & 3 & 0 & 7 \\
0 & 4 & 1 & 11
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix}=\mathbf{0}.\]

To find the solutions of the equation, we apply the Gauss-Jordan elimination.
The augmented matrix $[A\mid \mathbf{0}]$ can be reduced by elementary row operations as follows.
\begin{align*}
[A\mid \mathbf{0}]= \left[\begin{array}{rrrr|r}
1 & 1 & -1 & -2 & 0\\
0 &2 & -2 & -2 & 0 \\
-1 & 3 & 0 & 7& 0 \\
0 & 4 & 1 & 11& 0 \\
\end{array} \right] \xrightarrow{\substack{R_3+R_1 \\ \frac{1}{2}R_2}}
\left[\begin{array}{rrrr|r}
1 & 1 & -1 & -2 & 0\\
0 & 1 & -1 & -1 & 0 \\
0 & 4 & -1 & 5& 0 \\
0 & 4 & 1 & 11& 0 \\
\end{array} \right]\\[10pt] \xrightarrow{\substack{R_1-R_2\\ R_3-4R_2\\ R_4-4R_2}}
\left[\begin{array}{rrrr|r}
1 & 0 & 0 & -1 & 0\\
0 & 1 & -1 & -1 & 0 \\
0 & 0 & 3 & 9& 0 \\
0 & 0 & 5 & 15 & 0 \\
\end{array} \right] \xrightarrow{\substack{\frac{1}{3}R_3\\ \frac{1}{5}R_4}}
\left[\begin{array}{rrrr|r}
1 & 0 & 0 & -1 & 0\\
0 & 1 & -1 & -1 & 0 \\
0 & 0 & 1 & 3 & 0 \\
0 & 0 & 1 & 3 & 0 \\
\end{array} \right]\\[10pt] \xrightarrow{\substack{R_2+R_3\\R_4-R_3}}
\left[\begin{array}{rrrr|r}
1 & 0 & 0 & -1 & 0\\
0 & 1 & 0 & 2 & 0 \\
0 & 0 & 1 & 3 & 0 \\
0 & 0 & 0 & 0 & 0 \\
\end{array} \right].
\end{align*}
Thus, the general solution is given by
\begin{align*}
x_1&=x_4\\
x_2&=-2x_4\\
x_3&=-3x_4,
\end{align*}
where $x_4$ is a free variable.

If we let $x_4=1$, then we have a nonzero solution $x_1=1, x_2=-2, x_3=-3, x_4=1$.
Thus the set is linearly dependent.

Substituting these values into (*), we have
\[\begin{bmatrix}
1 \\
0 \\
-1 \\
0
\end{bmatrix}-2 \begin{bmatrix}
1 \\
2 \\
3 \\
4
\end{bmatrix}-3\begin{bmatrix}
-1 \\
-2 \\
0 \\
1
\end{bmatrix}+
\begin{bmatrix}
-2 \\
-2 \\
7 \\
11
\end{bmatrix}=\mathbf{0}.\] Solving this for the last vector we obtain the linear combination
\[ \begin{bmatrix}
-2 \\
-2 \\
7 \\
11
\end{bmatrix}
=
-\begin{bmatrix}
1 \\
0 \\
-1 \\
0
\end{bmatrix}+2 \begin{bmatrix}
1 \\
2 \\
3 \\
4
\end{bmatrix}+3\begin{bmatrix}
-1 \\
-2 \\
0 \\
1
\end{bmatrix}.\] Click here if solved 100

The post Determine Linearly Independent or Linearly Dependent. Express as a Linear Combination first appeared on Problems in Mathematics.

Let
\[A=\begin{bmatrix}
1 & 3\\
2& 4
\end{bmatrix}.\] Then

(a) Find all matrices
\[B=\begin{bmatrix}
x & y\\
z& w
\end{bmatrix}\] such that $AB=BA$.

(b) Use the results of part (a) to exhibit $2\times 2$ matrices $B$ and $C$ such that
\[AB=BA \text{ and } AC \neq CA.\]

Solution.

Find all matrices $B$ such that $AB=BA$

Since we want to find $B$ such that $AB=BA$, we need to find $x, y, z, w$ satisfying
\[\begin{bmatrix}
1 & 3\\
2& 4
\end{bmatrix}\begin{bmatrix}
x & y\\
z& w
\end{bmatrix}=\begin{bmatrix}
x & y\\
z& w
\end{bmatrix}\begin{bmatrix}
1 & 3\\
2& 4
\end{bmatrix}.\]

Comparing entries, we have the following system of linear equations.
\begin{align*}
x+3z&=x+2y\\
y+3w&=3x+4y\\
2x+4z&=z+2w\\
2y+4w&=3z+4w.
\end{align*}

Simplifying this, we obtain
\begin{align*}
2y-3z &=0\\
3x+3y-3w &=0\\
2x+3z-2w &=0\\
2y-3z &=0.
\end{align*}

Note that the first and the last equations are identical. So we omit the first equation from our consideration.

To solve this system, we use the Gauss-Jordan elimination method. Namely we form the augmented matrix of this system and apply elementary row operations as follows.
\begin{align*}
\left[\begin{array}{rrrr|r}
3 & 3 & 0 & -3 &0 \\
2 & 0 & 3 & -2 & 0 \\
0 & 2 & -3 & 0 & 0
\end{array}\right] \xrightarrow{\frac{1}{3}R_1}
\left[\begin{array}{rrrr|r}
1 & 1 & 0 & -1 &0 \\
2 & 0 & 3 & -2 & 0 \\
0 & 2 & -3 & 0 & 0
\end{array}\right] \xrightarrow{R_2-2R_1}\\[10pt] \left[\begin{array}{rrrr|r}
1 & 1 & 0 & -1 &0 \\
0 & -2 & 3 & 0 & 0 \\
0 & 2 & -3 & 0 & 0
\end{array}\right] \xrightarrow{\substack{R_1+\frac{1}{2}R_2 \\ R_2+R_3}}
\left[\begin{array}{rrrr|r}
1 & 0 & 3/2 & -1 &0 \\
0 & -2 & 3 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{array}\right]\\[10pt] \xrightarrow{-\frac{1}{2}R_2}
\left[\begin{array}{rrrr|r}
1 & 0 & 3/2 & -1 &0 \\
0 & 1 & -3/2 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{array}\right].
\end{align*}

The last matrix is in reduced row echelon form. From this, we obtain the general solution
\begin{align*}
x&=-\frac{3}{2}z+w\\
y&=\frac{3}{2}z,
\end{align*}
where $z$ and $w$ are free variables.

Therefore the matrix $B$ such that $AB=BA$ must be
\begin{align*}
B&=\begin{bmatrix}
-\frac{3}{2}z+w & \frac{3}{2}z\\
z& w
\end{bmatrix}\\
&=z\begin{bmatrix}
-\frac{3}{2} & \frac{3}{2}\\
1& 0
\end{bmatrix}
+w
\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix}
\end{align*}
for any $z$ and $w$.

(b) Find matrices $B, C$ such that $AB=BA$, $AC\neq CA$

By part (a), if
\[B=z\begin{bmatrix}
-\frac{3}{2} & \frac{3}{2}\\
1& 0
\end{bmatrix}
+w
\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix}\] for some $z, w$, then we have $AB=BA$.

For example, let $z=2, w=0$. Then the matrix
\[B=\begin{bmatrix}
-3 & 3\\
2& 0
\end{bmatrix}\] satisfies $AB=BA$.

To find a matrix $C$ such that $AC \neq CA$, the matrix $C$ must not be of the form of the formula of $B$.

For example, let
\[C=\begin{bmatrix}
0 & 0\\
1& 0
\end{bmatrix}.\] You may directly check that $AC\neq CA$.

Or, we can show that $C$ is never be the matrix of the form
\[\begin{bmatrix}
-\frac{3}{2}z+w & \frac{3}{2}z\\
z& w
\end{bmatrix}.\]

To see this, compare $(1,2)$ and $(2,1)$ entries. There is no $z$ such that
\[\frac{3}{2}z=0 \text{ and } z=1.\] (This is how I found the matrix $C$ as above.)

The post Find All Matrices $B$ that Commutes With a Given Matrix $A$: $AB=BA$ 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.

(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 $x_1=1, x_2=5$.

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.

• 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 1. Gauss-Jordan Elimination / Homogeneous System. Math 2568 Spring 2017. first appeared on Problems in Mathematics.

Let
\[A=\begin{bmatrix}
1 & 2 & 2 \\
2 &3 &2 \\
-1 & -3 & -4
\end{bmatrix} \text{ and }
B=\begin{bmatrix}
1 & 2 & 2 \\
2 &3 &2 \\
5 & 3 & 3
\end{bmatrix}.\]

Determine the null spaces of matrices $A$ and $B$.

Proof.

The null space of the matrix $A$

We first determine the null space of the matrix $A$.
By definition, the null space is
\[\calN(A):=\{\mathbf{x}\in \R^3 \mid A\mathbf{x}=\mathbf{0}\},\] that is, the null space of $A$ consists of the solution $\mathbf{x}$ of the linear system $A\mathbf{x}=\mathbf{0}$.

To solve the system $A\mathbf{x}=\mathbf{0}$, we apply the Gauss-Jordan elimination. We reduce the augmented matrix $[A|\mathbf{0}]$ by elementary row operations as follows.
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 2 & 2 & 0 \\
2 &3 & 2 & 0 \\
-1 & -3 & -4 & 0
\end{array} \right] \xrightarrow{\substack{R_2-2R_1\\R_3+R_1}}
\left[\begin{array}{rrr|r}
1 & 2 & 2 & 0 \\
0 & -1 & -2 & 0 \\
0 & -1 & -2 & 0
\end{array} \right] \xrightarrow{\substack{R_1+2R_2\\R_3-5R_2}}\\
\left[\begin{array}{rrr|r}
1 & 0 & -2 & 0 \\
0 & -1 & -2 & 0 \\
0 & 0 & 0 & 0
\end{array} \right] \xrightarrow{-R_2}
\left[\begin{array}{rrr|r}
1 & 0 & -2 & 0 \\
0 & 1 & 2 & 0 \\
0 & 0 & 0 & 0
\end{array} \right].
\end{align*}
The last matrix is in reduced row echelon form, and the solutions $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}$ of $A\mathbf{x}=\mathbf{0}$ satisfy
\[x_1=2x_3 \text{ and } x_2=-2x_3\] and $x_3$ is a free variable.
Thus we have
\[\mathbf{x}=\begin{bmatrix}
2x_3 \\
-2x_3 \\
x_3
\end{bmatrix}=x_3\begin{bmatrix}
2 \\
-2 \\
1
\end{bmatrix}\] for any $x_3$ are solutions.

Therefore the null space of the matrix $A$ is
\begin{align*}
\calN(A)&=\left\{\mathbf{x}\in \R^3 \quad \middle| \quad \mathbf{x}=x_3\begin{bmatrix}
2 \\
-2 \\
1
\end{bmatrix} \text{ for any } x_3\in \R \right\}\\
&=\Span\left(\begin{bmatrix}
2 \\
-2 \\
1
\end{bmatrix}\right),
\end{align*}
the subspace of $\R^3$ spanned by the vector $\begin{bmatrix}
2 \\
-2 \\
1
\end{bmatrix}$.

The null space of the matrix $B$

The same procedure works for the matrix $B$. Thus we omit some detail below.
For the matrix $B$, the augmented matrix is reduced to
\[ \left[\begin{array}{rrr|r}
1 & 2 & 2 & 0 \\
2 &3 & 2 & 0 \\
5 & 3 & 3 & 0
\end{array} \right] \to \cdots \to \left[\begin{array}{rrr|r}
1 & 0 & 0 & 0 \\
0 &1 & 0 & 0 \\
0 & 0 & 1 & 0
\end{array} \right].\] (Check the computation by yourself.)
This implies that $\mathbf{x}=\mathbf{0}$ is the only solution of the system $B\mathbf{x}=\mathbf{0}$.
Therefore, the null space $\calN(B)$ of the matrix $B$ consists of just the zero vector:
\[\calN(B)=\left\{\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix} \right\}.\]

Comment.

The dimension of the null space $\calN(A)$ is $1$, and the dimension of the null space $\calN(B)$ is $0$.
In other words, the nullity of the matrix $A$ is $1$, and the nullity of the matrix $B$ is $0$.
(Recall that the nullity of a matrix is just the dimension of the null space of the matrix.)

Related Question.

The proof of the fact that a null space of a matrix is a subspace is give in the post The null space (the kernel) of a matrix is a subspace of $\R^n$.

The post Determine Null Spaces of Two Matrices first appeared on Problems in Mathematics.

Let $C$ be a $4 \times 4$ matrix with all eigenvalues $\lambda=2, -1$ and eigensapces
\[E_2=\Span\left \{\quad \begin{bmatrix}
1 \\
1 \\
1 \\
1
\end{bmatrix} \quad\right \} \text{ and } E_{-1}=\Span\left \{ \quad\begin{bmatrix}
1 \\
2 \\
1 \\
1
\end{bmatrix},\quad \begin{bmatrix}
1 \\
1 \\
1 \\
2
\end{bmatrix} \quad\right\}.\]

Calculate $C^4 \mathbf{u}$ for $\mathbf{u}=\begin{bmatrix}
6 \\
8 \\
6 \\
9
\end{bmatrix}$ if possible. Explain why if it is not possible!

(The Ohio State University Linear Algebra Exam Problem)
 

Solution.

Let
\[\mathbf{v}_1=\begin{bmatrix}
1 \\
1 \\
1 \\
1
\end{bmatrix}, \mathbf{v}_2=\begin{bmatrix}
1 \\
2 \\
1 \\
1
\end{bmatrix} \mathbf{v}_3=\begin{bmatrix}
1 \\
1 \\
1 \\
2
\end{bmatrix}.\] The vector $\mathbf{v}_1$ is an eigenvector corresponding to eigenvalue $\lambda=2$ and the vectors $\mathbf{v}_2, \mathbf{v}_2$ are eigenvectors corresponding to $\lambda=-1$.

If the vector $\mathbf{u}$ is a linear combination of $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$, then we can compute $C\mathbf{u}$ and hence $C^4\mathbf{u}$.
So let us first determine whether
\[x_1 \mathbf{v}_1+x_2\mathbf{v}_2+x_3 \mathbf{v}_3=\mathbf{u}\] has a solution or not.

We use Gauss-Jordan elimination method to find the solution.
The augmented matrix for this system is
\[\left[\begin{array}{rrr|rrr}
1 & 1 & 1 & 6 \\
1 &2 & 1 & 8 \\
1 & 1 & 1 & 6 \\
1 & 1 & 2 & 9
\end{array}\right].\] By elementary row operations, this matrix reduces to
\[\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & 1 \\
0 &1 & 0 & 2 \\
0 & 0 & 1 & 3 \\
0 & 0 & 0 & 0
\end{array}\right].\] Thus the solution is
\[x_1=1, x_2=2, x_3=3\] and we have the linear combination
\[ \mathbf{v}_1+2\mathbf{v}_2+3 \mathbf{v}_3=\mathbf{u}.\]

Now we have
\begin{align*}
C^4\mathbf{u}&= C^4\mathbf{v}_1+2C^4\mathbf{v}_2+3C^4 \mathbf{v}_3\\
&=2^4\mathbf{v}_1+2(-1)^4\mathbf{v}_2+3(-1)^4 \mathbf{v}_3\\
&=16\begin{bmatrix}
1 \\
1 \\
1 \\
1
\end{bmatrix}+2\begin{bmatrix}
1 \\
2 \\
1 \\
1
\end{bmatrix}+3\begin{bmatrix}
1 \\
1 \\
1 \\
2
\end{bmatrix}\\
&=\begin{bmatrix}
21 \\
23 \\
21 \\
24
\end{bmatrix}.
\end{align*}
Therefore we obtain
\[C^4=\begin{bmatrix}
21 \\
23 \\
21 \\
24
\end{bmatrix}.\] Click here if solved 8

The post Given All Eigenvalues and Eigenspaces, Compute a Matrix Product first appeared on Problems in Mathematics.