Write the vector $\begin{bmatrix} 1 \\ 3 \\ -1 \end{bmatrix}$ as a linear combination of the vectors
\[\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} , \, \begin{bmatrix} 2 \\ -2 \\ 1 \end{bmatrix} , \, \begin{bmatrix} 2 \\ 0 \\ 4 \end{bmatrix}.\]

Solution.

We want to find real numbers $x_1 , x_2 , x_3$ which solve the equation
\[\begin{bmatrix} 1 \\ 3 \\ -1 \end{bmatrix} = x_1 \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + x_2 \begin{bmatrix} 2 \\ -2 \\ 1 \end{bmatrix} + x_3 \begin{bmatrix} 2 \\ 0 \\ 4 \end{bmatrix} = \begin{bmatrix} x_1 + 2 x_2 + 2 x_3 \\ – 2 x_2 \\ x_2 + 4 x_3 \end{bmatrix}.\]

Each row in this matrix now gives an equality, and so the three rows define a system of linear equations. To find a solution, we reduce its augmented matrix:
\begin{align*}
\left[\begin{array}{rrr|r} 1 & 2 & 2 & 1 \\ 0 & -2 & 0 & 3 \\ 0 & 1 & 4 & -1 \end{array} \right] \xrightarrow[R_2 + 2 R_3]{R_1 – 2 R_3} \left[\begin{array}{rrr|r} 1 & 0 & -6 & 3 \\ 0 & 0 & 8 & 1 \\ 0 & 1 & 4 & -1 \end{array} \right] \xrightarrow{ R_2 \leftrightarrow R_3 } \left[\begin{array}{rrr|r} 1 & 0 & -6 & 3 \\ 0 & 1 & 4 & -1 \\ 0 & 0 & 8 & 1 \end{array} \right] \\[6pt] \xrightarrow{ \frac{1}{8} R_3} \left[\begin{array}{rrr|r} 1 & 0 & -6 & 3 \\[3pt] 0 & 1 & 4 & -1 \\[3pt] 0 & 0 & 1 & \frac{1}{8} \end{array} \right] \xrightarrow[R_2 – 4 R_3 ]{ R_1 + 6 R_3} \left[\begin{array}{rrr|r} 1 & 0 & 0 & \frac{15}{4} \\[3pt] 0 & 1 & 0 & \frac{-3}{2} \\[3pt] 0 & 0 & 1 & \frac{1}{8} \end{array} \right].
\end{align*}

Now we can read off the solution $x_1 = \frac{15}{4}$, $x_2 = \frac{-3}{2}$, and $x_3 = \frac{1}{8}$.

The post Write a Vector as a Linear Combination of Three Vectors first appeared on Problems in Mathematics.

Determine whether the following augmented matrices are in reduced row echelon form, and calculate the solution sets of their associated systems of linear equations.

(a) $\left[\begin{array}{rrr|r} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 6 \end{array} \right]$.

(b) $\left[\begin{array}{rrr|r} 1 & 0 & 3 & -4 \\ 0 & 1 & 2 & 0 \end{array} \right]$.

(c) $\left[\begin{array}{rr|r} 1 & 2 & 0 \\ 1 & 1 & -1 \end{array} \right]$.
 

Solution.

(a) $\left[\begin{array}{rrr|r} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 6 \end{array} \right]$

The first matrix is in reduced row echelon form.

For this matrix, we can read off the solution $x_1 = 2, x_2 = -3, x_3 = 6$.

(b) $\left[\begin{array}{rrr|r} 1 & 0 & 3 & -4 \\ 0 & 1 & 2 & 0 \end{array} \right]$

The second matrix is in reduced row echelon form.

For this matrix, the variable $x_3$ is a free variable because there are no leading 1s in the 3rd column.

The solution set can be expressed as $x_1 = -4 – 3 x_3$, $x_2 = -2 x_3$, and $x_3$ can be any real number.

(c) $\left[\begin{array}{rr|r} 1 & 2 & 0 \\ 1 & 1 & -1 \end{array} \right]$

The third matrix is not in reduced echelon form because the bottom-left entry is $1$, not $0$, so we first use elementary row operations to put it in this form.
\begin{align*}
\left[\begin{array}{rr|r} 1 & 2 & 0 \\ 1 & 1 & -1 \end{array} \right] \xrightarrow{R_2 – R_1} \left[\begin{array}{rr|r} 1 & 2 & 0 \\ 0 & -1 & -1 \end{array} \right]\\[6pt] \xrightarrow{(-1) R_2 } \left[\begin{array}{rr|r} 1 & 2 & 0 \\ 0 & 1 & 1 \end{array} \right] \xrightarrow{R_1 – 2 R_2} \left[\begin{array}{rr|r} 1 & 0 & -2 \\ 0 & 1 & 1 \end{array} \right].
\end{align*}
With the reduced matrix, we can read off the solution $x_1 = -2$ and $x_2 = 1$.

The post Determine Whether Matrices are in Reduced Row Echelon Form, and Find Solutions of Systems first appeared on Problems in Mathematics.

Prove that if $A$ is an $n \times n$ matrix with rank $n$, then $\rref(A)$ is the identity matrix.

Here $\rref(A)$ is the matrix in reduced row echelon form that is row equivalent to the matrix $A$.
 

Proof.

Because $A$ has rank $n$, we know that the $n \times n$ matrix $\rref(A)$ has $n$ non-zero rows.
This means that all $n$ rows must have a leading 1.

Each leading 1 must be in a distinct column, so we must have that each of the $n$ columns has a leading 1.
In a row echelon matrix, these leading 1s must be arranged to lie on the diagonal.

In a reduced row echelon matrix, each column with a leading 1 has 0s above and below that 1.

These restrictions mean that $\rref(A)$ must be the identity matrix.

The post If a Matrix $A$ is Full Rank, then $\rref(A)$ is the Identity Matrix first appeared on Problems in Mathematics.

For each of the following matrices, find a row-equivalent matrix which is in reduced row echelon form. Then determine the rank of each matrix.

(a) $A = \begin{bmatrix} 1 & 3 \\ -2 & 2 \end{bmatrix}$.

(b) $B = \begin{bmatrix} 2 & 6 & -2 \\ 3 & -2 & 8 \end{bmatrix}$.

(c) $C = \begin{bmatrix} 2 & -2 & 4 \\ 4 & 1 & -2 \\ 6 & -1 & 2 \end{bmatrix}$.

(d) $D = \begin{bmatrix} -2 \\ 3 \\ 1 \end{bmatrix}$.

(e) $E = \begin{bmatrix} -2 & 3 & 1 \end{bmatrix}$.

Definition (Rank of a Matrix).

The rank of a matrix $A$ is the number of nonzero rows in the reduced row echelon form matrix $\rref(A)$ that is row equivalent to $A$.

Solution.

(a) $A = \begin{bmatrix} 1 & 3 \\ -2 & 2 \end{bmatrix}$

The matrix $A$ has rank 2, which can be seen by computing
\[ \begin{bmatrix} 1 & 3 \\ -2 & 2 \end{bmatrix} \xrightarrow{R_2 + 2 R_1} \begin{bmatrix} 1 & 3 \\ 0 & 8 \end{bmatrix} \xrightarrow{\frac{1}{8} R_2 } \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} \xrightarrow{R_1 – 3 R_2} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. \] Because the row-reduced matrix has two non-zero rows, the rank of $A$ is $2$.

(b) $B = \begin{bmatrix} 2 & 6 & -2 \\ 3 & -2 & 8 \end{bmatrix}$

The matrix $B$ has rank 2, which can be seen by computing
\begin{align*}
\begin{bmatrix} 2 & 6 & -2 \\ 3 & -2 & 8 \end{bmatrix} \xrightarrow{\frac{1}{2} R_1 } \begin{bmatrix} 1 & 3 & -1 \\ 3 & -2 & 8 \end{bmatrix} \xrightarrow{R_2 – 3 R_1} \begin{bmatrix} 1 & 3 & -1 \\ 0 & -11 & 11 \end{bmatrix} \\[6pt] \xrightarrow{\frac{-1}{11} R_2 } \begin{bmatrix} 1 & 3 & -1 \\ 0 & 1 & -1 \end{bmatrix} \xrightarrow{R_1 – 3 R_2} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & -1 \end{bmatrix}
\end{align*}

(c) $C = \begin{bmatrix} 2 & -2 & 4 \\ 4 & 1 & -2 \\ 6 & -1 & 2 \end{bmatrix}$

The matrix $C$ has rank 2, which can be seen by computing
\begin{align*}
\begin{bmatrix} 2 & -2 & 4 \\ 4 & 1 & -2 \\ 6 & -1 & 2 \end{bmatrix} \xrightarrow{\frac{1}{2} R_1 } \begin{bmatrix} 1 & -1 & 2 \\ 4 & 1 & -2 \\ 6 & -1 & 2 \end{bmatrix} \xrightarrow[R_3 – 6 R_1]{R_2 – 4 R_1} \begin{bmatrix} 1 & -1 & 2 \\ 0 & 5 & -10 \\ 0 & 5 & -10 \end{bmatrix} \\[6pt] \xrightarrow{R_3 – R_2} \begin{bmatrix} 1 & -1 & 2 \\ 0 & 5 & -10 \\ 0 & 0 & 0 \end{bmatrix} \xrightarrow{ \frac{1}{5} R_2 } \begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{bmatrix} \xrightarrow{R_1 + R_2} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{bmatrix} .
\end{align*}

(d) $D = \begin{bmatrix} -2 \\ 3 \\ 1 \end{bmatrix}$

The matrix $D$ has rank 1, which can be seen by calculating:
\[\begin{bmatrix} -2 \\ 3 \\ 1 \end{bmatrix} \xrightarrow{ \frac{-1}{2} R_1 } \begin{bmatrix} 1 \\ 3 \\1 \end{bmatrix} \xrightarrow{R_2 – 3 R_1 , R_3 – R_1} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}.\]

(e) $E = \begin{bmatrix} -2 & 3 & 1 \end{bmatrix}$

The matrix $E$ has rank 1, which can be seen by calculating:
\[\begin{bmatrix} -2 & 3 & 1 \end{bmatrix} \xrightarrow{ \frac{-1}{2} R_1 } \begin{bmatrix} 1 & \frac{-3}{2} & \frac{-1}{2} \end{bmatrix}.\] Click here if solved 134

The post Find a Row-Equivalent Matrix which is in Reduced Row Echelon Form and Determine the Rank first appeared on Problems in Mathematics.

For an $m\times n$ matrix $A$, we denote by $\mathrm{rref}(A)$ the matrix in reduced row echelon form that is row equivalent to $A$.
For example, consider the matrix $A=\begin{bmatrix}
1 & 1 & 1 \\
0 &2 &2
\end{bmatrix}$
Then we have
\[A=\begin{bmatrix}
1 & 1 & 1 \\
0 &2 &2
\end{bmatrix}
\xrightarrow{\frac{1}{2}R_2}
\begin{bmatrix}
1 & 1 & 1 \\
0 &1 & 1
\end{bmatrix}
\xrightarrow{R_1-R_2}
\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &1
\end{bmatrix}\] and the last matrix is in reduced row echelon form.
Hence $\mathrm{rref}(A)=\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &1
\end{bmatrix}$.

Find an example of matrices $A$ and $B$ such that
\[\mathrm{rref}(AB)\neq \mathrm{rref}(A) \mathrm{rref}(B).\]

Proof.

Let
\[A=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix} \text{ and } B=\begin{bmatrix}
0 & 0\\
1& 0
\end{bmatrix}.\] Then $A$ is already in reduced row echelon from.
So we have $\mathrm{rref}(A)=A$.

Applying the elementary row operations, we obtain
\[B=\begin{bmatrix}
0 & 0\\
1& 0
\end{bmatrix} \xrightarrow{R_1\leftrightarrow R_2}\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}.\] As the last matrix is in reduced row echelon from, we have $\mathrm{rref}(B)=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}$.
Therefore, we see that
\[\mathrm{rref}(A) \mathrm{rref}(B)=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}=\begin{bmatrix}
0 & 0\\
0& 0
\end{bmatrix}.\]

The product of $A$ and $B$ is
\[AB=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}
\begin{bmatrix}
0 & 0\\
1& 0
\end{bmatrix}=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}.\] It follows that $\mathrm{rref}(AB)=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}$.

In summary, we have
\[\mathrm{rref}(AB)=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix} \neq
\begin{bmatrix}
0 & 0\\
0& 0
\end{bmatrix}
=\mathrm{rref}(A) \mathrm{rref}(B),\] as required.

The post An Example of Matrices $A$, $B$ such that $\mathrm{rref}(AB)\neq \mathrm{rref}(A) \mathrm{rref}(B)$ first appeared on Problems in Mathematics.

Let $A=\begin{bmatrix}
1 & 0 & 1 \\
0 &1 &0
\end{bmatrix}$.

(a) Find an orthonormal basis of the null space of $A$.

(b) Find the rank of $A$.

(c) Find an orthonormal basis of the row space of $A$.

(The Ohio State University, Linear Algebra Exam Problem)
 

Solution.

First of all, note that $A$ is already in reduced row echelon form.

(a) Find an orthonormal basis of the null space of $A$.

Let us find a basis of null space of $A$.
The null space consists of the solutions of $A\mathbf{x}=0$.
Since $A$ is in reduced row echelon form, the solutions $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}$ satisfy
\[x_1=-x_3 \text{ and } x_2=0,\] hence the general solution is
\[\mathbf{x}=\begin{bmatrix}
-x_3 \\
0 \\
x_3
\end{bmatrix}=x_3\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}.\] Therefore, the set
\[\left\{\, \begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix} \,\right\}\] is a basis of the null space of $A$.
Since the length of the basis vector is $\sqrt{(-1)^2+0^2+1^2}=\sqrt{2}$, it is not orthonormal basis.
Thus, we divide the vector by its length and obtain an orthonormal basis
\[\left\{\, \frac{1}{\sqrt{2}}\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix} \,\right\}.\]  

(b) Find the rank of $A$.

From part (a), we see that the nullity of $A$ is $1$. The rank-nullity theorem says that
\[\text{rank of $A$} + \text{ nullity of $A$}=3.\] Hence the rank of $A$ is $2$.

The second way to find the rank of $A$ is to use the definition of the rank: The rank of a matrix $B$ is the number of nonzero rows in a reduced row echelon matrix that is row equivalent to $B$.
Since $A$ is in echelon form and it has two nonzero rows, the rank is $2$.

The third way to find the rank is to use the leading 1 method. By the leading 1 method, we see that the first two columns form a basis of the range, hence the rank of $A$ is $2$.
 

(c) Find an orthonormal basis of the row space of $A$.

By the row space method, the nonzero rows in reduced row echelon form a basis of the row space of $A$. Thus
\[ \left\{\, \begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}, \begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix} \,\right\}\] is a basis of the row space of $A$.
Since the dot (inner) product of these two vectors is $0$, they are orthogonal.
The length of the vectors is $\sqrt{2}$ and $1$, respectively.
Hence an orthonormal basis of the row space of $A$ is
\[ \left\{\, \frac{1}{\sqrt{2}} \begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}, \begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix} \,\right\}\]  

Linear Algebra Midterm Exam 2 Problems and Solutions

• True of False Problems and Solutions: True or False problems of vector spaces and linear transformations
• Problem 1 and its solution: See (7) in the post “10 examples of subsets that are not subspaces of vector spaces”
• Problem 2 and its solution: Determine whether trigonometry functions $\sin^2(x), \cos^2(x), 1$ are linearly independent or dependent
• Problem 3 and its solution (current problem): Orthonormal basis of null space and row space
• Problem 4 and its solution: Basis of span in vector space of polynomials of degree 2 or less
• Problem 5 and its solution: Determine value of linear transformation from $R^3$ to $R^2$
• Problem 6 and its solution: Rank and nullity of linear transformation from $R^3$ to $R^2$
• Problem 7 and its solution: Find matrix representation of linear transformation from $R^2$ to $R^2$
• Problem 8 and its solution: Hyperplane through origin is subspace of 4-dimensional vector space

The post Orthonormal Basis of Null Space and Row Space 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.

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

(a) Find a matrix $B$ in reduced row echelon form such that $B$ is row equivalent to the matrix $A$.

(b) Find a basis for the null space of $A$.

(c) Find a basis for the range of $A$ that consists of columns of $A$. For each columns, $A_j$ of $A$ that does not appear in the basis, express $A_j$ as a linear combination of the basis vectors.

(d) Exhibit a basis for the row space of $A$.

Hint.

In part (c), you may use the following theorem.

In part (d), you may use the following theorem.

Solution.

(a) Find a matrix $B$ in reduced roe echelon form such that $B$ is row equivalent to the matrix $A$.

We apply the elementary row operations to the matrix $A$ and obtain
\begin{align*}
A=\begin{bmatrix}
1 & 1 & 2 \\
2 &2 &4 \\
2 & 3 & 5
\end{bmatrix}
\xrightarrow{\substack{R_2-2R_1\\ R_3-2R_1}}
\begin{bmatrix}
1 & 1 & 2 \\
0 &0 &0 \\
0 & 1 & 1
\end{bmatrix}
\xrightarrow{R_2 \leftrightarrow R_3}\\
\begin{bmatrix}
1 & 1 & 2 \\
0 & 1 & 1\\
0 &0 &0
\end{bmatrix}
\xrightarrow{R_1-R_2}
\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 1\\
0 &0 &0
\end{bmatrix}.
\end{align*}
The last matrix is in reduced row echelon form that is row equivalent to $A$.
Thus we set
\[B=\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 1\\
0 &0 &0
\end{bmatrix}.\]

(b) Find a basis for the null space of $A$.

The null space $\calN(A)$ of the matrix is the set of solutions of the homogeneous system $A\mathbf{x}=\mathbf{0}$.
By part (a), the augmented matrix $[A\mid \mathbf{0}]$ is row equivalent to $[B \mid \mathbf{0}]$.
Thus, the solution $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}$ must satisfy
\[x_1=-x_3 \text{ and } x_2=-x_3,\] where $x_3$ is a free variable.
Thus the solutions are given by
\[\mathbf{x}=\begin{bmatrix}
-x_3 \\
-x_3 \\
x_3
\end{bmatrix}=x_3\begin{bmatrix}
-1 \\
-1 \\
1
\end{bmatrix}\] for any number $x_3$.
Hence we have
\begin{align*}
\calN(A)&=\{ \mathbf{x}\in \R^3 \mid \mathbf{x}=x_3\begin{bmatrix}
-1 \\
-1 \\
1
\end{bmatrix} \text{ for any } x_3\in \R \} \\
&=\Span\left\{ \quad\begin{bmatrix}
-1 \\
-1 \\
1
\end{bmatrix}\quad \right \}.
\end{align*}

From this, we deduce that the set
\[\left\{ \quad\begin{bmatrix}
-1 \\
-1 \\
1
\end{bmatrix}\quad \right \}\] is a basis for the null space of $A$.

(c) Find a basis for the range of $A$ that consists of columns of $A$. (Longer version)

Let us write
\[A=\begin{bmatrix}
1 & 1 & 2 \\
2 &2 &4 \\
2 & 3 & 5
\end{bmatrix}=[A_1, A_2, A_3],\] where $A_1, A_2, A_3$ are the column vectors of the matrix $A$.
The range $\calR(A)$ of $A$ is the same as the column space of $A$.
Thus, it suffices to find the maximal number of linearly independent vectors among the column vectors $A_1, A_2, A_3$.
Consider the linear combination
\[x_1A_1+x_2A_2+x_3A_3=\mathbf{0}. \tag{*} \] Then this is equivalent to the homogeneous system
\[A\mathbf{x}=\mathbf{0}\] and we already found the solutions in part (b):
\[x_1=-x_3 \text{ and } x_2=-x_3.\] This tells us the vectors $A_1, A_2, A_3$ are linearly dependent.
For example, $x_1=-1, x_2=-1, x_3=1$ is a nonzero solution of the system (*).
Thus, we have
\[\calR(A)=\Span(A_1, A_2, A_3)=\Span(A_1, A_2).\]

On the other hand, if we consider only $A_1$ and $A_2$, they are linearly independent.
(To see this, you just need to repeat the above argument without $A_3$. This amounts to just ignoring the third columns in the computations.)

Therefore, the set $\{A_1, A_2\}$ is a linearly independent spanning set for the range.
Hence a basis of the range consisting column vectors of $A$ is
\[\{A_1, A_2\}.\]

The only column vector which is not a basis vector is $A_3$.
We already found that $x_1=-1, x_2=-1, x_3=1$ is a nonzero solution of (*).
Thus we have
\[-A_1-A_2+A_3=\mathbf{0}.\] Solving this for $A_3$, we obtain the linear combination for $A_3$ of the basis vectors:
\[A_3=A_1+A_2.\]

(c) A shorter solution using the leading 1 method

Here is a shorter solution which uses the following theorem.

Theorem (leading-1 method): If $B$ is a matrix in reduced row echelon form that is row equivalent to $A$, then all the column vectors of $A$ whose corresponding columns in $B$ have leading 1’s form a basis of the range of $A$.

Looking at the matrix $B$, we see that the first and the second columns of $B$ have leading 1’s. Thus the first and the second column vectors of $A$ form a basis for the range of $A$.

(d) Exhibit a basis for the row space of $A$.

We use the following theorem.

Theorem (Row-space method): If $B$ is a matrix in (reduced) row echelon form that is row equivalent to $A$, then the nonzero rows of $B$ form a basis for the row space of $A$.

We have already found such $B$ in part (a), and the first and the second row vectors are nonzero. Thus they form a basis for the row space of $A$.
Hence a basis for the row space of $A$ is
\[\left\{\,\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}, \begin{bmatrix}
0 \\
1 \\
1
\end{bmatrix} \, \right \}.\]

Comment.

The longer solution of part (c) is essentially the proof of Theorem (leading 1 method).
It is good to know where the theorem came from, but when you solve a problem you may forget and just use the theorem like the shorter solution of (c).

The post Row Equivalent Matrix, Bases for the Null Space, Range, and Row Space of a Matrix first appeared on Problems in Mathematics.

Suppose that the following matrix $A$ is the augmented matrix for a system of linear equations.
\[A= \left[\begin{array}{rrr|r}
1 & 2 & 3 & 4 \\
2 &-1 & -2 & a^2 \\
-1 & -7 & -11 & a
\end{array} \right],\] where $a$ is a real number. Determine all the values of $a$ so that the corresponding system is consistent.

Solution.

We apply the elementary row operations to $A$ as follows.
\begin{align*}
A= \left[\begin{array}{rrr|r}
1 & 2 & 3 & 4 \\
2 &-1 & -2 & a^2 \\
-1 & -7 & -11 & a
\end{array} \right] \xrightarrow{\substack{R_2-2R_1\\ R_3+R_1}}
\left[\begin{array}{rrr|r}
1 & 2 & 3 & 4 \\
0 &-5 & -8 & a^2-8 \\
0 & -5 & -8 & a+4
\end{array} \right]\\
\xrightarrow{R_3-R_2}
\left[\begin{array}{rrr|r}
1 & 2 & 3 & 4 \\
0 &-5 & -8 & a^2-8 \\
0 & 0 & 0 & -a^2+a+12
\end{array} \right].
\end{align*}
The last matrix is in row echelon form. (It’s not in reduced row echelon form.)
Then the system is consistent if and only if $-a^2+a+12$ is zero. (See Remark below.)
Since we can factor
\[0=-a^2+a+12=-(a+3)(a-4),\] we see that the system is consistent if and only if $a=-3, 4$.

Remark

If $-a^2+a+12 \neq 0$, then we divide the third row by $-a^2+a+12$ and obtain
\[ \left[\begin{array}{rrr|r}
0 & 0 & 0 & 1
\end{array} \right]\] in the third row.
The corresponding linear equation is
\[0x_1+0x_2+0x_3=1,\] and clearly, there is no solution to this equation. Hence the system is inconsistent.

On the other hand, if $-a^2+a+12 = 0$, then the system has solutions. (You just need to reduce the above matrix further or use the back substitution.)

The post Find Values of $a$ so that Augmented Matrix Represents a Consistent System first appeared on Problems in Mathematics.