If $A, B$ have the same rank, can we conclude that they are row-equivalent?

If so, then prove it. If not, then provide a counterexample.

Solution.

Having the same rank does not mean they are row-equivalent.

For a simple counterexample, consider $A = \begin{bmatrix} 1 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 1 \end{bmatrix}$.

Both of these matrices have rank 1, but are not row-equivalent because they are already in reduced row echelon form.

Another solution.

The problem doesn’t specify the sizes of matrices $A$, $B$.

Note that if the sizes of $A$ and $B$ are distinct, then they can never be row-equivalent.
Keeping this in mind, let us consider the following two matrices.

\[A=\begin{bmatrix}
1 \\
0
\end{bmatrix} \text{ and } B=\begin{bmatrix}
1 & 0
\end{bmatrix}.\] Then both matrices are in reduced row echelon form and have rank $1$.
As noted above, they are not row-equivalent because the sizes are distinct.

The post If Two Matrices Have the Same Rank, Are They Row-Equivalent? first appeared on Problems in Mathematics.

For each of the following $3\times 3$ matrices $A$, determine whether $A$ is invertible and find the inverse $A^{-1}$ if exists by computing the augmented matrix $[A|I]$, where $I$ is the $3\times 3$ identity matrix.

(a) $A=\begin{bmatrix}
1 & 3 & -2 \\
2 &3 &0 \\
0 & 1 & -1
\end{bmatrix}$
 
(b) $A=\begin{bmatrix}
1 & 0 & 2 \\
-1 &-3 &2 \\
3 & 6 & -2
\end{bmatrix}$.

Elementary Row Operations and Inverse Matrices

Recall the following procedure of testing the invertibility of $A$ as well as finding the inverse matrix if exists.
If the augmented matrix $[A|I]$ is transformed into a matrix of the form $[I|B]$, then the matrix $A$ is invertible and the inverse matrix $A^{-1}$ is given by $B$.
If the reduced row echelon form matrix for $[A|I]$ is not of the form $[I|B]$, then the matrix $A$ is not invertible.

Solution.

(a) $A=\begin{bmatrix}
1 & 3 & -2 \\
2 &3 &0 \\
0 & 1 & -1
\end{bmatrix}$

We apply the elementary row operations as follows.
We have
\begin{align*}
&[A|I]= \left[\begin{array}{rrr|rrr}
1 & 3 & -2 & 1 &0 & 0 \\
2 & 3 & 0 & 0 & 1 & 0 \\
0 & 1 & -1 & 0 & 0 & 1 \\
\end{array} \right] \xrightarrow{R_2-2R_1}
\left[\begin{array}{rrr|rrr}
1 & 3 & -2 & 1 &0 & 0 \\
0 & -3 & 4 & -2 & 1 & 0 \\
0 & 1 & -1 & 0 & 0 & 1 \\
\end{array} \right]\\[6pt] &\xrightarrow{R_2\leftrightarrow R_3}
\left[\begin{array}{rrr|rrr}
1 & 3 & -2 & 1 &0 & 0 \\
0 & 1 & -1 & 0 & 0 & 1 \\
0 & -3 & 4 & -2 & 1 & 0 \\
\end{array} \right] \xrightarrow{\substack{R_1-3R_2\\ R_3+3R_2}}
\left[\begin{array}{rrr|rrr}
1 & 0& 1 & 1 &0 & -3 \\
0 & 1 & -1 & 0 & 0 & 1 \\
0 & 0 & 1 & -2 & 1 & 3 \\
\end{array} \right]\\[6pt] &\xrightarrow{\substack{R_1-R_3\\ R_2+R_3}}
\left[\begin{array}{rrr|rrr}
1 & 0& 0 & 3 & -1 & -6 \\
0 & 1 & 0 & -2 & 1 & 4 \\
0 & 0 & 1 & -2 & 1 & 3 \\
\end{array} \right].
\end{align*}

The left $3\times 3$ part of the last matrix is the identity matrix.
This implies that $A$ is invertible and the inverse matrix is given by the right $3\times 3$ matrix.
Hence
\[A^{-1}=\begin{bmatrix}
3 & -1 & -6 \\
-2 &1 &4 \\
-2 & 1 & 3
\end{bmatrix}.\]

(b) $A=\begin{bmatrix}
1 & 0 & 2 \\
-1 &-3 &2 \\
3 & 6 & -2
\end{bmatrix}$.

Now we consider the matrix $A=\begin{bmatrix}
1 & 0 & 2 \\
-1 &-3 &2 \\
3 & 6 & -2
\end{bmatrix}$.

Applying elementary row operations, we obtain
\begin{align*}
&[A|I]= \left[\begin{array}{rrr|rrr}
1 & 0 & 2 & 1 &0 & 0 \\
-1 & -3 & 2 & 0 & 1 & 0 \\
3 & 6 & -2 & 0 & 0 & 1 \\
\end{array} \right] \xrightarrow{\substack{R_2+R_1\\ R_3-3R_1}}
\left[\begin{array}{rrr|rrr}
1 & 0 & 2 & 1 &0 & 0 \\
0 & -3 & 4 & 1 & 1 & 0 \\
0 & 6 & -8 & -3 & 0 & 1 \\
\end{array} \right]\\[6pt] &\xrightarrow{R_3-2R_1}
\left[\begin{array}{rrr|rrr}
1 & 0 & 2 & 1 &0 & 0 \\
0 & -3 & 4 & 1 & 1 & 0 \\
0 & 0 & 0 & -5 & -2 & 1 \\
\end{array} \right] \xrightarrow{\frac{-1}{3}R_2}
\left[\begin{array}{rrr|rrr}
1 & 0 & 2 & 1 &0 & 0 \\
0 & 1 & -4/3 & -1/3 & -1/3 & 0 \\
0 & 0 & 0 & -5 & -2 & 1 \\
\end{array} \right].
\end{align*}

The last matrix is in reduced row echelon form but the left $3\times 3$ part is not the identity matrix $I$.
It follows that the matrix $A$ is not invertible.

Related Question.

This is a linear algebra exam problem at the Ohio State University.

The solution is given in the post↴
Find the Inverse Matrix of a $3\times 3$ Matrix if Exists

The post Find the Inverse Matrices if Matrices are Invertible by Elementary Row Operations 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.

Find the rank of the following real matrix.
\[ \begin{bmatrix}
a & 1 & 2 \\
1 &1 &1 \\
-1 & 1 & 1-a
\end{bmatrix},\] where $a$ is a real number.

 
(Kyoto University, Linear Algebra Exam)

Solution.

The rank is the number of nonzero rows of a (reduced) row echelon form matrix of the given matrix.

We apply elementary row operations as follows.
\begin{align*}
&\begin{bmatrix}
a & 1 & 2 \\
1 &1 &1 \\
-1 & 1 & 1-a
\end{bmatrix}
\xrightarrow{R_1 \leftrightarrow R_2}
\begin{bmatrix}
1 &1 &1 \\
a & 1 & 2 \\
-1 & 1 & 1-a
\end{bmatrix}
\xrightarrow[R_3+R_1]{R_2-aR_1}
\begin{bmatrix}
1 &1 &1 \\
0 & 1-a & 2-a \\
0 & 2 & 2-a
\end{bmatrix}\\[8pt] &\xrightarrow{R_2\leftrightarrow R_3}
\begin{bmatrix}
1 &1 &1 \\
0 & 2 & 2-a\\
0 & 1-a & 2-a
\end{bmatrix}
\xrightarrow{R_3-\frac{1-a}{2}R_2}
\begin{bmatrix}
1 &1 &1 \\
0 & 2 & 2-a\\
0 & 0 & (2-a) (a+1)/2
\end{bmatrix}.
\end{align*}

 
The last matrix is in row echelon form.
Therefore, if $a \neq -1, 2$, then $(3, 3)$-entry of the last matrix is not zero. From this we see that the rank is $3$ when $a \neq -1, 2$.

On the other hand, when $a=-1$ or $a=2$ the third row is a zero row, hence the rank is $2$.

The post Find the Rank of a Matrix with a Parameter first appeared on Problems in Mathematics.