# Condition that Two Matrices are Row Equivalent ## Problem 248

We say that two $m\times n$ matrices are row equivalent if one can be obtained from the other by a sequence of elementary row operations.

Let $A$ and $I$ be $2\times 2$ matrices defined as follows.
$A=\begin{bmatrix} 1 & b\\ c& d \end{bmatrix}, \qquad I=\begin{bmatrix} 1 & 0\\ 0& 1 \end{bmatrix}.$ Prove that the matrix $A$ is row equivalent to the matrix $I$ if $d-cb \neq 0$. Add to solve later

## Elementary row operations

We review the elementary row operations before the proof.
The three elementary row operations on a matrix are defined as follows.

• Interchanging two rows:
$R_i \leftrightarrow R_j$ interchanges rows $i$ and $j$.
• Multiplying a row by a non-zero scalar (a number):
$tR_i$ multiplies row $i$ by the non-zero scalar (number) $t$.
• Adding a multiple of one row to another row:
$R_j+tR_i$ adds $t$ times row $i$ to row $j$.

## Proof.

Suppose that $b-cd \neq 0$. Then we can obtain the matrix $I$ from the matrix $A$ by the following sequence of elementary row operations.
First, we apply $R_2-cR_1$ and get
\begin{align*}
A&=\begin{bmatrix}
1 & b\\
c& d
\end{bmatrix}
\xrightarrow{R_2-cR_1}
\begin{bmatrix}
1 & b\\
0& d-cb
\end{bmatrix}.
\end{align*}
The next step is $\frac{1}{d-cb}R_2$ and get
\begin{align*}
\begin{bmatrix}
1 & b\\
0& d-cb
\end{bmatrix}
\xrightarrow{\frac{1}{d-cb}R_2}\begin{bmatrix}
1 & b\\
0& 1
\end{bmatrix}.
\end{align*}
Note that this is where we need the assumption $d-cb \neq 0$ since $d-cb$ is in the denominator.
The last step is $R_1-bR_2$, and we obtain
\begin{align*}
\begin{bmatrix}
1 & b\\
0& 1
\end{bmatrix}
\xrightarrow{R_1-bR_2}
\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix}=I.
\end{align*}

In summary, we have the following sequence of elementary row operations from the matrix $A$ to the matrix I
$I$.
\begin{align*}
A &=\begin{bmatrix}
1 & b\\
c& d
\end{bmatrix}
\xrightarrow{R_2-cR_1}
\begin{bmatrix}
1 & b\\
0& d-cb
\end{bmatrix}\\[6pt] &\xrightarrow{\frac{1}{d-cb}R_2}\begin{bmatrix}
1 & b\\
0& 1
\end{bmatrix}
\xrightarrow{R_1-bR_2}
\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix}=I,
\end{align*}
and hence $A$ and $I$ are row equivalent. Add to solve later

### More from my site

#### You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

###### More in Linear Algebra ##### Determine Null Spaces of Two Matrices

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

Close