# 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