Condition that Two Matrices are Row Equivalent

Linear algebra problems and solutions

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 $d-cb \neq 0$.
 
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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.


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