# Homework - ERO with Unknown in Matrix

I was having a problem with how to properly perform elementary row operation (ERO) on a matrix.

In the question, we were given an augmented matrix $\begin{bmatrix}1&1&-1&1\\2&3&a&3\\1&a&3&2\end{bmatrix}$ and require us to determine all possible values of $a$ such that the system is

1. consistent with infinitely many solution
2. consistent with one and only one solution
3. inconsistent.

So I decided to make the matrix in row echelon form. Here are my steps:

1) $R_{3} - R_{1} \rightarrow R_{3}$

$\begin{bmatrix}1&1&-1&1\\2&3&a&3\\0&a-1&4&1\end{bmatrix}$

2) $R_{2} - 2R_{1} \rightarrow R_{2}$

$\begin{bmatrix}1&1&-1&1\\0&1&a+2&1\\0&a-1&4&1\end{bmatrix}$

3) $R_{3} - (a-1)R_{2} \rightarrow R_{3}$

$\begin{bmatrix}1&1&-1&1\\0&1&a+2&1\\0&0&4-(a+2)(a-1)&1-(a-1)\end{bmatrix}$

After I further simplify the matrix in step 3, I got $\begin{bmatrix}1 & 1 & -1 & 1 \\ 0 & 1 & a+2 & 1 \\ 0 & 0& (3+a)(2-a) & 2-a \end{bmatrix}$

With the simplified matrix, I have attempted to perform another ERO, $\frac{R_{3}}{2-a} \rightarrow R_{3}$ so that $\begin{bmatrix}1 & 1 & -1 & 1 \\ 0 & 1 & a+2 & 1 \\ 0 & 0& (3+a) & 1 \end{bmatrix}$. After this ERO, there will not be such a value that the system is consistent with infinitely many solutions, which is not same as the suggested answer. It appear that the ERO $\frac{R_{3}}{2-a} \rightarrow R_{3}$ is illegal.

I tried to check for the definition of ERO again, the textbook state that ERO include

• interchange any 2 equations of a system of linear equation
• multiply both sides of any equation in a system by a non-zero scalar
• add a multiple of one equation to another equation within the system

Under this definition, it seems that my step 3 is also illegal as it include a unknown $a$ in the ERO. However, using the result of step 3, I can manage to get the correct answer.

So I want to know that is that we can only perform ERO with a constant value (e.g. $R_{3} - 6R_{1} \rightarrow R_{3}$) or we can perform ERO with variable (e.g. $R_{3} - (a-8)R_{2} \rightarrow R_{3}$) but with certain restriction?

Hint: consider what you have after your step 3 and the simplification, that is,

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

1. If $a=2$, then …

2. If $a=-3$, then …

3. If $a\ne2$ and $a\ne 3$ then …

### Explanation of the error

You can't do $\frac{R_{3}}{2-a} \rightarrow R_{3}$ unless you know that $a\ne 2$. So if you perform that operation, you must also consider the case $a=2$ separately.

• Just to clarify, is that ERO of multiplying a row with unknown(like in step 3) is okay, but it is not ok if we divide the row with unknown and don't consider the value (in this case $a=2$) of that unknown which cause the ERO ($\frac{R_{3}}{2−a} \rightarrow R_{3}$) undefined? – Edison Sep 28 '13 at 8:56
• @Edison Doing, in your notation, $R_{3} - (a-1)R_{2} \rightarrow R_{3}$ is always possible, because you can undo this operation with $R_{3} + (a-1)R_{2} \rightarrow R_{3}$ (the “new” $R_3$, of course). Multiplying a row by an expression involving the unknown must be done with care, checking separately the cases where the expression doesn't make sense or is $0$. – egreg Sep 28 '13 at 17:23