In linear algebra we have been talking a lot about the three elementary row operations. What I don't understand is why we can't multiply by any column by a constant? Since a matrix is really just a grouping of column vectors, shouldn't we be able to multiply a whole column by a constant but maintain the same set of solutions for the original and resulting matrices?

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    $\begingroup$ A matrix isn't really just a grouping of column vectors. $\endgroup$ Commented Sep 14, 2011 at 1:09

8 Answers 8


Yes we can, but the question is what properties of the matrix we preserve when we do so. The row operations (which correspond to multiplication by an invertible matrix on the left) preserve the row space (the linear span of the rows) and the null space. The column operations preserve the column space (the linear span of the columns) and the null space of the transpose, but they don't preserve the row space or the null space.


Let's say we have some relation, like

$\left( \begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array} \right) \left( \begin{array}{ccc} x \\ y \\ z\end{array}\right) = \left( \begin{array}{ccc} 10 \\ 11 \\ 12\end{array} \right)$

This has a lot of information here. For instance, it says that $x + 2y + 3z = 10$. In fact, one can uniquely solve for $x, y, z$ from this relationship. But suppose we multiply the first column by 3:

$\left( \begin{array}{ccc} 3 & 2 & 3 \\ 12 & 5 & 6 \\ 21 & 8 & 9 \end{array} \right) \left( \begin{array}{ccc} x \\ y \\ z\end{array}\right) = \left( \begin{array}{ccc} 10 \\ 11 \\ 12\end{array} \right)$

Both of these cannot be correct (you can fully solve it out if you want, in fact, I encourage it). For instance, knowing that $x + 2y + 3z = 10$ and that $3x + 2y + 3z = 10$ tells us that $x = 0$. That's not so good - clearly, it is not always the case (and it's not here, either). Further, this would result in 3 non-dependent equations in 2 variables - no solution exists.

The main idea is that matrices hide linear equations, and the way in which they hide them is dependent on the laws of matrix multiplication. It's linear with respect to row operations, because that's the same as redundant operations. But it's not the same with respect to column operations. If we wanted to change rows, then we could use matrices... if we multiply on the right instead of the left. Again, it's an artifact of the way in which matrices multiply.

  • $\begingroup$ OK, I worked this out and see what you are saying. But it seems that there is still a valid solution for the second matrix, but you are saying this may not always be the case? $\endgroup$ Commented Sep 14, 2011 at 1:35
  • $\begingroup$ @weezy: No, it has solution (sometimes). But they're different. The good thing about row operations is that the solutions don't change. I mean, we care about how we change matrices, right? $\endgroup$
    – davidlowryduda
    Commented Sep 14, 2011 at 1:39
  • $\begingroup$ Ok, so the column operations will change the solution for a given b , but they will not change the possible b 's for which there are solutions? $\endgroup$ Commented Sep 14, 2011 at 1:43
  • $\begingroup$ @weezy: no, they might. It's as if you're coming up with a completely unrelated, new, matrix almost. So they're not really related. $\endgroup$
    – davidlowryduda
    Commented Sep 14, 2011 at 1:44
  • $\begingroup$ but I thought that asking if some vector b was a solution for a matrix consisting of the vectors ($v_1$, $v_2$...) was equivalent to asking if you could produce that vector through a multiplication of the vectors by come constants and then add them together to produce b $\endgroup$ Commented Sep 14, 2011 at 1:51

We can perform elementary column operations: if you multiply a matrix on the right by an elementary matrix, you perform an "elementary column operation".

However, elementary row operations are more useful when dealing with things like systems of linear equations, or finding inverses of matricces. And anything you want to do with columns you can do with rows... in the transpose.

You can develop the entire theory with elementary column operations instead; you just have to set up everything "the other way around". You would look at systems of linear equations as systems of the form $$(b_1,b_2,\ldots,b_m) = (x_1,\ldots,x_m)A$$ (that is, $\mathbf{b} = \mathbf{x}A$), with row vectors; the coefficients of the $m$th equation would become the $m$th column of the coefficient matrix $A$. And you would perform elementary column operations instead of elementary row operations.

When you perform elementary row operations with systems of the form $A\mathbf{x}=\mathbf{b}$, your operations do not respect the columnspace, but do respect the rowspace and the nullspace. If you perform elementary column operations to these systems, you respect the columnspace, but you do not respect the rowspace of the nullspace.

That said, matrices are not "really just a grouping of column vectors". They are a lot more.


I want to point out that you can solve a linear system with column operations. Column operations do change the solution... but the new solution is in terms of different variables! A column operation is a change of variables!

(This is unsurprising of course; row operations corresponds to multiplying by elementary matrices on the left, $\cdots E_2 E_1 A x = y$, ie. we're transforming the result of applying $A$, while column operations correspond to multiplying on the right, $A E_1 E_2\cdots x = y$, ie. we're transforming the variables before they get fed into $A$.)

If you have $\begin{align}2X+3Y&=1\\6X-4Y&=2\end{align}$

  • Multiplying the first column by $k$ corresponds to the substitution $X = kX'$.


  • Exchanging the columns corresponds to the substitution $X = Y', Y = X'$


  • Adding $k$ times the second column to the first corresponds to the substitution $Y = Y'+kX$

    $$\begin{align}2X+3(Y' + kX)&=(2+3k)X+3Y'&=1\\6X-4(Y' + kX)&=(6-4k)X-4Y'&=2\end{align}$$

So to solve this equation

  1. Multiply the first column by $1/2$: $X = X'/2$


  2. Add -3 times the first column to the second column: $X' = X''-3Y$

    $$\begin{align}X''& &=1\\3X''&-13Y&=2\end{align}$$

  3. Multiply the second column by $-1/13$: $Y=-Y'/13$

    $$\begin{align}X''& &=1\\3X''&+Y'&=2\end{align}$$

  4. Add -3 times the second column to the first column: $Y' = Y''-3X''$


Now as has been noted, $(1,2)$ is not a solution to the original equation. We need to run all our substitutions backwards to re-express it in terms of $X$ and $Y$.

4. $Y' = Y''-3X''=2-3=-1$
3. $Y = -Y'/13=1/13$
2. $X' = X''-3Y =1-3/13=10/13$
1. $X = X'/2 = (10/13)/2 = 5/13$

Giving the correct solution $X=5/13, Y=1/13$.

This also shows why people prefer row operations. You get the answer in terms of the original variables!

Also note that there was no reason to consider augmented matrices when working this way.


Doing something similar to what was mentioned in another answer, but this time in a situation where there is a unique solution rather than infinitely many solutions, first consider the following system: $$ \begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} 10 \\ 11 \end{bmatrix}$$ The solution is $x= -28/3$, $y= 29/3$.

If we multiply the elements in the first column by $3$, we get $$ \begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} 10 \\ 11 \end{bmatrix}$$ and now the solution is $x=-10/3$, $y=29/3$.

So the solution did not remain the same.

Now remember a couple of things: (1) You can multiply both sides of $u=v$ by the same number, and if what you had was true, then what you get is still true. That justifies row operations, but not column operations. (2) When you do a row operation, you multiply both sides on the left by the same matrix. Suppose we want to add $-4$ times the first row to the second in the first problem above. Then we get $$ \begin{bmatrix} 1 & 0 \\ -4 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -4 & 1 \end{bmatrix}\begin{bmatrix} 10 \\ 11 \end{bmatrix}. $$ We're multiplying a matrix on the left by something, and on the other side of the equation we're multiplying the SAME matrix on the left by the same thing. If we want to multiply on the right rather than on the left, we need a $1\times m$ matrix, for some $m$. Thus $$ \begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} x \\ y\end{bmatrix} \begin{bmatrix} \bullet & \bullet & \bullet & \cdots \end{bmatrix} = \begin{bmatrix} 10 \\ 11 \end{bmatrix}\begin{bmatrix} \bullet & \bullet & \bullet & \cdots \end{bmatrix}$$

This does not accomplish a column operation. That would be done like this: $$ \begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} \bullet & \bullet \\ \bullet & \bullet \end{bmatrix}$$ But you'd have to multiply by the same thing on the right side of the equation, and there's no way to do that.


One area where column and row ops can be used are when finding the determinant of a matrix. In this case, the determinant property is not affected by the choice of ops because we are not multiplying it by any specific vector.

Example: $$ det\left( \begin{array}{ccc} 1 & 0 & 0 & 3 \\ 2 & 7 & 0 & 6 \\ 0 & 6 & 3 & 0 \\ 7 & 3 & 1 & -5 \end{array} \right) $$

This one is a bit time consuming to solve by using cofactor expansion, but we can also use elementary row ops on this one.

You could take the time to work through it just like you were taught in Linear Algebra I or you could take another look at it. What happens if we multiply the first column by -3 and add it to the fourth column? We get:

$$ det\left( \begin{array}{ccc} 1 & 0 & 0 & 3-3 \\ 2 & 7 & 0 & 6-6 \\ 0 & 6 & 3 & 0 \\ 7 & 3 & 1 & -5-21 \end{array} \right) = det\left( \begin{array}{ccc} 1 & 0 & 0 & 0 \\ 2 & 7 & 0 & 0 \\ 0 & 6 & 3 & 0 \\ 7 & 3 & 1 & -26 \end{array} \right) = 1*7*3*-26 = -546 $$

Bingo! An upper triangular matrix. No correction is needed for this operation so the answer is the product of the diagonal entries: -546;

So to sum up, there are some times when you can use both column operations and row operations. It works as long as you don't need to preserve a multiplication relationship. If you start looking at both column and row operations, it will open up your mind to the power of the matrix. Just be VERY CAREFUL.

-- techdude


You might find Building generalized inverses of matrices using only row and column operations interesting. As others have mentioned, "column operations" are multiplying on the right by an elementary matrix. You can use them to push further than reduced row echelon form and, given $A \in \mathbb R^{m \times n}$, you can compute $P$ and $Q$ such that $$ PAQ = \left[ \begin{array}{c|c} I_r & \mathbf{0}_{r \times (n-r)} \\ \hline \mathbf{0}_{(m-r) \times r} & \mathbf{0}_{(m-r) \times (n-r)} \\ \end{array} \right] $$ where $r = \mathrm{rank}\left(A\right)$ and $P$ and $Q$ are invertible. You can get $P$ with just row operations but you need column operations to get $Q$.


i think that the reason which causes "elementary" column operations incorrect is because of our rules toward the form of linear equations. That is, we define linear equations to be written as what we get used to writing now! You can try to write down theses equations straight and apply column operations to them, and see what happens.


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