# Row/column operation on matrices and determinants

1. Feb 9, 2015

### Raghav Gupta

How we cannot apply row and column operation simultaneously on matrix when finding its inverse by elementary transformation but can apply it in determinant?
I think kernel and image gets disturbed in a matrix, though I don't know what it actually is.
Why not in determinant case?

2. Feb 10, 2015

### Svein

You can - but only when you do it on the right hand side simultaneously. Assume you want to find $B$ such that $A \cdot B = I$. You do the same row operations on $A$ and $I$. After a while (excluding singularities etc.) you end up with $U \cdot B = R$, where $U$ is upper triangular. Now substitute back until you end up with $I \cdot B = H$ and you are done.

3. Feb 10, 2015

### Raghav Gupta

But doesn't the kernel or image gets disturbed?

4. Feb 10, 2015

### Svein

I am not an expert on matrices so, I cannot answer that. What I know, is how to find the inverse (as detailed above). Useful facts:
• The matrix can be inverted if its determinant is different from 0.
• The determinant of the inverse matrix is the inverse of the determinant of the original matrix.

5. Feb 10, 2015

### Hawkeye18

Row operations are equivalent to left multiplication by corresponding elementary matrices, column operations are equivalent to right multiplication. So, when you perform row operations on a (square) matrix $A$, you get a matrix $E$ (the product of corresponding elementary matrices) such that $EA=I$. For square matrices that means that $E=A^{-1}$; performing the same row operations on the right hand side $I$ you get $EI=E=A^{-1}$.

Again, you can find the inverse performing only column operations: you get a matrix $E$ (the product of corresponding elementary matrices) such that $AE=I$, which again means that $E=A^{-1}$. Performing the same column operations on the right hand side $I$ you get there $IE=E=A^{-1}$.

If you perform both row and column operations you get two matrices $E_1$ and $E_2$ such that $E_1AE_2=I$, so in the right hand side you get $E_1IE_2=E_1E_2$. For $E_1E_2$ to be the inverse of $A$ you need $AE_1E_2=I$ (or equivalently $E_1E_2A =I$), and you have $E_1AE_2=I$. Matrix multiplication is not commutative, so there is no reason for the latter identity imply the former.

"No reason" is not a formal proof that the method does not work, for the formal proof you need a counterexample. You can find it just by playing with applying row and column operations to $2\times 2$ matrices. There are also more scientific method of constructing a counterexample.

The reason that applying both row and column operations is that while $E_1AE_2=I$ does not imply that $E_1E_2=A^{-1}$ it implies that $\operatorname{det} (E_1E_2)=\operatorname{det} A^{-1}$. Namely, $$1=\operatorname{det}(E_1AE_1) = \operatorname{det} E_1 \operatorname{det} A \operatorname{det}E_2,$$ so $$\operatorname{det}(E_1E_2) \operatorname{det}E_1\operatorname{det}E_2= (\operatorname{det}A)^{-1} =\operatorname{det}A^{-1}.$$

Finally, for invertible matrices the image is all $\mathbb R^n$ and the kernel is always trivial (i.e. $\{\mathbf 0\}$), they cannot be "disturbed" by row/column operations, they will remain the same.

6. Feb 10, 2015

### Staff: Mentor

If you're talking about a matrix you get by applying a row operation to another matrix, the answer is no. If you start with a matrix A, and apply one of the three row operations to it to get A1, the matrices A and A1 are equivalent. They have exactly the same kernel and image.
Edit: The image can change. See my later post in this thread.

If you're talking about applying column operations, I don't know -- I have never needed to apply column operations to reduce a matrix. However, if you swap the columns of a matrix, you are swapping the roles of the variables these columns represent.

Last edited: Feb 10, 2015
7. Feb 10, 2015

### Hawkeye18

That is not true. Row operations preserve kernel, column operations preserve image (column space).

In the case of invertible matrices, however, the image is always all $\mathbb R^n$ and the kernel is always $\{\mathbf 0\}$, so we can say that in this case the image and kernel are "preserved" user row and column operations.

8. Feb 10, 2015

### Hawkeye18

A bit of specification. Row operations preserve kernel (but generally not image), column operations preserve image (but generally not kernel).

9. Feb 10, 2015

### Staff: Mentor

I mispoke. Row operations don't necessarily preserve the range, as I said. A simple example shows this:
$$A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 1\end{bmatrix}$$
Using row reduction, we get an equivalent matrix.
$$B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0\end{bmatrix}$$
Although the dimensions of the column spaces of A and B are equal (2), they span different subspaces of R3.

The columns of A represent a plane in R3 that is perpendicular to <-1, -1, 1>. The columns of B represent a different plane in R3 that is perpendicular to the z-axis.

10. Feb 11, 2015

### Raghav Gupta

I see you have applied R3--------> R3 - R1
But there should be 1 in row 3 column 2 or more simply B32 should be equal to 1?

11. Feb 11, 2015

### Staff: Mentor

No. Here's what I did: -R1 + R3 --> R3 and then -R2 + R3 --> R3.

IOW, add -R1 to R3, and then add -R2 to R3.

12. Feb 11, 2015

### Raghav Gupta

Got it. Thanks to all of you - HawkEye 18, Mark44 and Svein.