3.3 Row-operation matrices

Row operations first appeared as procedural moves on a matrix. We swapped rows, scaled a row by a nonzero number, and added a multiple of one row to another. Those moves are already legitimate for solving systems, because they preserve solution sets.

There is a second, more structural way to read the same moves: each elementary row operation is the same as left-multiplication by a special square matrix. This matters because it converts a sequence of row operations into an ordinary matrix equation. Later, this is exactly the bridge between row reduction, invertibility, rank, determinants, and basis arguments.

The idea: do the row operation to the identity first

Suppose a row operation $\rho$ is meant to act on matrices with p rows. Start with the identity matrix $I_p$, apply the same row operation to $I_p$, and call the result $E_\rho$.

The reason this definition is useful is the following theorem.

The multiplication must be on the left. Row operations change rows by mixing rows. Left multiplication forms new rows of $A$ as linear combinations of old rows of $A$. Right multiplication would instead mix columns.

Three basic examples

The three allowed row operations give three corresponding types of row-operation matrices.

A sequence of row operations becomes one matrix product

The real payoff is not just representing one row operation. A whole sequence of row operations becomes one product of row-operation matrices.

The order in this formula is important. The first row operation appears closest to $A_1$, because it is applied first:

$$A_2 = E_{\rho_1}A_1, \qquad A_3 = E_{\rho_2}A_2 = E_{\rho_2}E_{\rho_1}A_1.$$

Reading a longer row-operation product

In assignment-style questions, the row operations are often given as a long chain. The goal is not to multiply many matrices blindly. The goal is to keep three objects separate:

• the matrices $A_1,A_2,\ldots,A_{k+1}$ being transformed;
• the row-operation matrices $H_1,H_2,\ldots,H_k$;
• the single combined left multiplier $J=H_k\cdots H_2H_1$.

Here is a typical four-row example. Suppose

$$\begin{aligned} \rho_1 &: R_2 \leftarrow -\frac12 R_2,\\ \rho_2 &: R_1 \leftrightarrow R_2,\\ \rho_3 &: R_3 \leftarrow R_3-2R_1,\\ \rho_4 &: R_4 \leftarrow R_4-2R_3,\\ \rho_5 &: R_1 \leftarrow R_1-R_3,\\ \rho_6 &: R_1 \leftarrow R_1-R_2. \end{aligned}$$

If

$$A_1 \xrightarrow{\rho_1} A_2 \xrightarrow{\rho_2} A_3 \xrightarrow{\rho_3} A_4 \xrightarrow{\rho_4} A_5 \xrightarrow{\rho_5} A_6 \xrightarrow{\rho_6} A_7,$$

then

$$A_7 = H_6H_5H_4H_3H_2H_1A_1.$$

The corresponding row-operation matrices are

$$\begin{aligned} H_1&= \begin{bmatrix} 1&0&0&0\\ 0&-\frac12&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{bmatrix}, & H_2&= \begin{bmatrix} 0&1&0&0\\ 1&0&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{bmatrix},\\[0.8em] H_3&= \begin{bmatrix} 1&0&0&0\\ 0&1&0&0\\ -2&0&1&0\\ 0&0&0&1 \end{bmatrix}, & H_4&= \begin{bmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&-2&1 \end{bmatrix},\\[0.8em] H_5&= \begin{bmatrix} 1&0&-1&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{bmatrix}, & H_6&= \begin{bmatrix} 1&-1&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{bmatrix}. \end{aligned}$$

Their product is

$$J=H_6H_5H_4H_3H_2H_1= \begin{bmatrix} -1&-\frac32&-1&0\\ 1&0&0&0\\ 0&1&1&0\\ 0&-2&-2&1 \end{bmatrix}.$$

The efficient way to obtain $J$ is to apply the six operations to $I_4$, not to expand all six factors by hand. The matrix $J$ records the total effect of the chain on rows:

$$A_7=JA_1.$$

This equation is also a good check on the order. If the first operation were placed on the far left, the product would describe a different chain.

Reverse row operations and inverses

Every elementary row operation has a reverse operation:

• the reverse of $R_j \leftarrow R_j + cR_i$ is $R_j \leftarrow R_j - cR_i$;
• the reverse of $R_i \leftarrow cR_i$, with $c \ne 0$, is R_i \leftarrow (1/c)R_i;
• the reverse of a row swap is the same row swap.

This gives a clean matrix statement.

For example, if

$$E = \begin{bmatrix} 1 & 0 & 0\\ 3 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$$

performs $R_2 \leftarrow R_2 + 3R_1$, then

$$E^{-1} = \begin{bmatrix} 1 & 0 & 0\\ -3 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$$

performs $R_2 \leftarrow R_2 - 3R_1$.

This is the algebraic reason row operations are reversible, and it explains why row reduction is so closely connected to invertible matrices.

Why this viewpoint matters later

If $B$ is row-equivalent to $A$, then there is a finite sequence of row operations taking $A$ to $B$. Therefore there is a product of row-operation matrices $E$ such that

$$B = EA.$$

Because each row-operation matrix is invertible, the product $E$ is invertible. So row equivalence can be discussed either procedurally, by listing row operations, or algebraically, by writing an equation with an invertible matrix on the left.

This is useful in several later arguments:

• a square matrix row-equivalent to $I_n$ is a product of row-operation matrices;
• row operations preserve homogeneous solution information because they amount to multiplying by invertible matrices;
• determinant rules for row operations can be expressed through elementary matrices;
• rank and basis arguments can use row reduction without pretending the original columns themselves have not changed.

Common mistakes

Quick checks

Exercises

Related notes

This page builds on 2.2 Augmented matrices and row operations and 3.1 Matrix multiplication and identity matrices. It prepares the algebraic row-reduction viewpoint used in 5.1 Invertible matrices and 7.2 Row operations, products, and invertibility.