Elementary Operations for Matrices
Elementary operations for matrices play a crucial role in finding the inverse
or solving linear systems. They may also be used for other calculations. On this
page, we will discuss these type of operations. Before we define an elementary
operation, recall that to an nxm matrix A, we can associate n rows and m
columns. For example, consider the matrix
Its rows are
Its columns are
Let us consider the matrix transpose of A
Its rows are
As we can see, the transpose of the columns of A are the rows of AT.
So the transpose operation interchanges the rows and the columns of a matrix.
Therefore many techniques which are developed for rows may be easily translated
to columns via the transpose operation. Thus, we will only discuss elementary
row operations, but the reader may easily adapt these to columns.
Elementary Row Operations.
- 1.
- Interchange two rows.
- 2.
- Multiply a row with a nonzero number.
- 3.
- Add a row to another one multiplied by a number.
Definition. Two matrices are row equivalent if and only if one
may be obtained from the other one via elementary row operations.
Example. Show that the two matrices
are row equivalent.
Answer. We start with A. If we keep the second row and add the
first to the second, we get
We keep the first row. Then we subtract the first row from the second one
multiplied by 3. We get
We keep the first row and subtract the first row from the second one. We get
which is the matrix B. Therefore A and B are row
equivalent.
One powerful use of elementary operations consists in finding solutions to
linear systems and the inverse of a matrix. This happens via Echelon Form
and Gauss-Jordan Elimination. In order to appreciate these two
techniques, we need to discuss when a matrix is row elementary equivalent to a
triangular matrix. Let us illustrate this with an example.
Example. Consider the matrix
First we will transform the first column via elementary row operations into one
with the top number equal to 1 and the bottom ones equal 0. Indeed, if we
interchange the first row with the last one, we get
Next, we keep the first and last rows. And we subtract the first one multiplied
by 2 from the second one. We get
We are almost there. Looking at this matrix, we see that we can still take care
of the 1 (from the last row) under the -2. Indeed, if we keep the first two rows
and add the second one to the last one multiplied by 2, we get
We can't do more. Indeed, we stop the process whenever we have a matrix which
satisfies the following conditions
- 1.
- any row consisting of zeros is below any row that contains at least one
nonzero number;
- 2.
- the first (from left to right) nonzero entry of any row is to the left
of the first nonzero entry of any lower row.
Now if we make sure that the first nonzero entry of every row is 1, we get a
matrix in row echelon form. For example, the matrix above is not in
echelon form. But if we divide the second row by -2, we get
This matrix is in echelon form.
An application of this, namely to solve linear systems via Gaussian
elimination may be found on another page.
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