248 THE HANDBOOK OF PORTFOLIO MATHEMATICS
Three types of row operations can be performed:1.Any two rows may be interchanged.
2.Any row may be multiplied by any nonzero constant.
3.Any row may be multiplied by any nonzero constant and added to the
corresponding entries of any other row.
Using these three operations, we seek to transform the coefficients
matrix to an identity matrix, which we do in a very prescribed manner.
The first step, of course, is to simply start out by creating the augmented
matrix. Next, we perform the first elementary transformation by invoking
row operations rule 2. Here, we take the value in the first row, first column,
which is .095, and we want to convert it to the number 1. To do so, we
multiply each value in the first row by the constant 1/.095. Since any number
times 1 divided by that number yields 1, we have obtained a 1 in the first
row, first column. We have also multiplied every entry in the first row by
this constant, 1/.095, as specified by row operations rule 2. Thus, we have
obtained elementary transformation number 1.
Our next step is to invoke row operations rule 3 for all rows except the
one we have just used rule 2 on. Here, for each row, we take the value of that
row corresponding to the column we just invoked rule 2 on. In elementary
transformation number 2, for row 2, we will use the value of 1, since that is
the value of row 2, column 1, and we just performed rule 2 on column 1. We
now make this value negative (or positive if it is already negative). Since
our value is 1, we make it−1. We now multiply by the corresponding entry
(i.e., same column) of the row we just performed rule 2 on. Since we just
performed rule 2 on row 1, we will multiply this−1 by the value of row 1,
column 1, which is 1, thus obtaining−1. Now we add this value back to the
value of the cell we are working on, which is 1, and obtain 0.
Now on row 2, column 2, we take the value of that row corresponding to
the column we just invoked rule 2 on. Again we will use the value of 1, since
that is the value of row 2, column 1, and we just performed rule 2 on column
- We again make this value negative (or positive if it is already negative).
Since our value is 1, we make it−1. Now multiply by the corresponding
entry (i.e., same column) of the row we just performed rule 2 on. Since we
just performed rule 2 on row 1, we will multiply this−1 by the value of row 1,
column 2, which is 1.3684, thus obtaining−1.3684. Again, we add this value
back to the value of the cell we are working on, row 2, column 2, which is 1,
obtaining 1+(−1.3684)=−.3684. We proceed likewise for the value of every
cell in row 2, including the value of the right-hand side vector of row 2. Then
we do the same for all other rows until the column we are concerned with,
column 1 here, is all zeros. Notice that we need not invoke row operations
rule 3 for the last row, since that already has a value of zero for column 1.