Classical Portfolio Construction 249
When we are finished, we will have obtained elementary transformation
number 2. Now the first column is already that of the identity matrix. Now we
proceed with this pattern, and in elementary transformation 3 we invoke row
operations rule 2 to convert the value in the second row, second column to a
- In elementary transformation number 4, we invoke row operations rule 3
to convert the remainder of the rows to zeros for the column corresponding
to the column we just invoked row operations rule 2 on.
We proceed likewise, converting the values along the diagonals to ones
per row operations rule 2, then converting the remaining values in that
column to zeros per row operations rule 3 until we have obtained the identity
matrix on the left. The right-hand side vector will then be our solution
set.
Starting Augmented Matrix
X 1 X 2 X 3 X 4 L 1 L 2 | Answer Explanation
0.095 0.13 0.21 0.085 0 0 | 0.14
1 11100| 1
0.1 −0.023 0.01 0 0.095 1 | 0
−0.023 0.25 0.079 0 0.13 1 | 0
0.01 0.079 0.4 0 0.21 1 | 0
0 0 0 0 0.085 1 | 0
Elementary Transformation Number 1
1 1.3684 2.2105 0.8947 0 0 | 1.47368 row 1∗(1/.095)
111100 | 1
0.1 −0.023 0.01 0 0.095 1 | 0
−0.023 0.25 0.079 0 0.13 1 | 0
0.01 0.079 0.4 0 0.21 1 | 0
0 0 0 0 0.085 1 | 0
Elementary Transformation Number 2
X 1 X 2 X 3 X 4 L 1 L 2 | Answer Explanation
1 1.3684 2.2105 0.8947 0 0 | 1.47368
0 −0.368 −1.210 0.1052 0 0 |−0.4736 row 2+(− 1 ∗row 1)
0 −0.160 −0.211 −0.089 0.095 1 |−0.1473 row 3+(−.1∗row 1)
0 0.2824 0.1313 0.0212 0.13 1 | .03492 row 4+(.0237∗row 1)
0 0.0653 0.3778−0.008 0.21 1 |−0.0147 row 5+(−.01∗row 1)
0 0 0 0 0.085 1 | 0