252 THE HANDBOOK OF PORTFOLIO MATHEMATICS
Elementary Transformation Number 11
X 1 X 2 X 3 X 4 L 1 L 2 | Answer Explanation
1 0 0 0 0 4.9826 | 0.68283
0 1 0 0 0 1.7682 | 0.32622
00 1 0 0 −1.035 | 0.26795
0 0.0000 −0.000 1.0000 −0.000 −5.715 |−0.2769
0 0 0 0 1 5.8312 |−0.6655
00 0 0 0 1 | 0.11217 row 6∗(1/.50434)
Elementary Transformation Number 12
100000 | 0.12391 row 1+(−4.98265∗row 6)
010000 | 0.12787 row 2+(−1.76821∗row 6)
001000 | 0.38407 row 3+(1.0352∗row 6)
000100 | 0.36424 row 4+(5.7158∗row 6)
000010 |−1.3197 row 5+(−5.83123∗row 6)
000001 | 0.11217
Identity Matrix Obtained
100000 | 0.12391 =X 1
010000 | 0.12787 =X 2
001000 | 0.38407 =X 3
000100 | 0.36424 =X 4
000010 |−1.3197/.5 =−2.6394=L 1
000001 | 0.11217/.5=.22434=L 2
Interpreting the Results
Once we have obtained the identity matrix, we can interpret its meaning.
Here, given the inputs of expected returns and expected variance in returns
for all of the components under consideration, and given the linear corre-
lation coefficients of each possible pair of components, for an expected
yield of 14% this solution set is optimal.Optimal, as used here, means
that this solution set will yield the lowest variance for a 14% yield. In a
moment, we will determine the variance, but first we must interpret the
results.