Page 139 - Numerical Analysis Using MATLAB and Excel

The Adjoint of a Matrix

5v + 2v = 9 (4.36)
1
3
Next, we multiply the third equation of (4.35) by 3, and we add it with the second to eliminate
v 2 . Then, we obtain the following equation.

5v – 5v = 20 (4.37)
1
3
Subtraction of (4.37) from (4.36) yields

------
7v = – 11 or v = – 11 (4.38)
3
3
7
Now, we can find the unknown v 1 from either (4.36) or (4.37). By substitution of (4.38) into
(4.36) we obtain

5v + 2 ⋅ ⎛ ⎝ – 11 ⎞ ------ ⎠ = 9 or v = 17 (4.39)
------
1
1
7
7
Finally, we can find the last unknown v 2 from any of the three equations of (4.35). By substitu-
tion into the first equation we obtain

------ –
v = 2v + 3v – 5 = 34 33 35 – 34 (4.40)
------ =
------ –
------
2
1
3
7
7
7
7
These are the same values as those we found in Example 4.10.
The Gaussian elimination method works well if the coefficients of the unknowns are small inte-
gers, as in Example 4.11. However, it becomes impractical if the coefficients are large or fractional
numbers.
The Gaussian elimination is further discussed in Chapter 14 in conjunction with the LU factor-
ization method.

4.8 The Adjoint of a Matrix

Let us assume that is an square matrix and α ij is the cofactor of a ij . Then the adjoint of ,
A
A
n
n
denoted as adjA , is defined as the square matrix shown on the next page.

Numerical Analysis Using MATLAB® and Excel®, Third Edition 4−21
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