Linear Algebra

328 Chapter 4. Determinants
by expanding along the first row, as in Example 1.1.
�
�
|T | =1· (+1) �5
�8 � �
6�
�
�
9�
+2· (−1) �4
�7 � �
6�
�
�
9�
+3· (+1) �4
�7 �
5�
�
8�
= −3+12−9=0 Alternatively, we can expand down the second column.
�
�
|T | =2· (−1) �4
�7 � �
6�
�
�
9�
+5· (+1) �1
�7 � �
3�
�
�
9�
+8· (−1) �1
�4 �
3�
�
6�
=12−60 + 48 = 0
1.7 Example A row or column with many zeroes suggests a Laplace expansion.
������
1
2
3
5
1
−1
�
0�
�
� �
1�
� =0· (+1) �2
�
0�
3
� �
1 � �
�
−1�
+1· (−1) �1
�3 � �
5 � �
�
−1�
+0· (+1) �1
�2 �
5�
�
1�
=16
We finish by applying this result to derive a new formula for the inverse
of a matrix. With Theorem 1.5, the determinant of an n × n matrix T can
be calculated by taking linear combinations of entries from a row and their
associated cofactors.
ti,1 · Ti,1 + ti,2 · Ti,2 + ···+ ti,n · Ti,n = |T | (∗)
Recall that a matrix with two identical rows has a zero determinant. Thus, for
any matrix T , weighing the cofactors by entries from the “wrong” row — row k
with k �= i — gives zero
ti,1 · Tk,1 + ti,2 · Tk,2 + ···+ ti,n · Tk,n =0 (∗∗)
because it represents the expansion along the row k of a matrix with row i equal
to row k. This equation summarizes (∗) and (∗∗).
⎛
⎜
⎝
t1,1
t2,1
t1,2
t2,2
.
...
...
t1,n
t2,n
⎞ ⎛
T1,1
⎟ ⎜T1,2
⎟ ⎜
⎟ ⎜
⎠ ⎝
T2,1
T2,2
.
...
...
⎞
Tn,1
Tn,2 ⎟
⎠ =
⎛
|T |
⎜ 0
⎜
⎝
0
|T |
.
...
...
⎞
0
0 ⎟
⎠
0 0 ... |T |
tn,1 tn,2 ... tn,n
T1,n T2,n ... Tn,n
Note that the order of the subscripts in the matrix of cofactors is opposite to
the order of subscripts in the other matrix; e.g., along the first row of the matrix
of cofactors the subscripts are 1, 1 then 2, 1, etc.
1.8 Definition The matrix adjoint to the square matrix T is
⎛
T1,1
⎜T1,2
⎜
adj(T )= ⎜
⎝
T2,1
T2,2
.
...
...
⎞
Tn,1
Tn,2 ⎟
⎠
where Tj,i is the j, i cofactor.
T1,n T2,n ... Tn,n