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5 A
−A
A . 15. The adjoint formula for matrix inverse.
∙ adj( ) = det( ) = adj( ) ∙ . The (1,1)entry in A adj(A) is equal to
−
+
which is the same as det(A). Similarly, we can see that the diagonal entries
of A adj(A) are all equal to det(A). On the other hand, the (2,1)entry in the
product A adj(A) is
−
+
−
.
By the Laplace expansion on the second row, this is equal to the following
3x3 determinant
.
The first and second rows are identical, and hence by (13C), this
determinant is equal to zero. By similarly calculations, we can show that for
i not equal to j, the (i,j)entry of A adj(A) is equal to a 3x3 determinant, in
which two rows are identical, and hence equal to zero. This proves that A
adj(A) = det(A) I3. From the property that det(M) = det(MT), we can prove
that adj(A) A is equal to det(A) I3.
If det(A) is nonzero, we can divide both sides of ∙ adj( ) = det( )
by det(A) and obtain the following adjoint formula for matrix inverse = 1
adj( )
det( ) This proves that reverse direction of (*) for 3x3 matrix.
(The Kreyszig ’s book, the above formula for matrix inverse is written in
terms of the cofactors, which is defined as (1)i+j Aij.)
16. For the forward direction of (*), we need to establish the multiplicative
property of the determinant function for 3x3 matrices.
Now, we consider the determinant as a function of the three columns. To
emphasize this view point, we let a1, a2, and a3 be the three columns of A,
and write the determinant as
6 .
det( ) = det[ is a linear function in each
From (13B), we know that det[
input column vector. For any real number p and q, we have
det[
= det[
det[ + + det[
, det[ + det[
,
= + det[
= + det[ + det[
. Let A and B be 3x3 matrices. The three columns in AB are respectively
=
+
+
,
=
+
+
,
=
+
+
,
where aj and bj represent respectively the jth row of A and B, for j=1,2,3. =
= det (
det [
det [ )
+ + + + + + Using the linear property (13B) of determinant, we expand det(AB) to
det[ The above calculation is essentially the same as the expansion
(
)(
)(
+
+
+
+
+
=∑ ∑ ∑
. + ) Among the twenty seven terms, lots of them are zero. Because if any two are identical, the determinant is zero. There
columns of det[
are six remaining terms, one for each permutation of 1, 2 and 3,+
det[
det[+
+
det[
det[+.
+
det[
det[
The determinant in each term is very “close” to the determinant of A,
except that the order of the columns is not right. For the second term, we
exchange column 2 and column 3, and by the exchange property in (13D),
7 = −.
det[
det[
After exchanging column 1 and column 2 in the third term, we obtain= −.
det[
det[
Repeat the same procedure for the last three terms, we finally get−
det( ) =
det[
det[+
−
det[
det[−.
+
det[
det[ is a common factor, we can factor it out, and
Because det[
what remains inside the parenthesis is precisely the determinant of B,
det ( ) = ( − = − +
( ) det ( )
det − +
) det ( ) This completes the proof of the multiplicative property of determinant for
3x3 matrix. Once we...
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This document was uploaded on 09/16/2013.
 Spring '13

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