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Title: Matrix inverse by cofactors and Cramer's rule
Description: Linear algebra course

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Matrix Inverse by Cofactors
and Cramer's Rule

5
...
Determinant calculations are generally much longer than
row reduction
...
However, it is useful in some theoretical applications
because it provides a formula for A-1 in terms of the entries of A
...

[False Expansion Theorem]

Theorem l

If A is an n

x

n matrix and i t= k, then

Proof: Let B be the matrix obtained from A by replacing (not swapping) the k-th row
of A by the i-th row of A
...
2
...
Since the cofactors c;j of B are equal to the cofactors

CkJ of A, and the coefficients bkJ of the k-th row of B are equal to the coefficients aiJ

of the i-th row of A, we get

as required
...
We define the cofactor matrix of A, denoted cof A, by

(cof A)iJ

=

CiJ

EXAMPLE 1
Let A=

2[0 4 -1 1
-2

1
...

5

3

6
Solution: The nine cofactors of A are

=Cl)1-� �I 17 12 I� �I=
1
1
2
C21 14_2 1= -18 C22= 1 1=
-11=7 C32=(-1)012 -11 =-2
=

ell

= (-1)

5

c

= c-1)
( 1)

C13=(1)I� _;1= -18
C23= I� -�I= 28
C33 (1)1� �I

6

5

6

1

(-1)

16

=

l

=6

Hence,

= [-1

EXERCISE 1

Calculate the cofactor matrix of A

�n

Observe that the cofactors of the i-th row of A form the i-th row of cofA, so they
form the i-th column of (cof Al
...
Moreover, by the False Expansion
Theorem, the dot product of the i-th row of A and the }-th column of cof A equals
if i * j
...

/

[c1

c,!]

-+T
a

i11 ·en

n

=

a11



C1

detA

0
=
0

where I is the identity matrix
...
en

detA

(detA)/

detA

(de;A) (cofAl= I, and, therefore,

A-1

=( � )
de A

ccofA)7

If detA

=

0, then, by Theorem 5
...
4, A is not invertible
...
(Some people refer to the trans­
pose of the cofactor matrix as the adjugate matrix and therefore call this the adjugate
method)
...


i

2

4

-1

0

3

1

0

-14

8

2(24 + 14)

=

=

76

Thus, A is invertible
...
This is manageable, but it is more work than would be required by the
row reduction method
...


Cramer's Rule
Consider the system of

n

linear equations in

n

variables, Ax

=

that A is invertible, then the solution may be written in the form

x

=

A

-

1

b

X1

X;

Xn

=

detA

1
=

detA

(cof A)

7

b

C11

C 21

C111

b1

C1;

C2;

c,li

b;

C111

C211

Cnn

bn

b
...
+bn C)
Ill
I

I

detA

·

multiplied by the dot product of the vector

b with the i-th row of (cof Af
...
So

x; is the dot product of the vector

b with the i-th column of cof A divided by detA
Title: Matrix inverse by cofactors and Cramer's rule
Description: Linear algebra course