Understanding Engineering Mathematics

Problem 13.12

Find the determinant and adjoint of

A=

[− 123

011

− 102

]

and evaluateAAdjA

|A|=

∣

∣

∣

∣∣

− 123

011

− 102

∣

∣

∣

∣∣=−^1 (^2 )−^2 (^1 )+^3 (^1 )

=− 1

AdjA=

      

∣

∣∣

∣

11

02

∣

∣∣

∣ −

∣

∣∣

∣

01

− 12

∣

∣∣

∣

∣

∣∣

∣

01

− 10

∣

∣∣

∣

−

∣

∣

∣

∣

23

02

∣

∣

∣

∣

∣

∣

∣

∣

− 13

− 12

∣

∣

∣

∣ −

∣

∣

∣

∣

− 12

− 10

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

23

11

∣

∣

∣

∣ −

∣

∣

∣

∣

− 13

01

∣

∣

∣

∣

∣

∣

∣

∣

− 12

01

∣

∣

∣

∣

      

T

Note, for example, that the element−

∣

∣

∣

∣

− 12

− 10

∣

∣

∣

∣in the 2, 3 position of the matrix about to

be transposed is indeed the cofactor of the elementa 23 inAand check the other elements
similarly.

=

[ 2 − 11

− 41 − 2

− 11 − 1

]T

=

[ 2 − 4 − 1

− 111

1 − 2 − 1

]

We t h e n fi n d

AAdjA=

[− 123

011

− 102

][ 2 − 4 − 1

− 111

1 − 2 − 1

]

=

[− 100

0 − 10

00 − 1

]

=− 1

[ 100

010

001

]

=− 1 I

=|A|I

This is an example of the general result

A

AdjA

|A|

=I

which we will return to below.