The theory of matrices and determinants 239
- Evaluate
∣ ∣ ∣ ∣ ∣ ∣
3 ∠ 60 ◦ j 21
0 ( 1 +j) 2 ∠ 30 ◦
02 j 5
∣ ∣ ∣ ∣ ∣ ∣ [
26. 94 ∠− 139. 52 ◦or
(− 20. 49 −j 17. 49 )
]
- Find the eigenvaluesλthat satisfy the follow-
ing equations:
(a)
∣
∣
∣
∣
( 2 −λ) 2
− 1 ( 5 −λ)
∣
∣
∣
∣=^0
(b)
∣ ∣ ∣ ∣ ∣ ∣
( 5 −λ) 7 − 5
0 ( 4 −λ) − 1
28 (− 3 −λ)
∣ ∣ ∣ ∣ ∣ ∣
= 0
(You may need to refer to chapter 1, pages
8–12, for the solution of cubic equations).
[(a)λ=3or4 (b)λ=1 or 2 or 3]
22.7 The inverse or reciprocal of a 3 by 3 matrix
Theadjointof a matrixAis obtained by:
(i) forming a matrixBof the cofactors ofA,and
(ii) transposingmatrixBto giveBT,whereBTis the
matrix obtained by writing the rows ofBas the
columns ofBT.ThenadjA=BT.
Theinverse of matrixA,A−^1 is given by
A−^1 =
adjA
|A|
where adjAis the adjoint of matrixAand|A|is the
determinant of matrixA.
Problem 17. Determine the inverse of the matrix
⎛
⎜
⎝
34 − 1
207
1 − 3 − 2
⎞
⎟
⎠
The inverse of matrixA,A−^1 =
adjA
|A|
The adjoint ofAis found by:
(i) obtaining the matrix of the cofactors of the ele-
ments, and
(ii) transposing this matrix.
The cofactor of element 3 is+
∣
∣
∣
∣
07
− 3 − 2
∣
∣
∣
∣=21.
The cofactor of element 4 is−
∣
∣
∣
∣
27
1 − 2
∣
∣
∣
∣=11, and so on.
The matrix of cofactors is
⎛
⎝
21 11 − 6
11 − 513
28 − 23 − 8
⎞
⎠
The transpose of the matrix of cofactors, i.e. the adjoint
ofthematrix,isobtainedbywritingtherowsascolumns,
and is
⎛
⎝
21 11 28
11 − 5 − 23
− 613 − 8
⎞
⎠
From Problem 14, the determinant of
∣ ∣ ∣ ∣ ∣ ∣
34 − 1
207
1 − 3 − 2
∣ ∣ ∣ ∣ ∣ ∣
is 113.
Hence the inverse of
⎛
⎝
34 − 1
207
1 − 3 − 2
⎞
⎠is
⎛
⎝
21 11 28
11 − 5 − 23
− 613 − 8
⎞
⎠
113
or
1
113
⎛
⎝
21 11 28
11 − 5 − 23
− 613 − 8
⎞
⎠
Problem 18. Find the inverse of
⎛
⎜
⎝
15 − 2
3 − 14
− 36 − 7
⎞
⎟
⎠
Inverse=
adjoint
determinant
The matrix of cofactors is
⎛
⎝
− 17 9 15
23 − 13 − 21
18 − 10 − 16
⎞
⎠
The transpose of the matrix of cofactors (i.e. the
adjoint) is
⎛
⎝
−17 23 18
9 − 13 − 10
15 − 21 − 16
⎞
⎠