Inverse of a Matrix Using the Cayley-Hamilton Theorem
Jeff Melmed
Eastern Maine Community College
jmelmed@emcc.edu
The determinant of an (n x n) matrix can be written in terms of its trace and the trace of powers of the matrix up to the nth.
n=2
u 0
Let A  

0
v


so that
u2
A  
0
2
0

v 2 
det A  uv
trA  u  v
trA2  u 2  v 2
(trA) 2  trA2  (u  v) 2  (u 2  v 2 )  2uv
Therefore det A 
1
[(trA) 2  trA2 ]
2
This result is generally true, even if the matrix is not diagonal:
1
1 i j
k
l
i
j
k
l
det A   ij kl Ai A j  ( k  l   l  k ) Ai A j
2
2
1
 (trA) 2  trA 2 
2
n=3
u2
u 0 0 



2
Let A   0 v 0  so that A   0
0
 0 0 w



0
v2
0
0
u3


0  , A3   0
0
w 2 

0
v3
0
0

0  and det A  uvw
w3 
trA  u  v  w
(trA) 2  (u  v  w) 2  u 2  v 2  w2  2(uv  uw  vw)
(trA)3  (u  v  w)3  u 3  v 3  w3  3(uv2  uw2  vu 2  vw2  wu 2  wv2 )  6uvw
trA2  u 2  v 2  w2
trA3  u 3  v 3  w3
(trA)(trA2 )  (u  v  w)(u 2  v 2  w2 )  u 3  v 3  w3  (uv2  uw2  vu 2  vw2  wu 2  wv2 )
(trA) 3  2(trA3 )  3(trA)(trA2 )  6uvw
1
6
Therefore det A  [(trA)  2(tr A )  3(trA)(trA )]
3
3
2
This result is generally true, even if the matrix is not diagonal:
1 ijk
l
m
n
  lmn Ai A j Ak
3!
1 i j k
i
j
k
i
j
k
i
j
k
i
j
k
i
j
k
l
m
n
 ( l  m  n   n  l  m   m  n  l   l  n  m   m  l  n   n  m  l ) Ai A j Ak
6
1
 (trA) 3  2(trA 3 )  3(trA)(trA 2 )
6
det A 
To obtain a formula for the inverse consider the Characteristic Equation for a matrix M (n=3)
Let A  M  I so that
A2  M 2  2M  2 I
A3  M 3  3M 2  32 M  3 I
trA  trM  3
trA2  trM 2  2 trM  32
trA3  trM 3  3 trM 2  32 trM  33
The Characteristic Equation is
1
0  det A  [( trM  3 ) 3  2(tr M 3  3 trM 2  32 trM  33 )  3(trM  3 )(trM 2  2 trM  32 )]
6

1
 [( trM ) 3  2(tr M 3 )  3(trM )(trM 2 )  3 ( trM 2  ( trM ) 2 )  62 trM  63 ]


6 
6 det M
Invoking the Cayley-Hamilton Theorem, replace the scalar with the matrix:   M
1
(det M ) I  [( trM ) 2  trM 2 ]M  (trM )M 2  M 3  0
2
Multiply through by M
M 1 
1
and solve:
1 1
2
2
2
[(
tr
M
)

tr
M
]
I

(
tr
M
)
M

M

det M  2