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SEEMOUS
2023 SEEMOUS
P1
Strange matrix identity
Strange matrix identity
Source: 17th SEEMOUS 2023, Problem 1
March 9, 2023
linear algebra
matrix
determinant
Seemous
Problem Statement
Prove that if
A
A{}
A
and
B
B{}
B
are
n
×
n
n\times n
n
×
n
matrices with complex entries which satisfy
A
=
A
B
−
B
A
+
A
2
B
−
2
A
B
A
+
B
A
2
+
A
2
B
A
−
A
B
A
2
,
A=AB-BA+A^2B-2ABA+BA^2+A^2BA-ABA^2,
A
=
A
B
−
B
A
+
A
2
B
−
2
A
B
A
+
B
A
2
+
A
2
B
A
−
A
B
A
2
,
then
det
(
A
)
=
0
\det(A)=0
det
(
A
)
=
0
.
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