Download A Method for finding a Square Root of a 2x2 Matrix

A Method for finding a Square Root of a 2x2 Matrix
By: P. C. Somayya.
[email protected]
a11
a12
a21
a22
Let A =
and B = ± √ A.
a11+ T
a12
then B = ± (1/R)
a21
Where
a22 + T
T= ±√|A|
R = a11 + a12 + 2 T
Ref: Paper entitled “ Root of a 2x2 Matrix” Published in The Mathematics Education Vol..
XXXI. No. 1, March 1997. Siwan, Bihar State. INDIA, is given below :
The Mathematics Education
Vol. XXXI, No. 1, March 1997.
Root of a 2 x 2 Matrix
By P. C. Somayya, Vice Principal, S.B.E.S College of Science, Aurangabad 431 001, M.S. India.
[ Received March 28, 1995 ]
Summary: A Method for finding root of a 2x2 matrix is proposed in this paper.
Method : Suppose
A11
A12
A=
A21
A22
And
B = A1/2
Case 1 :
(2)
If A12 = A21 = 0
±√A11
Obviously we get,
0
B=
(3)
0
Case 2:
(1)
±√A22
If A12=0 & A21 ≠ 0
It can be easily verified that
±√A11
,
0
B=
(4)
± A21 /(√A11 + √A22)
±√A22
Similarly, we get,
±√A11
,
± A12/(√A11 + √A22)
B=
(5)
0
±√A22
If A12 ≠ 0 and A21 = 0.
Case 3: If A12 ≠ 0 and A21 ≠ 0
then
,
(A11 + T)
B=
±(1/R)
(6)
A21
Where
A12
(A22 + T)
T = ± | A | ½ = ± √ A11 A22 - A12 A21
{ 53 }
(7)
{ 54 }
And R2 = A11 + A22 + 2 T , R ≠ 0
Proof : B * B = ( 1/ R2 )
(A11 + T)
A12
A21
= ( 1/ R2 )
(8)
(A11 + T)
(A22 + T)
A21
A12(A11+T)+A12(A22+T)
A21(A11+T)+A21(A22+T)
(A22+T)2 + A12 A21
A12 (A11+A22+ 2T)
(A22)2+2A22T +(T2+ A12 A21)
A21(A11+A22+2T)
(A11) 2+ 2 A11 T + A11 A22
A12 R2
A21R2
= ( 1/ R2 )
(A22 + T)
(A11 + T)2 +A12 A21
= ( 1/ R2 ) (A11) 2+ 2 A11 T + (T2+A12 A21)
= ( 1/ R2 )
A12
(A22)2+2A22T +A11 A22
(A11) R2
A12 R2
A21 R2
(A22) R2
= A.
Example: (1)
1
3
5
6
A=
,
|A| = -9 , T= ± 3i
(1± 3i )
and R2 = A11+A22+2T= 1+6±6i = 7±6i
3
B= [± 1/ √(7±6i)]
5
(2)
11
-3
2
6.
(6± 3i ).
A =
|A| = 72 and T= ±6√2 , R2 = 17 ±12 √2
11±6√2
B
-3
= [± 1/ √(17±12√2)]
2
6 (1±√2)