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 Let A = a21 a22

and B = ± √ A.

a11+ T a12 then B = ± (1/R) a21 a22 + T

Where 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 = (1) A21 A22

And B = A1/2 (2)

Case 1 : If A12 = A21 = 0

Obviously we get, ±√A11 0 B = (3) 0 ±√A22

Case 2: 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

, (A11 + T) A12 B = ±(1/R) (6) A21 (A22 + T)

½ Where T = | A | = √ A11 A22 - A12 A21 (7)

{ 53 }

{ 54 }

2 And R = A11 + A22 + 2 T , R ≠ 0 (8)

2 Proof : B * B = ( 1/ R ) (A11 + T) A12 (A11 + T) A12

A21 (A22 + T) A21 (A22 + T)

2 2 = ( 1/ R ) (A11 + T) +A12 A21 A12(A11+T)+A12(A22+T)

2 A21(A11+T)+A21(A22+T) (A22+T) + A12 A21

2 2 2 = ( 1/ R ) (A11) + 2 A11 T + (T +A12 A21) A12 (A11+A22+ 2T)

2 2 A21(A11+A22+2T) (A22) +2A22T +(T + A12 A21)

2 2 2 = ( 1/ R ) (A11) + 2 A11 T + A11 A22 A12 R

2 2 A21R (A22) +2A22T +A11 A22

2 2 2 = ( 1/ R ) (A11) R A12 R

2 2 A21 R (A22) R

Example: (1) 1 3 A = 2 5 6 , |A| = -9 , T= 3i and R = A11+A22+2T= 1+6+6i = 7+6i

(1+ 3i ) 3 B= [± 1/ √(7+6i)] 5 (6+ 3i ).

(2) 11 -3 A = 2 6. |A| = 72 and T= 6√2 , R2 = 17 + 12 √2

11+6√2 -3 B = [± 1/ √(17+12√2)] 2 6 (1+√2)