Question

17
1 0
0
3. If A =
and B =
1
show that (aA + bB) (aA - bB) = (a2 + b2)A.
0 1
- 1
1 0​

Answer 1

Given :- If A = [ 1 0 and B = [ 0 1

0 1 ] -1 0 ]

show that (aA + bB)(aA - bB) = (a² + b²)A ?

Answer :-

solving LHS,

→ (aA + bB)(aA - bB)

→ (a[1 0 + b[0 1 )(a[1 0 - b[0 1 )

0 1] -1 0] 0 1] -1 0]

→ ([a 0 + [0 b )( [a 0 - [0 b )

0 a] -b 0] 0 a] -b 0]

→ [a b * [a - b

-b a] b a]

→ [ a²+b² -ab+ba

-ab+ba a²+b² ]

→ [a² + b² 0

0 a² + b²]

now, solving RHS,

→ (a² + b²)A

→ (a² + b²)[ 1 0

0 1 ]

→ [a² + b² 0

0 a² + b²]

therefore,

→ LHS = RHS .