17
1 0
0
3. If A =
and B =
1
show that (aA + bB) (aA - bB) = (a2 + b2)A.
0 1
- 1
1 0
Answers
Answered by
2
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 .
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