If A=[cos alpha sin alpha]
[-sin alpha cos alpha] , A–¹= A' , then find alpha
Answers
Answer:
Solution: We have, <br> COs a sin a A sin a cos a and COs a – sin a A' sin a cos a <br> Also, A-1 = A '<br> 1 » AA AA <br> COs a sin a sin a cos a <br> cos? a + sin? a 0 sin? a <br> By using equality of matrices, we get <br> cos a + sin a = 1 <br> which is true for all real values of ||
Given : A=[cos alpha sin alpha]
[-sin alpha cos alpha]
To find : A⁻¹ = A , alpha
Solution:
A = \left[\begin{array}{ccc}Cos \alpha&Sin \alpha \\-Sin \alpha&Cos \alpha \end{array}\right]
A = IA
\left[\begin{array}{ccc}Cos \alpha&Sin \alpha \\-Sin \alpha&Cos \alpha \end{array}\right] = \left[\begin{array}{ccc}1&0 \\0&1 \end{array}\right] \left[\begin{array}{ccc}Cos \alpha&Sin \alpha \\-Sin \alpha&Cos \alpha \end{array}\right]
R₁ → R₁Cosα - SinαR₂
1 0 Cosα -Sinα
-Sinα Cosα 0 1
R₂ → R₂+ R₁Sinα
1 0 Cosα -Sinα
0 Cosα CosαSinα 1 - Sin²α
1 0 Cosα -Sinα
0 Cosα CosαSinα Cos²α
R₂ → R₂ / Cosα
1 0 Cosα -Sinα
0 1 Sinα Cosα
I = A⁻¹A
=> A⁻¹ =
A⁻¹ = A
Hence its true for all real Value of α
Another method
A⁻¹ = A' given
AA⁻¹ = I
=> AA' = I
=> * = I
=> Cos²α + Sin²α -CosαSinα + SinαCosα = 1 0
-SinαCosα + CosαSinα Sin²α + Cos²α 0 1
=> 1 0 = 1 0
0 1 0 1
independent of alpha α
Hence true for all real value of α
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