Class 12 Matrices · Long Answer3 Marksrotation_matrices

Class 12 Maths · Matrices · Long Answer #66

For f(α)=\(\begin{bmatrix}\cos\alpha & -\sin\alpha & 0\\\sin\alpha & \cos\alpha & 0\\0 & 0 & 1\end{bmatrix}\), prove f(α)f(−β)=f(α−β).

✅ Step-by-Step Solution

Final Answer: Identity proved.

Use cos(−β)=cosβ and sin(−β)=−sinβ to write f(−β)=[[cosβ,sinβ,0],[−sinβ,cosβ,0],[0,0,1]].
The upper-left 2×2 product has entries cosαcosβ+sinαsinβ, cosαsinβ−sinαcosβ, sinαcosβ−cosαsinβ, sinαsinβ+cosαcosβ.
The angle-difference formulas turn this block into [[cos(α−β),−sin(α−β)],[sin(α−β),cos(α−β)]]. The other off-block entries are 0 and the last diagonal entry is 1.
Thus the full product is f(α−β), as required.

[0.5 marks] Use cos(−β)=cosβ and sin(−β)=−sinβ to write f(−β)=[[cosβ,sinβ,0],[−sinβ,cosβ,0],[0,0,1]].
[1 mark] The upper-left 2×2 product has entries cosαcosβ+sinαsinβ, cosαsinβ−sinαcosβ, sinαcosβ−cosαsinβ, sinαsinβ+cosαcosβ.
[1 mark] The angle-difference formulas turn this block into [[cos(α−β),−sin(α−β)],[sin(α−β),cos(α−β)]]. The other off-block entries are 0 and the last diagonal entry is 1.
[0.5 marks] Thus the full product is f(α−β), as required.

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