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$VersionNumber
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13.2
ESM 5314, HW3, Saim Ehtesham Ali
ESM 5314, HW3, Saim Ehtesham Ali
Answer 1a
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ClearAll["Global`*"];
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R1={{1,0,0},{0,Cos[ϕ],Sin[ϕ]},{0,-Sin[ϕ],Cos[ϕ]}};R2={{Cos[θ],0,-Sin[θ]},{0,1,0},{Sin[θ],0,Cos[θ]}};
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R1R2=R1.R2;R1R2//MatrixForm
Out[]//MatrixForm=
Cos[θ] | 0 | -Sin[θ] |
Sin[θ]Sin[ϕ] | Cos[ϕ] | Cos[θ]Sin[ϕ] |
Cos[ϕ]Sin[θ] | -Sin[ϕ] | Cos[θ]Cos[ϕ] |
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R2R1=R2.R1;R2R1//MatrixForm
Out[]//MatrixForm=
Cos[θ] | Sin[θ]Sin[ϕ] | -Cos[ϕ]Sin[θ] |
0 | Cos[ϕ] | Sin[ϕ] |
Sin[θ] | -Cos[θ]Sin[ϕ] | Cos[θ]Cos[ϕ] |
Clearly, R1R2 ≠ R2R1, but I will try using values of θ and ϕ to further prove
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Aa=R1R2/.{θ->Pi/2,ϕ->Pi/2};Aa//MatrixForm
Out[]//MatrixForm=
0 | 0 | -1 |
1 | 0 | 0 |
0 | -1 | 0 |
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Ba=R2R1/.{θ->Pi/2,ϕ->Pi/2};Ba//MatrixForm
Out[]//MatrixForm=
0 | 1 | 0 |
0 | 0 | 1 |
1 | 0 | 0 |
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Aa===Ba(*Comparisonofthe2matrices,"True"indicatestheyarethesame,and"False"indicatesthattheyarenot*)
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False
Answer 1b
For small angles, we know Sin[x] = x and Cos[x] = 1, so we use this substitution
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Ab=N[R1R2/.{θ->1,ϕ->1,Cos[θ]1,Cos[ϕ]1}];Ab//MatrixForm
-10
10
-10
10
Out[]//MatrixForm=
1. | 0. | -1.× -10 10 |
1.× -20 10 | 1. | 1.× -10 10 |
1.× -10 10 | -1.× -10 10 | 1. |
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Bb=N[R2R1/.{θ->1,ϕ->1,Cos[θ]1,Cos[ϕ]1}];Bb//MatrixForm
-10
10
-10
10
Out[]//MatrixForm=
1. | 1.× -20 10 | -1.× -10 10 |
0. | 1. | 1.× -10 10 |
1.× -10 10 | -1.× -10 10 | 1. |
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result=Ab-Bb;result//MatrixForm
Out[]//MatrixForm=
0. | -1.× -20 10 | 0. |
1.× -20 10 | 0. | 0. |
0. | 0. | 0. |