Inspire - Lent Term 2022

Page 23

1 βˆ’1 ) and Firstly, I have done a leftward version of the original transformation which described by the matrix ( 0 1 then I have stretched out space by 2 units in the π‘₯ direction and 3 units in the 𝑦 direction. This is described by the 2 0 ). Can you see why? (It might help if you draw it out) matrix ( 0 3 We say that the outcome of applying both of these tranformations is their composition. What you are effectively doing, is multiplying the matrices. To do this, you have to imagine they are functions (which they are) and so function notation applies (reading from right to left). For example: 𝑔(𝑓(π‘₯)) This means to calculate 𝑓(π‘₯) then put it into 𝑔(π‘₯). The same applies for matrices: 𝑀1 𝑀2 This means do 𝑀2 first and then 𝑀1. With what I said earlier in mind, what do you think is the matrix composition of: (

2 0 1 βˆ’1 )( ) 0 3 0 1

The question is really saying β€œWhat is the overall effect of applying a rightward shear (name of the first transformation) followed by a stretching of space?” It could be put even more simply as β€œRecord the position of 𝑖̂ and 𝑗̂ after both transformations” Let’s run through this step by step. Firstly, let’s find where 𝑖̂ lands after both transformations: 1 ( ) 𝑖𝑠 π‘€β„Žπ‘’π‘Ÿπ‘’ 𝑖𝑑 π‘™π‘Žπ‘›π‘‘π‘  π‘Žπ‘“π‘‘π‘’π‘Ÿ π‘‘β„Žπ‘’ π‘“π‘–π‘Ÿπ‘ π‘‘ π‘‘π‘Ÿπ‘Žπ‘›π‘ π‘“π‘œπ‘Ÿπ‘šπ‘Žπ‘‘π‘–π‘œπ‘› 0 𝑀𝑒 π‘‘β„Žπ‘’π‘› 𝑓𝑖𝑛𝑑 π‘œπ‘’π‘‘ π‘€β„Žπ‘’π‘Ÿπ‘’ π‘‘β„Žπ‘–π‘  π‘£π‘’π‘π‘‘π‘œπ‘Ÿ π‘”π‘œπ‘’π‘  π‘Žπ‘“π‘‘π‘’π‘Ÿ π‘Žπ‘π‘π‘™π‘¦π‘–π‘›π‘” π‘‘β„Žπ‘’ π‘ π‘’π‘π‘œπ‘›π‘‘ π‘‘π‘Ÿπ‘Žπ‘›π‘ π‘“π‘œπ‘Ÿπ‘šπ‘Žπ‘‘π‘–π‘œπ‘› 𝑒𝑠𝑖𝑛𝑔 π‘‘β„Žπ‘’ π‘ π‘Žπ‘šπ‘’ π‘šπ‘’π‘‘β„Žπ‘œπ‘‘ 𝑀𝑒 β„Žπ‘Žπ‘‘ π‘’π‘Žπ‘Ÿπ‘™π‘–π‘’π‘Ÿ: 2 0 2 1βˆ™( )+0βˆ™( )= ( ) 0 3 0 𝑀𝑒 π‘‘β„Žπ‘’π‘› π‘Ÿπ‘’π‘π‘’π‘Žπ‘‘ π‘‘β„Žπ‘–π‘  π‘šπ‘’π‘‘β„Žπ‘œπ‘‘ π‘‘π‘œ 𝑓𝑖𝑛𝑑 π‘€β„Žπ‘’π‘Ÿπ‘’ 𝑗̂ π‘™π‘Žπ‘›π‘‘π‘  π‘Žπ‘“π‘‘π‘’π‘Ÿ π‘π‘œπ‘‘β„Ž π‘‘π‘Ÿπ‘Žπ‘›π‘ π‘“π‘œπ‘Ÿπ‘šπ‘Žπ‘‘π‘–π‘œπ‘›π‘  π‘˜π‘›π‘œπ‘€π‘–π‘›π‘” 1 π‘‘β„Žπ‘Žπ‘‘ 𝑖𝑑 π‘™π‘Žπ‘›π‘‘π‘’π‘‘ π‘Žπ‘‘ ( ) π‘Žπ‘“π‘‘π‘’π‘Ÿ π‘‘β„Žπ‘’ π‘“π‘–π‘Ÿπ‘ π‘‘ π‘‘π‘Ÿπ‘Žπ‘›π‘ π‘“π‘œπ‘Ÿπ‘šπ‘Žπ‘‘π‘–π‘œπ‘›: 1 2 0 βˆ’2 βˆ’1 βˆ™ ( ) + 1 βˆ™ ( ) = ( ) 0 3 3

π‘‘β„Žπ‘’π‘›, 𝑀𝑒 π‘π‘Žπ‘› π‘π‘Ÿπ‘’π‘Žπ‘‘π‘’ π‘Ž π‘šπ‘Žπ‘‘π‘Ÿπ‘–π‘₯ 𝑒𝑠𝑖𝑛𝑔 π‘‘β„Žπ‘–π‘  π‘–π‘›π‘“π‘œπ‘Ÿπ‘šπ‘Žπ‘‘π‘–π‘œπ‘› π‘€β„Žπ‘–π‘β„Ž π‘ π‘’π‘šπ‘  𝑒𝑝 π‘π‘œπ‘‘β„Ž π‘œπ‘“ π‘‘β„Žπ‘’π‘ π‘’ 2 βˆ’2 ) 0 3

π‘‘π‘Ÿπ‘Žπ‘›π‘ π‘“π‘œπ‘šπ‘Žπ‘‘π‘–π‘œπ‘›π‘ : (

If you don’t understand how we got to this point, go back to make sure you really understand how to find where a vector has moved to after a linear transformation. This is all we have done here: we have looked at where the (already transformed) 𝑖̂ and 𝑗̂ have gone to after the second transformation. Then, we have stored this information in a matrix which perfectly sums up both transformations.

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