1. Circle the figure you would use to make the Escher‐like tessellation given.
2 answers:
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The given operation involves just swapping the two rows, so carrying out 
on the matrix gives
![\left[\begin{array}{cc|c}5&5&5\\1&-\dfrac23&5\end{array}\right]\to\left[\begin{array}{cc|c}1&-\dfrac23&5\\5&5&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Cc%7D5%265%265%5C%5C1%26-%5Cdfrac23%265%5Cend%7Barray%7D%5Cright%5D%5Cto%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Cc%7D1%26-%5Cdfrac23%265%5C%5C5%265%265%5Cend%7Barray%7D%5Cright%5D)
Answer:
yes
Step-by-step explanation:
The graph is of a function (one output for every input) that also passes the horizontal line test (one input for every output). Hence it is one-to-one.
For a <em>function</em>, the key for it being one-to-one is that it pass the horizontal line test.
