Students are making ornaments and each takes 1 2 of a piece of Bristol board. They go to the cupboard and find 1 1 2 pieces of B
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]Eigenvectors are found by the equation 
implying that 
. We then can write:
![A-\lambda I = \left [ \begin{array}{cc} 4-\lambda & 2 \\ 5 & 1-\lambda \end{array}\right ]](https://tex.z-dn.net/?f=%20A-%5Clambda%20I%20%3D%20%5Cleft%20%5B%20%5Cbegin%7Barray%7D%7Bcc%7D%204-%5Clambda%20%26%202%20%5C%5C%205%20%26%201-%5Clambda%20%5Cend%7Barray%7D%5Cright%20%5D%20)
And:
Gives us the characteristic polynomial:
So, solving for each eigenvector subspace:
![\left [ \begin{array}{cc} 4 & 2 \\ 5 & 1 \end{array} \right ] \left [ \begin{array}{c} x \\ y \end{array} \right ] = \left [ \begin{array}{c} -x \\ -y \end{array} \right ]](https://tex.z-dn.net/?f=%5Cleft%20%5B%20%5Cbegin%7Barray%7D%7Bcc%7D%204%20%26%202%20%5C%5C%205%20%26%201%20%5Cend%7Barray%7D%20%5Cright%20%5D%20%5Cleft%20%5B%20%5Cbegin%7Barray%7D%7Bc%7D%20x%20%5C%5C%20y%20%5Cend%7Barray%7D%20%5Cright%20%5D%20%3D%20%5Cleft%20%5B%20%5Cbegin%7Barray%7D%7Bc%7D%20-x%20%5C%5C%20-y%20%5Cend%7Barray%7D%20%5Cright%20%5D%20)
Gives us the system of equations:
Producing the subspace along the line
We can see then that 3 is the answer.
And:
Gives us the characteristic polynomial:
So, solving for each eigenvector subspace:
Gives us the system of equations:
Producing the subspace along the line
We can see then that 3 is the answer.