]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.
Answer: y=17-5x/-9
Step-by-step explanation: 5x-9y=17
-9y=17-5x
Y=17-5x/-9
You can’t go any further than that because you can’t combine unlike Terms on the right side of the equation.
Sorry I am late but the I think it is this, I don’t know the answer but here is what I know. answer is: Imagine a rectangle that has one vertex at the origin and the opposite vertex is A. Now that you can see the image of A(3,4) under the rotation is A’(-4,3). It is easier to rotate the points that lie on the axes, and these help us find the image of A.
POINT: (3,0) (0,4) (3,4)
IMAGE (3,0) (-4,0) (-4,3)
Answer:
8.4*10^-3
Step-by-step explanation:
Move the decimal point 3 places in order to get a number in the one's place.
<span><span>C=(<span>x0</span>,<span>y0</span>,<span>z0</span>)</span><span>C=(<span>x0</span>,<span>y0</span>,<span>z0</span>)</span></span><span> and radius </span><span>rr</span>.
<span><span>(x−<span>x0</span><span>)2</span>+(y−<span>y0</span><span>)2</span>+(z−<span>z0</span><span>)2</span>=<span>r2</span></span><span>(x−<span>x0</span><span>)2</span>+(y−<span>y0</span><span>)2</span>+(z−<span>z0</span><span>)2</span>=<span>r2</span></span></span><span> </span>
