]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:
Either, (9+√69)/6 or (9-√69)/6
Step wise:
3x²-9x+1=0-------------------(i)
comparing equation onw with (ax²+bx+c=0), we get,
a=3, b=-9, c=1
now,
using Quadratic Formula,
(-b±√b²-4ac)/2a=x
{-(-9)±√(-9)²-4.3.1}/2.3=x
(9±√81-12)/6=x
(9±√69)/6=x
Taking +(ve) sign Taking -(ve) sign
(9+√69)/6=0 (9-√69)/6=0
∴(9+√69)/6=0 ∴(9-√69)/6=0
[∵They cannot be further solved]
-0.24 should be the answer. idk i might be wrong. im not to great at math just i just used a calculator
We are given with the expression or equation s<span>in2x cosx + cos 2x sin x = √3/2. The expression can be patterned from the trigonometric identity sin (a + b) = sin a cos B + cos A sin B. In this case, the expression is equal to sin 3x = sqrt 3 /2. using arc sign, x is equal to pi/9 </span>
2.7272727 snd it just keeps going on and on but and i am just here to help
