Hello, let's note A the matrix, we need to find
such that A
=
I, where I is the identity matrix, so the determinant is 0, giving us the characteristic equation as

We just need to solve this equation using the discriminant.

And then the eigenvalues are.

To find the basis, we have to solve the system of equations.
![A\lambda_1-\lambda_1 I=\left[\begin{array}{cc}3i&3\\-3&3i\end{array}\right] \\\\=3\left[\begin{array}{cc}i&1\\-1&i\end{array}\right] \\\\\text{For a vector (a,b), we need to find a and b such that.}\\\\\begin{cases}ai+b=0\\-a+bi=0\end{cases}\\\\\text{(1,-i) is a base of this space, as i-i=0 and -1-}i^2\text{=-1+1=0.}](https://tex.z-dn.net/?f=A%5Clambda_1-%5Clambda_1%20I%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3i%263%5C%5C-3%263i%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%3D3%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Di%261%5C%5C-1%26i%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5Ctext%7BFor%20a%20vector%20%28a%2Cb%29%2C%20we%20need%20to%20find%20a%20and%20b%20such%20that.%7D%5C%5C%5C%5C%5Cbegin%7Bcases%7Dai%2Bb%3D0%5C%5C-a%2Bbi%3D0%5Cend%7Bcases%7D%5C%5C%5C%5C%5Ctext%7B%281%2C-i%29%20is%20a%20base%20of%20this%20space%2C%20as%20i-i%3D0%20and%20-1-%7Di%5E2%5Ctext%7B%3D-1%2B1%3D0.%7D)
![A\lambda_2-\lambda_2 I=\left[\begin{array}{cc}-3i&3\\-3&-3i\end{array}\right] \\\\=3\left[\begin{array}{cc}-i&1\\-1&-i\end{array}\right]\\\\\text{For a vector (a,b), we need to find a and b such that.}\\\\\begin{cases}-ai+b=0\\-a-bi=0\end{cases}\\\\\text{(1,i) is a base of this space as -i+i=0 and -1-i*i=0.}](https://tex.z-dn.net/?f=A%5Clambda_2-%5Clambda_2%20I%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-3i%263%5C%5C-3%26-3i%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%3D3%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-i%261%5C%5C-1%26-i%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5Ctext%7BFor%20a%20vector%20%28a%2Cb%29%2C%20we%20need%20to%20find%20a%20and%20b%20such%20that.%7D%5C%5C%5C%5C%5Cbegin%7Bcases%7D-ai%2Bb%3D0%5C%5C-a-bi%3D0%5Cend%7Bcases%7D%5C%5C%5C%5C%5Ctext%7B%281%2Ci%29%20is%20a%20base%20of%20this%20space%20as%20-i%2Bi%3D0%20and%20-1-i%2Ai%3D0.%7D)
Thank you
Answer:
Option (B)
Step-by-step explanation:
Since the linear graph is passing through the origin, there is a proportional relationship between the amount of chocolate (C) and amount of peanuts (P) used in the package of snacks mix,
P ∝ C
P = kC
k = 
Here k = proportionality constant
Since, the given line passes through a point (3, 2),
k = 
Therefore, ratio of chocolate and peanuts mixed is
.
Option (B) is the correct option.
I believe the change in temperature by minute is 9
Area of rhombus = 0.5 x D1 x D2
Thus,
Area = 0.5 x 30 x 18
= 270 units ^2
Answer:
Step-by-step explanation:
(a + b)² = a² + 2ab +b²
a = y & b = 10
(y + 10)² = y² + 2 *y*10 + (10)²
= y² + 20y + 100