Answer:
0.07017
Step-by-step explanation:
Letter C
Use the Distance formula
sq rt [ (x2 - x1)^2 + (y2 - y1)^2 ]
sq rt [ (10-5)^2 + (2 - 4)^2 ]
sq rt [ (5)^2 + (-2)^2 ]
sq rt ( 25 + 4)
Square root of 29
Answer:when in doubt go with c
Step-by-step explanation:
just do it
Answer:
a) -4
b) 1
c) 1
Step-by-step explanation:
a) The matrix A is given by:
![A=\left[\begin{array}{ccc}-3&0&1\\2&-4&2\\-3&-2&1\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%260%261%5C%5C2%26-4%262%5C%5C-3%26-2%261%5Cend%7Barray%7D%5Cright%5D)
to find the eigenvalues of the matrix you use the following:
![det(A-\lambda I)=0](https://tex.z-dn.net/?f=det%28A-%5Clambda%20I%29%3D0)
where lambda are the eigenvalues and I is the identity matrix. By replacing you obtain:
![A-\lambda I=\left[\begin{array}{ccc}-3-\lambda&0&1\\2&-4-\lambda&2\\-3&-2&1-\lambda\end{array}\right]](https://tex.z-dn.net/?f=A-%5Clambda%20I%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3-%5Clambda%260%261%5C%5C2%26-4-%5Clambda%262%5C%5C-3%26-2%261-%5Clambda%5Cend%7Barray%7D%5Cright%5D)
and by taking the determinant:
![[(-3-\lambda)(-4-\lambda)(1-\lambda)+(0)(2)(-3)+(2)(-2)(1)]-[(1)(-4-\lambda)(-3)+(0)(2)(1-\lambda)+(2)(-2)(-3-\lambda)]=0\\\\-\lambda^3-6\lambda^2-12\lambda-16=0](https://tex.z-dn.net/?f=%5B%28-3-%5Clambda%29%28-4-%5Clambda%29%281-%5Clambda%29%2B%280%29%282%29%28-3%29%2B%282%29%28-2%29%281%29%5D-%5B%281%29%28-4-%5Clambda%29%28-3%29%2B%280%29%282%29%281-%5Clambda%29%2B%282%29%28-2%29%28-3-%5Clambda%29%5D%3D0%5C%5C%5C%5C-%5Clambda%5E3-6%5Clambda%5E2-12%5Clambda-16%3D0)
and the roots of this polynomial is:
![\lambda_1=-4\\\\\lambda_2=-1+i\sqrt{3}\\\\\lambda_3=-1-i\sqrt{3}](https://tex.z-dn.net/?f=%5Clambda_1%3D-4%5C%5C%5C%5C%5Clambda_2%3D-1%2Bi%5Csqrt%7B3%7D%5C%5C%5C%5C%5Clambda_3%3D-1-i%5Csqrt%7B3%7D)
hence, the real eigenvalue of the matrix A is -4.
b) The multiplicity of the eigenvalue is 1.
c) The dimension of the eigenspace is 1 (because the multiplicity determines the dimension of the eigenspace)