Answer:
No, it cannot have a unique solution. Because there are more variables than equations, there must be at least one free variable. If the linear system is consistent and there is at least one free variable, the solution set contains infinitely many solutions. If the linear system is inconsistent, there is no solution.
Step-by-step explanation:
the questionnaire options are incomplete, however the given option is correct
We mark this option as correct because in a linear system of equations there can be more than one solution, since the components of the equations, that is, the variables are multiple, leaving free variables which generates more alternative solutions, however when there is no consistency there will be no solution
Answer:
A function with a positive constant other than 1 raised to a variable exponent.
Step-by-step explanation:
D because a rational number can be converted into a fraction and the square root of 5 cannot be covert into a fraction because it equals 2.2
The fourth root? There is no simplifying to be done.
The only way to continue work on this is converting to decimal. To four places the answer is 3.3437.
If you meant to ask for the cube root, the answer is 5.
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:
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:
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)
