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)
Substitute x and y with the value from table. The correct answer is A. 7=-3(-1)+4.
Answer:
3 > n - 1 = n < 4
t − 3 > −2 = t > 1
4 < p - 2 = p > 6
2 ≥ h − 5 = h ≤ 7
v − 5 > −9 = v > 4
p + 3 ≤ 4 = p ≤ 1
- 7 < 7 + t = t > 14
6 + k > 5 = k > 1
12 ≤ r + 5 = r ≥ 7
w − (−4) < 8 = w < 4
Step-by-step explanation:
Hope this helps