Answer:
We have the matrix ![A=\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&8&4\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%26-4%26-4%5C%5C0%26-8%26-4%5C%5C0%268%264%5Cend%7Barray%7D%5Cright%5D)
To find the eigenvalues of A we need find the zeros of the polynomial characteristic 
Then
![p(\lambda)=det(\left[\begin{array}{ccc}-4-\lambda&-4&-4\\0&-8-\lambda&-4\\0&8&4-\lambda\end{array}\right] )\\=(-4-\lambda)det(\left[\begin{array}{cc}-8-\lambda&-4\\8&4-\lambda\end{array}\right] )\\=(-4-\lambda)((-8-\lambda)(4-\lambda)+32)\\=-\lambda^3-8\lambda^2-16\lambda](https://tex.z-dn.net/?f=p%28%5Clambda%29%3Ddet%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4-%5Clambda%26-4%26-4%5C%5C0%26-8-%5Clambda%26-4%5C%5C0%268%264-%5Clambda%5Cend%7Barray%7D%5Cright%5D%20%29%5C%5C%3D%28-4-%5Clambda%29det%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-8-%5Clambda%26-4%5C%5C8%264-%5Clambda%5Cend%7Barray%7D%5Cright%5D%20%29%5C%5C%3D%28-4-%5Clambda%29%28%28-8-%5Clambda%29%284-%5Clambda%29%2B32%29%5C%5C%3D-%5Clambda%5E3-8%5Clambda%5E2-16%5Clambda)
Now, we fin the zeros of 
.
Then, the eigenvalues of A are 
of multiplicity 1 and 
of multiplicity 2.
Let's find the eigenspaces of A. For : 
.Then, we use row operations to find the echelon form of the matrix
![A=\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&8&4\end{array}\right]\rightarrow\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&0&0\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%26-4%26-4%5C%5C0%26-8%26-4%5C%5C0%268%264%5Cend%7Barray%7D%5Cright%5D%5Crightarrow%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%26-4%26-4%5C%5C0%26-8%26-4%5C%5C0%260%260%5Cend%7Barray%7D%5Cright%5D)
We use backward substitution and we obtain
1.
2.
Therefore,
For 
: 
.Then, we use row operations to find the echelon form of the matrix
![A+4I_3=\left[\begin{array}{ccc}0&-4&-4\\0&-4&-4\\0&8&8\end{array}\right] \rightarrow\left[\begin{array}{ccc}0&-4&-4\\0&0&0\\0&0&0\end{array}\right]](https://tex.z-dn.net/?f=A%2B4I_3%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%26-4%26-4%5C%5C0%26-4%26-4%5C%5C0%268%268%5Cend%7Barray%7D%5Cright%5D%20%5Crightarrow%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%26-4%26-4%5C%5C0%260%260%5C%5C0%260%260%5Cend%7Barray%7D%5Cright%5D)
We use backward substitution and we obtain
1.
Then,
D. Plug in the options to both equations and find which makes both true.
Answer:
Step-by-step explanation:
The quadratic formula is:
The discriminant of a quadratic is just the expression under the square root, or 
. This can tell us the number of solutions a quadratic has.
If the discriminant is:
- Positive = 2 real solutions
- Equal to Zero = 1 real double/repeated solution
- Negative = 0 real solutions, but 2 imaginary solutions
Our quadratic equation has a discriminant of 5, which is positive. Therefore, it has 2 real solutions.
Jane walked a quarter of the distance she walked on Monday, so that means she walked 4 times less hoped this helped. :]
The amount of times that 14 goes into 78 is; 5 times
<h3>How to carry out division of numbers?</h3>
We want to find how many 14 goes into 78.
Now, it is pertinent to note that this expression is used in our common daily lives for basic things of life. For example, if i want to know how many 10 dollars can go into 100 dollars, i will just divide 100 by 10.
Similarly another example is if i want to know how many 5 liters container of fuel would go into a 60 liters tank, i will just divide 60 by 5 to get the number.
In a similar vein, we can solve this question by dividing 78 by 14 to get;
78/14 = 5 ⁸/₁₄
This means 14 can only go in 5 full times as we will not count the fraction.
Read more about division of numbers at; brainly.com/question/1622425
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