9514 1404 393
Answer:
(a, b) = (-2, -1)
Step-by-step explanation:
The transpose of the given matrix is ...
![A^T=\left[\begin{array}{ccc}1&2&a\\2&1&2\\2&-2&b\end{array}\right]](https://tex.z-dn.net/?f=A%5ET%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%26a%5C%5C2%261%262%5C%5C2%26-2%26b%5Cend%7Barray%7D%5Cright%5D)
Then the [3,1] term of the product is ...
![(A\cdot A^T)_{31}=\left[\begin{array}{ccc}a&2&b\end{array}\right]\cdot\left[\begin{array}{ccc}1&2&2\end{array}\right]=a+2b+4](https://tex.z-dn.net/?f=%28A%5Ccdot%20A%5ET%29_%7B31%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%262%26b%5Cend%7Barray%7D%5Cright%5D%5Ccdot%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%262%5Cend%7Barray%7D%5Cright%5D%3Da%2B2b%2B4)
and the [3,2] term is ...
![(A\cdot A^T)_{32}=\left[\begin{array}{ccc}a&2&b\end{array}\right]\cdot\left[\begin{array}{ccc}2&1&-2\end{array}\right]=2a-2b+2](https://tex.z-dn.net/?f=%28A%5Ccdot%20A%5ET%29_%7B32%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%262%26b%5Cend%7Barray%7D%5Cright%5D%5Ccdot%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%261%26-2%5Cend%7Barray%7D%5Cright%5D%3D2a-2b%2B2)
Both of these terms in the product matrix are 0. We can solve the system of equations by adding these two terms:
(a +2b +4) +(2a -2b +2) = (0) +(0)
3a +6 = 0
a = -2
Substituting for 'a' in term [3,1] gives ...
-2 +2b +4 = 0
b = -1
The ordered pair (a, b) is (-2, -1).
6•4=24 so he spent a total of $24
Answer:
17
Step-by-step explanation
Using the formula: A= base1 + base 2/ 2 *h
The base you are looking for will be named a;
a = 2 a/h - b 2*112/8 - 11 =17
first a is base a. = 2 times area divided by height minus base b which is 2 times 112 over 8 minus 11 which gives you the answer of 17
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