Answer:
its B
Step-by-step explanation:
Answer:
x^y=p
Step-by-step explanation:
x=?
y=?
p=x^y
what is x and y?
Yes, 2 + 2 = 4 - 1 does equal 3
N = d - 5
2n/(d + 16) = n/d - 1/3 n/d
2n/(d + 16) = 2/3 n/d Divide by 2n
1 / (d + 16) = 1/3 d
Here's the tricky part.
d + 16 = 3d the two denominators are equal. the numerators are both 1.
16 = 2d
d = 8
so the numerator is d - 5
n = 8 - 5
n = 3
Let's see if it checks out.
n = 3
d = 8
2*3 = 6
8 + 16 = 24
New fraction 6/24 = 1/4
(3/8 - 1/3 ) = 1/8
3/8 - 1/8 = 1/4 so it checks with the original conditions put on it.
Answer:
![B)\,\,A^{-1}=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right]](https://tex.z-dn.net/?f=B%29%5C%2C%5C%2CA%5E%7B-1%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%26-1%26-1%5C%5C-3%261%260%5C%5C-3%260%261%5Cend%7Barray%7D%5Cright%5D%20)
Step-by-step explanation:
Expanding with first row
![det(A) = \left\Bigg|\begin{array}{ccc}1&1&1\\3&4&3\\3&3&4\end{array}\right\Bigg|\\\\\\det(A)= (1)\left\Big|\begin{array}{cc}4&3\\3&4\end{array}\right\Big|-(1)\left\Big|\begin{array}{cc}3&3\\3&4\end{array}\right\Big|+(1)\left\Big|\begin{array}{cc}3&4\\3&3\end{array}\right\Big|\\\\det(A)=1[16-9]-1[12-9]+1[9-12]\\\\det(A)=7-3-3\\\\det(A)=1](https://tex.z-dn.net/?f=det%28A%29%20%3D%20%5Cleft%5CBigg%7C%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%5C%5C3%264%263%5C%5C3%263%264%5Cend%7Barray%7D%5Cright%5CBigg%7C%5C%5C%5C%5C%5C%5Cdet%28A%29%3D%20%281%29%5Cleft%5CBig%7C%5Cbegin%7Barray%7D%7Bcc%7D4%263%5C%5C3%264%5Cend%7Barray%7D%5Cright%5CBig%7C-%281%29%5Cleft%5CBig%7C%5Cbegin%7Barray%7D%7Bcc%7D3%263%5C%5C3%264%5Cend%7Barray%7D%5Cright%5CBig%7C%2B%281%29%5Cleft%5CBig%7C%5Cbegin%7Barray%7D%7Bcc%7D3%264%5C%5C3%263%5Cend%7Barray%7D%5Cright%5CBig%7C%5C%5C%5C%5Cdet%28A%29%3D1%5B16-9%5D-1%5B12-9%5D%2B1%5B9-12%5D%5C%5C%5C%5Cdet%28A%29%3D7-3-3%5C%5C%5C%5Cdet%28A%29%3D1)
To find inverse we first find cofactor matrix
Cofactor matrix is
![C=\left[\begin{array}{ccc}7&-3&3\\-1&1&0\\-1&0&1\end{array}\right] \\\\Adj(A)=C^{T}\\\\Adj(A)=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] \\\\\\A^{-1}=\frac{adj(A)}{det(A)}\\\\A^{-1}=\frac{\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] }{1}\\\\A^{-1}=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right]](https://tex.z-dn.net/?f=C%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%26-3%263%5C%5C-1%261%260%5C%5C-1%260%261%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5CAdj%28A%29%3DC%5E%7BT%7D%5C%5C%5C%5CAdj%28A%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%26-1%26-1%5C%5C-3%261%260%5C%5C-3%260%261%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5CA%5E%7B-1%7D%3D%5Cfrac%7Badj%28A%29%7D%7Bdet%28A%29%7D%5C%5C%5C%5CA%5E%7B-1%7D%3D%5Cfrac%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%26-1%26-1%5C%5C-3%261%260%5C%5C-3%260%261%5Cend%7Barray%7D%5Cright%5D%20%7D%7B1%7D%5C%5C%5C%5CA%5E%7B-1%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%26-1%26-1%5C%5C-3%261%260%5C%5C-3%260%261%5Cend%7Barray%7D%5Cright%5D)
