We want to find the values of a, b, c, and d such that the given matrix product is equal to a 2x2 identity matrix. We will solve a system of equations to find:
<h3>
Presenting the equation:</h3>
Basically, we want to solve:
![\left[\begin{array}{cc}-1&2\\a&1\end{array}\right]*\left[\begin{array}{cc}b&c\\1&d\end{array}\right] = \left[\begin{array}{cc}1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-1%262%5C%5Ca%261%5Cend%7Barray%7D%5Cright%5D%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Db%26c%5C%5C1%26d%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
The matrix product will be:
![\left[\begin{array}{cc}-b + 2&-c + 2d\\a*b + 1&a*c + d\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-b%20%2B%202%26-c%20%2B%202d%5C%5Ca%2Ab%20%2B%201%26a%2Ac%20%2B%20d%5Cend%7Barray%7D%5Cright%5D)
Then we must have:
-b + 2 = 1
This means that:
b = 2 - 1 = 1
We also need to have:
a*b + 1 = 0
we know the value of b, so we just have:
a*1 + b = 0
Now the two remaining equations are:
-c + 2d = 0
a*c + d = 1
Replacing the value of a we get:
-c + 2d = 0
-c + d = 1
Isolating c in the first equation we get:
c = 2d
Replacing that in the other equation we get:
-(2d) + d = 1
-d = 1
Then:
c = 2d = 2*(-1) = -2
So the values are:
If you want to learn more about systems of equations, you can read:
brainly.com/question/13729904
Answer:
473 books to each class
Step-by-step explanation:
2365÷5=473
Answer:
choice 1
Step-by-step explanation:
Given
V = πr²h ( isolate r² by dividing both sides by πh )
= r² ( take the square root of both sides )
= r ← the positive value of r
Step-by-step explanation:
Rhombus since the other 3 have lines that go accross the graph horizontally for 3 squares.
This can solved using the cosine law which is:
c² = a² + b² - 2ab cos θ
Using the values given from the problem
6² = b² + b² - 2bb cos 112.62
And solving for b
36 = 2b² - 2b² cos 112.62
b = 3.6
The answer is the 3rd option.