![\left[ \begin{matrix} 2 & a \\ -1 & -2 \end{matrix} \right] + \left[ \begin{matrix} b & 4 \\ -2 & 1 \end{matrix} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%202%20%26%20a%20%5C%5C%20-1%20%26%20-2%20%5Cend%7Bmatrix%7D%20%5Cright%5D%20%2B%20%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%20b%20%26%204%20%5C%5C%20-2%20%26%201%20%5Cend%7Bmatrix%7D%20%5Cright%5D)
This addition of matrices can be combined into one matrix.
To add matrices, add the corresponding components of each matrix.
After adding, we'll have the following
![\left[ \begin{matrix} 2+b & a+4 \\ -3 & -1 \end{matrix} \right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%202%2Bb%20%26%20a%2B4%20%5C%5C%20-3%20%26%20-1%20%5Cend%7Bmatrix%7D%20%5Cright%5D%20)
This matrix should be equal to the matrix on the right-hand side of the equation. This means that each corresponding component of this matrix and the other matrix should be equivalent.
This means that

AND

Solving these one-step equations will give the values of a = -4 and b = -1. That's answer choice D.
Answer:
11 feet
Step-by-step explanation:
3 yards equals 9 feet because there are 2 feet in one yard. 2*3=9
9+2= 11
4a - 2b = 10
a + c = 3 so a = 3 - c and c = 3 - a
4(3 - c) - 2b = 10
12 - 4c - 2b = 10
-4c -2b = -2
2c - b = 1
That’s all I have so far :/
Yep they do, because if you plug in y for both equations, you get x=0. To find y, plug in 0 for either equation to get -7. (0,-7) should work for both equations since the intersection point is the point where both equations meet.