Answer:
B=![\left[\begin{array}{ccc}0&0\\0&1\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%260%5C%5C0%261%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
Let's do the multiplication AB.
If A=![\left[\begin{array}{ccc}1&0\\0&0\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C0%260%5C%5C%5Cend%7Barray%7D%5Cright%5D)
then the first row of A is= (1 0) by the first column of B= (0 0) is equal to zero.
the first row of A is= (1 0) by the second column of B= (0 1) is equal to zero too because 1.0+0.1=0.
the second row of A is= (0 0) by any colum of B is equal to zero too.
So we have found an example that works!
X + Y + Z = 8
2x = Z - 2
X + Z = 5
X = 1
1 + Y + Z = 8
Y + Z = 6
1 + Z = 5
Z = 4
1 + Y + 4 = 8
5 + Y = 8
Y = 3
The variables are 1, 3, 4.
I believe the answer is b because u need to combine like terms and 6c+2c+5c is 13c.
Answer: 4h/5
Step-by-step explanation:
5x + y = -13.....multiply by -3
4x + 3y = -17
-----------------
-15x - 3y = 39 (result of multiplying by -3)
4x + 3y = -17
-----------------add
-11x = 22
x = -22/11
x = -2
5x + y = -13
5(-2) + y = -13
-10 + y = -13
y = -13 + 10
y = -3
solution (-2,-3)
