1 is b
2 is c
3 is c
4 is a
5 is c
6 is b
7 is a
8 is d
9 is d
10 is d
The first thing you must remember is that the product of matrices is done by multiplying rows by columns.
For this case, the AB matrix is
AB = [1 0 0; 0 1 0; (x + 1) 0 y]
Equalizing the identity matrix we have
AB = I
[1 0 0; 0 1 0; (x + 1) 0 y] = [1 0 0; 0 1 0; 0 0 1]
Therefore we have
x + 1 = 0
y = 1
The values of x and y are
x = -1
y = 1
the answer is
X=-1
Y=1
f(x) = sqrt [( x - 3) / (x - 5)]
well x cannot be 5 because that would make x - 5 = 0.
Also the fraction x - 3 / x - 5 cannot be negative because theres no real square root of a negative.
So x must be <= 3 or > 5
In interval notation the domain is ( -∞,3] or (5 , ∞)
