You didn't say what you wanted to find.  But I'll assume you want
to find the maximum height above the base that the plane could
fly parallel to the ground and clear the archway by 2 feet on
each side.

To make things easier I'm going to work in units of 10 feet
instead of 1 foot.  It's just a matter of moving the decimal.
So 160 ft will be represented by 16 units and 100 feet by 10
units.

I'll assume the arch is a parabola with a maximum point at (0,16)
with x intercepts at (-5,0) and (5,0), so that they will be 10 units
(100 feet) apart.

You say you want to use matrices.  So we will use the quadratic
equation



We will substitute those three points in that equation:

Substitute (x,y) = (0,16)





Substitute (x,y) = (5,0)





Substitute (x,y) = (-5,0)





So we have the system of equations



You can solve that by matrices or by any method you
want and get 

, , 

I'll assume you know how. If you don't post again
asking how.

So the equation of the arch is



Its graph is this:



The plane is 39 feet wide and we want it to have 2 feet
clearance on both sides.  So we want to know at what height
the graph is 39+2+2 or 43 feet wide.  Half of that 43 feet will
be on the right of the y-axis and the other half on the left, so
we want to find the two points on that parabola which have 
x-coordinate of ±half of 43 feet, or ±21.5 feet. That will
be ±2.15 of our units. so we substitute x=±2.15 into the
equation.  I'll just use the positive, because it'll be
the same since we square the x. 







That's 13.0416 ten-foot units or 130.416 feet

In other words, to clear the edges of the arch by 2 feet,
the plane will have to fly 130.416 feet above the base.

I'll draw a horizontal line across the arch to show where
that will be:
 
 

Comment:
Whoever made up this problem should have made the height of
the arch the same as the width of the base if he wanted the
arch to look like the one in St. Louis, for that one is 630 feet
high and 630 feet across the base and it would look more like 
this:



Edwin