If our Solar System was compressed by a factor of 10 billion such that the Sun is about the size of a grapefruit here in downtown Atlanta, then the nearest star (also about the size of a grapefruit) would be approximately at the distance of

3,602.4 km

Explanation:

Ok, the distance between the sun and the nearest star is 4.22 light-years.

1 light-year = 9.461x10^12 km

4.22 light-years. = 4.4*(9.461x10^12 km) = 41.6284x10^12km.

Now, the radius of the Sun is:

r1 = 693,340 km

The radius of the grapefruit is around:

r2 = 6cm

Then in this new scale, 6 centimeters are equivalent to  693,340 km.

Now, how many times there are 693,340 km in 41.6284x10^12km?

We can take the quotient:

N = (41.6284x10^12km/ 693,340 km)  = 60,040,384.

Then, the distance between the sun-grapefuit and the nearest star will be  60,040,384 times 6 centimeters:

Distance = 60,040,384*6cm = 360,242,304 cm ≈ 3,602.4 km

i think it because equal matrices have equal dimensions, only square matrices can be symmetric. for all indices and . every square diagonal matrix is symmetric, since all off-diagonal elements are zero. in linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space.