About jRuby and clone

As you know, in ruby ​​everything and everything is objects. This is a plus, but there are some difficulties in understanding the interaction with complex objects. Standard example:
person1 = "Tim"
person2 = person1
person1[0] = 'J'
person1 ->"Jim"
person2 -> "Jim"
There is a standard solution to this problem:

person1 = "Tim"
person2 = person1.clone
person1[0] = "J"
person1 -> "Jim"
person2 -> "Tim"
But let we have an object of the following form:
a= [[1,2,3],[4,5,6],[7,8,9]]
Apply the standard approach to it:
a = [[1,2,3],[4,5,6],[7,8,9]]
a1 = a.clone
a1[0] = [0,0,0]
puts a[0]
puts a1[0]
Let's complicate the task: it is required to build the array elements - 3, 6, 9 in a square. We will act by the standard method:
a = [[1,2,3],[4,5,6],[7,8,9]]
a1 = a.clone
a1.each{|item| item[2]=item[2]**2}
puts a[0]
puts a1[0]

My mistake in this case is that I do not take into account that everything in Ruby is an object and I work with links to the created objects.
In this case, I cloned the array itself a, but the subarrays of a1 remain references to the massives of a, respectively.
How I correct my mistake:
a = [[1,2,3],[4,5,6],[7,8,9]]
a1=[]
a.each { |item| a1[a.index(item)]=item.clone }
a1.each{|item| item[2]=item[2]**2}
puts a[0]
puts a1[0]

That is, it turns out that I have cloned each of the subarrays of object a, and therefore the entire array a is obtained.
Check it out:
a = [[1,2,3],[4,5,6],[7,8,9]]
a1=[]
a.each { |item| a1[a.index(item)]=item.clone }
a1[0]=[0,0,0]
puts a[0]
puts a1[0]