C++ resize 2d vector where function is const - c++

Question

I create a 2D vector (0,0) and would like to resize it (n,m), however, my resize function must remain const
I have tried doing
void resize(int row, int col) const
{
array.resize(row, vector<int>(col));
}
but keep getting
passing ‘const std::vector<std::vector<int>, std::allocator<std::vector<int> > >’ as ‘this’ argument discards qualifiers
How can I do this?
Matrix.h
#pragma once
#include <vector>
using namespace std;
template <typename Object>
class matrix
{
public:
matrix(int rows, int cols) : array{ rows } {
for (auto& thisRow : array)
thisRow.resize(cols);
}
matrix( initializer_list<vector<Object>> lst ) : array( lst.size( ) )
{
int i = 0;
for( auto & v : lst )
array[ i++ ] = std::move( v );
}
matrix( const vector<vector<Object>> & v ) : array{ v } {}
matrix( vector<vector<Object>> && v ) : array{ std::move( v ) } {}
matrix() {}
const vector<Object> & operator[]( int row ) const
{
return array[ row ];
}
vector<Object> & operator[]( int row )
{
return array[ row ];
}
int numrows() const
{
return array.size( );
}
int numcols() const
{
return numrows( ) ? array[ 0 ].size( ) : 0;
}
void resize(int row, int col) const
{
array.resize(row, vector<int>(col));
}
private:
vector<vector<Object>> array;
};
main.cpp
matrix<int> mat = matrix<int>();
cout << "Zero-parameter matrix (rows,cols) = (" << mat.numrows() << "," << mat.numcols() << ")" << endl;
mat.resize(4, 3);
cout << "Resized matrix to 4x3" << endl;
cout << mat << endl;
mat[2][1] = 12;
cout << "Modified (2,1)" << endl;
cout << mat << endl;

Answer 1

When you put const on the end of the function there, you're saying the implicit this is const. That is, you're promising not to modify the state of the object this function is being called on.
But isn't the whole point of calling resize() to modify the state of the object? If I were you, I'd take the const off of there.
In other words, you have two options: either keep your promise to not change the state of the object (perhaps by returning a resized copy?), or lose the const.