# How to find the mirror of an identity matrix without using numpy?

Here I have an identity matrix that goes from top left to bottom right. I’m trying to flip it so I can get a row of 1’s going from top right to bottom left but I don’t want to use numpy. But I just cant work out how to do it…

``````num = int(input("enter your number"))
for i in range(0, num):
for j in range(0, num):
if (i == j):
print(1, sep=" ", end=" ")
else:
print(0, sep=" ", end=" ")
print()
``````

Example:
Input: 4
Output:

1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1

### >Solution :

There is simple relation between `column`, `row` and `num`

``````if i + j + 1 == num:
``````

Full code:

``````num = int(input("enter your number"))
for i in range(0, num):
for j in range(0, num):
if i + j + 1 == num:
print(1, sep=" ", end=" ")
else:
print(0, sep=" ", end=" ")
print()
``````

EDIT:

Other idea is to revers one range

``````for j in range(num-1, -1, -1):
``````

Full code:

``````num = int(input("enter your number"))
for i in range(0, num):
for j in range(num-1, -1, -1):
if i == j:
print(1, sep=" ", end=" ")
else:
print(0, sep=" ", end=" ")
print()
``````