Organize the logic to transform given matrix into required:

Givien[1 2 3

4 5 6

7 8 9]

Reqried[-1 2 3

0 -1 6

0 0 -1]

Answer 1

[_]

A = reshape(1:9,3,[]).'

A = 3×3

1 2 3 4 5 6 7 8 9

B = triu(A);
B(1:size(A,1)+1:end) = -1

B = 3×3

-1 2 3 0 -1 6 0 0 -1

[Muhammad Taha]

@Benjamin

B(1:size(A,1)+1:end) = -1

Can you please Explain this logic?

[_]

That statement sets the elements along the diagonal of B to be -1.

C = magic(6)

C = 6×6

35 1 6 26 19 24 3 32 7 21 23 25 31 9 2 22 27 20 8 28 33 17 10 15 30 5 34 12 14 16 4 36 29 13 18 11

C(1:size(C,1)+1:end)

ans = 1×6

35 32 2 17 14 11

C(1:size(C,1)+1:end) = 1000

C = 6×6

1000 1 6 26 19 24 3 1000 7 21 23 25 31 9 1000 22 27 20 8 28 33 1000 10 15 30 5 34 12 1000 16 4 36 29 13 18 1000

It is using linear indexing, which in MATLAB goes down the columns first. Starting with index 1 (the upper-left), incrementing by one more than the number of rows gives you the index of each element along the diagonal.

C = magic(3)

C = 3×3

8 1 6 3 5 7 4 9 2

C(1:end) % all elements by linear index

ans = 1×9

8 3 4 1 5 9 6 7 2

C([1 5 9]) % diagonal elements

ans = 1×3

8 5 2

C(1:4:end) % same

ans = 1×3

8 5 2

You could also use eye(), the identity matrix function:

eye(3)

ans = 3×3

1 0 0 0 1 0 0 0 1

C(logical(eye(3)))

ans = 3×1

8 5 2

C(logical(eye(3))) = -1

C = 3×3

-1 1 6 3 -1 7 4 9 -1

[John D'Errico]

You could also have done in a slightly simpler way:

A = reshape(1:9,3,[]).'

A = 3×3

1 2 3 4 5 6 7 8 9

B = triu(A,1) - eye(size(A))

B = 3×3

-1 2 3 0 -1 6 0 0 -1

Thus triu (and tril) with a second argument, allows you to control which diagonal to go to.