Write out the 3 x 3 matrix A when Aij = ii + jj - ij

Question

Write out the 3 x 3 matrix A when Aij = ii + jj - ij

Answer 1

Aij = ii + jj - ij

 

Let us do this row by row.

 

First row:

 

A₁₁ = 1*1 + 1*1 - 1*1 = 1

A₁₂ = 1*1 + 2*2 - 1*2 = 3

A₁₃ = 1*1 + 3*3 - 1*3 = 7

 

Second row:

 

A₂₁ = 2*2 + 1*1 - 2*1 = 3

A₂₂ = 2*2 + 2*2 - 2*2 = 4

A₂₃ = 2*2 + 3*3 - 2*3 = 7

 

Third row:

 

A₃₁ = 3*3 + 1*1 - 3*1 = 7

A₃₂ = 3*3 + 2*2 - 3*2 = 7

A₃₃ = 3*3 + 3*3 - 3*3 = 9

 

The 3 rows are

 

(1 3 7)

(3 4 7)

(7 7 9)

 

It looks symmetric about the main diagonal.

 

Answer 2

 I don't have the means by which to show the sides of the matrices

 

            i/j       0       0

 

A =       0       j/i      0

 

             0       0      -1

 

In order to multiply matrix A and get:  ii + JJ - ij  for an 

answer, you must be multiplying by a 3 by 1 matrix:

 

           ij

 

           ij

 

           ij

 

This is as much as I can read into the problem as it is stated.