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Entries of a matrix (Posted on 2021-01-29) Difficulty: 3 of 5
Let be a natural number n≥2 and the n×n matrix whose entries at the i-th line and j-th column is min(i,j). Calculate:

a) its determinant
b) its inverse

No Solution Yet Submitted by Danish Ahmed Khan    
Rating: 4.0000 (1 votes)

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Solution Solution | Comment 4 of 5 |
The 5x5 example case of this matrix is illustrated:
[  1  1  1  1  1 ]
[  1  2  2  2  2 ]
[  1  2  3  3  3 ]
[  1  2  3  4  4 ]
[  1  2  3  4  5 ]

A common method of calculating a matrix inverse is to augment an identity matrix to the right of the matrix to invert and then apply row operations until the original matrix becomes an identity matrix. Then the augmented matrix is now the inverse of the original matrix.

Example of the augmented matrix (before row operations):
[  1  1  1  1  1 |  1  0  0  0  0 ]
[  1  2  2  2  2 |  0  1  0  0  0 ]
[  1  2  3  3  3 |  0  0  1  0  0 ]
[  1  2  3  4  4 |  0  0  0  1  0 ]
[  1  2  3  4  5 |  0  0  0  0  1 ]

Let r_k and r_k+1 be the kth and (k+1)th rows of the matrix.  Then apply the row operation -1*r_k + r_k+1 -> r_k+1, applied to each pair k,k+1 starting from the bottom and working up.  n-1 row operations will be applied and none of the row operations alter the value of the determinant of the original matrix.

After this pass of row operations the augmented matrix then becomes:
[  1  1  1  1  1 |  1  0  0  0  0 ]
[  0  1  1  1  1 | -1  1  0  0  0 ]
[  0  0  1  1  1 |  0 -1  1  0  0 ]
[  0  0  0  1  1 |  0  0 -1  1  0 ]
[  0  0  0  0  1 |  0  0  0 -1  1 ]

At this point the left matrix is a triangular matrix, whose determinant is easy to evaluate as 1*1*1*1*1 = 1.  This is also the determinant of the original matrix.  So the answer to part A is the determinant equals 1 for all size n.

Now again let r_k and r_k+1 be the kth and (k+1)th rows of the matrix.  Then apply the row operation -1*r_k+1 + r_k -> r_k, applied to each pair k,k+1 this time starting from the top and working down.  This will finish the row operations to invert the matrix since the left matrix will now be an identity matrix:
[  1  0  0  0  0 |  2 -1  0  0  0 ]
[  0  1  0  0  0 | -1  2 -1  0  0 ]
[  0  0  1  0  0 |  0 -1  2 -1  0 ]
[  0  0  0  1  0 |  0  0 -1  2 -1 ]
[  0  0  0  0  1 |  0  0  0 -1  1 ]

The inverse of the original matrix is then described as a matrix whose main diagonal is all 2s except for entry (n,n) which is 1; whose first upper and first lower diagonals are all -1; and has all other entries equal to 0.

  Posted by Brian Smith on 2021-01-30 21:32:06
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