There is no need to perform any elementary row operations on the identity matrix; it is already in its reduced row echelon form. (There is a leading one in each row, and each column).

A matrix  is diagonal if and only if  - that is, the element in column , row  is equal to zero - for all . The given matrix violates this condition, since  and five other elements are equal to nonzero numbers.

A matrix  is diagonal if and only if  - that is, the element in column , row  is equal to zero - for all . The given matrix fits this criterion.

A matrix  is diagonal if and only if  - that is, the element in column , row  is equal to zero - for all . The given matrix satisfies this condition, since its only nonzero elements are the first element in Column 1, the second element in Column 2, and so forth.

 is a diagonal matrix - its only nonzero entries are in its main diagonal, which comprises the elements in row , column , for .

The inverse  of a such a matrix can be found simply by replacing each element in the main diagonal with its reciprocal. Rewriting the elements in the diagonal matrix as fractions,  

,

or

.

Replace each diagonal element with its reciprocal to find :

A diagonal matrix has only zeroes as entries off of its main (upper-left to lower-right) diagonal.  has three nonzero entries off this diagonal (Row 1, column 2, for example), so it is not a diagonal matrix.

An  matrix is strictly diagonally dominant if, for each , it holds that on Row , the absolute value of the element in Column  is greater than the sum of the absolute values of the other elements in that row. Therefore, for  to be strictly diagonally dominant, the following must hold:

For Row 1: 

or 

For Row 2: 

or 

For Row 3: ,

or 

For all three conditions to hold, it is necessary that . This is the correct choice.

 is a diagonal matrix, and its dimensions are , so, for some complex  and  (the problem did not specify that the entries were real),

.

To raise a diagonal matrix to a power, simply raise each nonzero element to that power. It holds that 

,

and, consequently, .

Equivalently, both  and  must be a one-hundredth root of 2, of which there are 100. Therefore, the number of possible matrices for  is .

This question does not make sense since there is no such thing as the  identity matrix, and it is not possible to take the trace of a matrix that is not square. This question is mostly meant to test your ability to think critically when reading certain mathematics problems.

A matrix  is diagonalizable if it can be written as , where  is any invertible matrix, and  is any diagonal matrix. If  is already a diagonal matrix, we can of course write it as . Hence any diagonal matrix is diagonalizable.