2 votes 2 votes The diagonal elements of a square matrix $M$ are odd integers while the off-diagonals are even integers. Then $M$ must be singular $M$ must be nonsingular There is not enough information to decide the singularity of $M$ $M$ must have a positive eigenvalue. Linear Algebra isi2017-mma engineering-mathematics linear-algebra matrix + – jjayantamahata asked Mar 27, 2018 • recategorized May 11, 2019 by akash.dinkar12 jjayantamahata 1.4k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes $A = \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix}$ determinant of A $\neq 0$ $B = \begin{bmatrix} 1 &0 &0 \\ 0& 1 &0 \\ 0& 0 &1 \end{bmatrix}$ determinant of B $\neq 0$ $C = \begin{bmatrix} 3 & 2 \\ 4 & 3 \end{bmatrix}$ determinant of C $\neq 0$ If we take any configuration with such condition, matrix will be always non-singular. https://math.stackexchange.com/questions/2327643/non-singularity-of-a-square-matrix/2327667 pankaj_vir answered Mar 27, 2018 pankaj_vir comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes Always nonsingular(B) jjayantamahata answered Mar 27, 2018 jjayantamahata comment Share Follow See 1 comment See all 1 1 comment reply bunty choudhary commented Apr 17, 2018 reply Follow Share But how it is possible. If diagonal has even elements then off diagonal have odd int or vice versa . Central element always contradict this statement....Plz explain 0 votes 0 votes Please log in or register to add a comment.