Question

After deleting the \(i\)-th row and \(j\)-th column from a square matrix of order \(n\), what is the order of the remaining matrix used to calculate \(M_{ij}\)?

Options

• \((n-1)\times(n-1)\)

• \(n\times n\)

• \((n+1)\times(n+1)\)

• \(i\times j\)

Answer 1

Deleting one row and one column reduces each dimension by one. Therefore, the remaining matrix has order \((n-1)\times(n-1)\).