Using properties of determinants prove that: |((m+n)^2, l^2,mn),((n+1)^2,m^2, ln)((l+m)^2, n^2, lm)| = (l^2 +m^2 + n^2)(l-m)(m-n)(n-1)

Question

Using properties of determinants prove that:

\(\begin{vmatrix} (m+n)^2 & 1^2 & mn \\[0.3em] (n+1)^2 & m^2 & ln \\[0.3em] (1+m)^2 & n^2 & lm \end{vmatrix}\) 

|((m+n)², l²,mn),((n+1)²,m², ln)((l+m)², n², lm)| = (l² +m² + n²)(l-m)(m-n)(n-1)

Answer 1

\(\begin{vmatrix} (m+n)^2 & 1^2 & mn \\[0.3em] (n+1)^2 & m^2 & ln \\[0.3em] (1+m)^2 & n^2 & lm \end{vmatrix}\)

= (l² + m² + n ²)(l - m)(m - n){0 + 0 - l(l + m) + n(m + n)} [expansion by first row] 

= (l² + m² + n²)(l - m)(m - n){0 + 0 - l(l + m) + n(m + n)} 

= (l² + m² + n²)(l - m)(m - n)( - l² - ml + mn + n²) 

= (l² + m² + n²)(l - m)(m - n){(n² - l²) + m(n - l)} 

= (l² + m² + n²)(l - m)(m - n)(n - l)(l + m + n)