Determinant Matrix Problems – I

Determinant Matrix Problems – I

1. Show that \(\left| \begin{matrix}a & b & c \\{{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\{{a}^{3}} & {{b}^{3}} & {{c}^{3}}  \\\end{matrix} \right|\) = abc (a – b) (b – c) (c – a)

Solution: Given matrix

\(\left| \begin{matrix}a & b & c  \\{{a}^{2}} & {{b}^{2}} & {{c}^{2}}  \\{{a}^{3}} & {{b}^{3}} & {{c}^{3}}  \\\end{matrix} \right|\)

Take a common abc three rows

\(=abc\left| \begin{matrix}1 & 1 & 1  \\a & b & c  \\{{a}^{2}} & {{b}^{2}} & {{c}^{2}}  \\\end{matrix} \right|\)

We can apply elementary column transformations

C₁ → C₁ – C₂

C₂ → C₂ – C₃

\(=abc\left| \begin{matrix}0 & 0 & 0  \\a-b & b-c & c  \\{{a}^{2}}-{{b}^{2}} & {{b}^{2}}-{{c}^{2}} & {{c}^{2}}  \\\end{matrix} \right|\)

Take a common (a – b) (b – c)

\(=abc(a-b)(b-c)\left| \begin{matrix}0 & 0 & 0  \\1 & 1 & 1  \\a+b & b+c & {{c}^{2}}  \\\end{matrix} \right|\)

= abc (a – b) (b – c) [b + c – a – b]

= abc (a – b) (b – c) (c – a)

Hence proved  \(\left| \begin{matrix}a & b & c  \\{{a}^{2}} & {{b}^{2}} & {{c}^{2}}  \\{{a}^{3}} & {{b}^{3}} & {{c}^{3}}  \\\end{matrix} \right|\) = abc (a – b) (b – c) (c – a)

2. Without expanding the determinant, show that \(\left| \begin{matrix}a & {{a}^{2}} & bc \\b & {{b}^{2}} & ca \\c & {{c}^{2}} & ab  \\\end{matrix} \right|=\left| \begin{matrix}1 & {{a}^{2}} & {{a}^{3}}  \\1 & {{b}^{2}} & {{b}^{3}}  \\1 & {{c}^{2}} & {{c}^{3}}  \\\end{matrix} \right|\)

Solution: Given matrix

\(\left| \begin{matrix}a & {{a}^{2}} & bc  \\b & {{b}^{2}} & ca  \\c & {{c}^{2}} & ab  \\\end{matrix} \right|\)

Take a common abc 3rd column

\(\left| \begin{matrix}a & {{a}^{2}} & bc  \\b & {{b}^{2}} & ca  \\c & {{c}^{2}} & ab  \\\end{matrix} \right|=abc\left| \begin{matrix}a & {{a}^{2}} & \frac{1}{a}  \\b & {{b}^{2}} & \frac{1}{b}  \\c & {{c}^{2}} & \frac{1}{c}  \\\end{matrix} \right|\)

\(=\left| \begin{matrix}{{a}^{2}} & {{a}^{3}} & 1  \\{{b}^{2}} & {{b}^{3}} & 1  \\{{c}^{2}} & {{c}^{3}} & 1  \\\end{matrix} \right|\)

\(=\left| \begin{matrix}1 & {{a}^{2}} & {{a}^{3}}  \\1 & {{b}^{2}} & {{b}^{3}}  \\1 & {{c}^{2}} & {{c}^{3}}  \\\end{matrix} \right|\)

Hence proved \(\left| \begin{matrix}a & {{a}^{2}} & bc  \\b & {{b}^{2}} & ca  \\c & {{c}^{2}} & ab  \\\end{matrix} \right|=\left| \begin{matrix}1 & {{a}^{2}} & {{a}^{3}}  \\1 & {{b}^{2}} & {{b}^{3}}  \\1 & {{c}^{2}} & {{c}^{3}}  \\\end{matrix} \right|\)