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