Question

evaluate ¦1 a a^2¦ without expanding.
¦1 b b^2¦
¦1 c c^2¦​

Answer 1

$\large\underline{\sf{Solution-}}$

Given determinant is

$\begin{gathered}\sf \left | \begin{array}{ccc}1&a& {a}^{2} \\1&b& {b}^{2} \\1&c& {c}^{2} \end{array}\right | \end{gathered}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\tt{OP \: R_2 \: \to \: R_2 - R_1 }}$

$\rm \: = \: \begin{gathered}\sf \left | \begin{array}{ccc}1&a& {a}^{2} \\0&b - a& {b}^{2} - {a}^{2} \\1&c& {c}^{2} \end{array}\right | \end{gathered}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\tt{OP \: R_3 \: \to \: R_3 - R_1 }}$

$\rm \: = \: \begin{gathered}\sf \left | \begin{array}{ccc}1&a& {a}^{2} \\0&b - a& {b}^{2} - {a}^{2} \\0&c - a& {c}^{2} - {a}^{2} \end{array}\right | \end{gathered}$

We know,

$\boxed{\tt{ {x}^{2} - {y}^{2} = (x + y)(x - y) \: }}$

So, using this identity, we get

$\rm \: = \: \begin{gathered}\sf \left | \begin{array}{ccc}1&a& {a}^{2} \\0&b - a& (b - a)(b + a) \\0&c - a&(c - a)(c + a) \end{array}\right | \end{gathered}$

Take out (b - a) and (c - a) common from second and third row respectively, we get

$\rm \: = \: (b - a)(c - a)\begin{gathered}\sf \left | \begin{array}{ccc}1&a& {a}^{2} \\0&1& b + a \\0&1&c + a \end{array}\right | \end{gathered}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\tt{OP \: R_3 \: \to \: R_3 - R_2 }}$

$\rm \: = \:(b - a)(c - a) \begin{gathered}\sf \left | \begin{array}{ccc}1&a& {a}^{2} \\0&1& b + a \\0&0&c - b \end{array}\right | \end{gathered}$

$\rm \: = \: (b - a)(c - a)(c - b)$

$\rm \: = \: [ - (a - b)](c - a)[ - (b - c)]$

$\rm \: = \: (a - b)(b - c)(c - a)$

Hence,

$\rm\implies \:\boxed{\tt{ \begin{gathered}\sf \left | \begin{array}{ccc}1&a& {a}^{2} \\1&b& {b}^{2} \\1&c& {c}^{2} \end{array}\right | \end{gathered} = (a - b)(b - c)(c - a)}}$

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ADDITIONAL INFORMATION

1. The determinant value remains unaltered if rows and columns are interchanged.

2. The determinant value is 0, if two rows or columns are identical.

3. The determinant value is multiplied by - 1, if successive rows or columns are interchanged.

4. The determinant value remains unaltered if rows or columns are added or subtracted.

5. The determinant value is 0, if elements of any row or column all are 0.

6. The determinant of skew - symmetric matrix of odd order is 0

Answer 2

Solution:-

Given determinant is

$\begin{gathered}\begin{gathered}\sf \left | \begin{array}{ccc}1&a& {a}^{2} \\1&b& {b}^{2} \\1&c& {c}^{2} \end{array}\right | \end{gathered}\end{gathered} </p><p>$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\tt{OP \: R_2 \: \to \: R_2 - R_1 }} </p><p>$

$\begin{gathered}\rm \: = \: \begin{gathered}\sf \left | \begin{array}{ccc}1&a& {a}^{2} \\0&b - a& {b}^{2} - {a}^{2} \\1&c& {c}^{2} \end{array}\right | \end{gathered}\end{gathered} </p><p>$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\tt{OP \: R_3 \: \to \: R_3 - R_1 }} </p><p>$

$\begin{gathered}\rm \: = \: \begin{gathered}\sf \left | \begin{array}{ccc}1&a& {a}^{2} \\0&b - a& {b}^{2} - {a}^{2} \\0&c - a& {c}^{2} - {a}^{2} \end{array}\right | \end{gathered}\end{gathered} </p><p>$

We know,

$\begin{gathered}\boxed{\tt{ {x}^{2} - {y}^{2} = (x + y)(x - y) \: }} \\ \end{gathered} </p><p>$

So, using this identity, we get

$\begin{gathered}\rm \: = \: \begin{gathered}\sf \left | \begin{array}{ccc}1&a& {a}^{2} \\0&b - a& (b - a)(b + a) \\0&c - a&(c - a)(c + a) \end{array}\right | \end{gathered}\end{gathered} </p><p>$

Take out (b - a) and (c - a) common from second and third row respectively, we get

$\begin{gathered}\rm \: = \: (b - a)(c - a)\begin{gathered}\sf \left | \begin{array}{ccc}1&a& {a}^{2} \\0&1& b + a \\0&1&c + a \end{array}\right | \end{gathered}\end{gathered} </p><p>$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\tt{OP \: R_3 \: \to \: R_3 - R_2 }} </p><p>$

$\begin{gathered}\rm \: = \:(b - a)(c - a) \begin{gathered}\sf \left | \begin{array}{ccc}1&a& {a}^{2} \\0&1& b + a \\0&0&c - b \end{array}\right | \end{gathered}\end{gathered} </p><p>$

$\rm \: = \: (b - a)(c - a)(c - b)$

$\rm \: = \: [ - (a - b)](c - a)[ - (b - c)]$

$\rm \: = \: (a - b)(b - c)(c - a)$

Hence,

$\begin{gathered}\rm\implies \:\boxed{\tt{ \begin{gathered}\sf \left | \begin{array}{ccc}1&a& {a}^{2} \\1&b& {b}^{2} \\1&c& {c}^{2} \end{array}\right | \end{gathered} = (a - b)(b - c)(c - a)}} \\ \end{gathered} </p><p>$