Prove that|b²c² bc b+c||c^2a^2 ca c+a||a^2b^2 ab a+b| =0

1.	Prove that\|b²c² bc b+c\|\|c^2a^2 ca c+a\|\|a^2b^2 ab a+b\| =0
Answer» Given Question PROVE that $\rm \: \: \: \: \begin{gathered}\sf \left \| \begin{array}{ccc} {b}^{<klux>2</klux>} {c}^{2} &bc&b + c\\ {c}^{2} {a}^{2} &<klux>CA</klux>&c + a\\ {a}^{2} {b}^{2} &<klux>AB</klux>&a + b\end{array}\right \| \end{gathered} = 0$ Solution :- Consider, $\rm :\longmapsto\:\begin{gathered}\sf \left \| \begin{array}{ccc} {b}^{2} {c}^{2} &bc&b + c\\ {c}^{2} {a}^{2} &ca&c + a\\ {a}^{2} {b}^{2} &ab&a + b\end{array}\right \| \end{gathered}$ $\red{\rm :\longmapsto\:OP \: R_1 \to \: aR_1} \\ \red{\rm :\longmapsto\:OP \: R_2 \to \: bR_2} \\ \red{\rm :\longmapsto\:OP \: R_3 \to \: cR_3}$ $\: \: \rm = \: \: \begin{gathered}\sf \left \| \begin{array}{ccc} a{b}^{2} {c}^{2} &abc&ab + ac\\ b{c}^{2} {a}^{2} &bca&bc + ba\\ c{a}^{2} {b}^{2} &cab&ca + cb\end{array}\right \| \end{gathered}$ Take out abc common from Column 1 and column 2, we get $\: \: \rm = \: \: {a}^{2} {b}^{2} {c}^{2} \begin{gathered}\sf \left \| \begin{array}{ccc} {b}{c} &1&ab + ac\\ {c}{a} &1&bc + ab\\ {a}{b} &1&ac + bc\end{array}\right \| \end{gathered}$ $\red{\rm :\longmapsto\:OP \: C_3 \to \: C_3 + C_1}$ $\: \: \rm = \: \: {a}^{2} {b}^{2} {c}^{2} \begin{gathered}\sf \left \| \begin{array}{ccc} {b}{c} &1&ab + ac + bc\\ {c}{a} &1&bc + ab + ac\\ {a}{b} &1&ac + bc + ab\end{array}\right \| \end{gathered}$ Take out ab + bc + ca common from column 3, we get $\: \: \rm = \: \: {a}^{2} {b}^{2} {c}^{2}(ab + bc + ca) \begin{gathered}\sf \left \| \begin{array}{ccc} {b}{c} &1&1\\ {c}{a} &1&1\\ {a}{b} &1&1\end{array}\right \| \end{gathered}$ $\: \: \rm = \: \: {a}^{2} {b}^{2} {c}^{2}(ab + bc + ca) \times 0$ $\: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \: \because \: C_2 \: and \: C_3 \: are \: identical}$ $\: \: \rm = \: \: 0$ ${\boxed{\boxed{\bf{Hence, Proved}}}}$ Additional Information :- 1. The determinant remains unaltered if its rows are changed into columns and the columns into rows. This is known as the property of reflection. 2. If all the elements of a row (or column) are zero, then the determinant is zero. 3. If the all elements of a row (or column) are proportional (identical) to the elements of some other row (or column), then the determinant is zero. 4. The interchange of any two rows (or columns) of the determinant changes its sign. 5. If all the elements of a determinant above or below the main diagonal consist of zeros, then the determinant is EQUAL to the product of diagonal elements

Answer»

Given Question

$\rm \: \: \: \: \begin{gathered}\sf \left | \begin{array}{ccc} {b}^{<klux>2</klux>} {c}^{2} &bc&b + c\\ {c}^{2} {a}^{2} &<klux>CA</klux>&c + a\\ {a}^{2} {b}^{2} &<klux>AB</klux>&a + b\end{array}\right | \end{gathered} = 0$

Solution :-

Consider,

$\rm :\longmapsto\:\begin{gathered}\sf \left | \begin{array}{ccc} {b}^{2} {c}^{2} &bc&b + c\\ {c}^{2} {a}^{2} &ca&c + a\\ {a}^{2} {b}^{2} &ab&a + b\end{array}\right | \end{gathered}$

$\red{\rm :\longmapsto\:OP \: R_1 \to \: aR_1} \\ \red{\rm :\longmapsto\:OP \: R_2 \to \: bR_2} \\ \red{\rm :\longmapsto\:OP \: R_3 \to \: cR_3}$

$\: \: \rm = \: \: \begin{gathered}\sf \left | \begin{array}{ccc} a{b}^{2} {c}^{2} &abc&ab + ac\\ b{c}^{2} {a}^{2} &bca&bc + ba\\ c{a}^{2} {b}^{2} &cab&ca + cb\end{array}\right | \end{gathered}$

Take out abc common from Column 1 and column 2, we get

$\: \: \rm = \: \: {a}^{2} {b}^{2} {c}^{2} \begin{gathered}\sf \left | \begin{array}{ccc} {b}{c} &1&ab + ac\\ {c}{a} &1&bc + ab\\ {a}{b} &1&ac + bc\end{array}\right | \end{gathered}$

$\red{\rm :\longmapsto\:OP \: C_3 \to \: C_3 + C_1}$

$\: \: \rm = \: \: {a}^{2} {b}^{2} {c}^{2} \begin{gathered}\sf \left | \begin{array}{ccc} {b}{c} &1&ab + ac + bc\\ {c}{a} &1&bc + ab + ac\\ {a}{b} &1&ac + bc + ab\end{array}\right | \end{gathered}$

Take out ab + bc + ca common from column 3, we get

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

$\: \: \rm = \: \: {a}^{2} {b}^{2} {c}^{2}(ab + bc + ca) \times 0$

$\: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \: \because \: C_2 \: and \: C_3 \: are \: identical}$

$\: \: \rm = \: \: 0$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

1. The determinant remains unaltered if its rows are changed into columns and the columns into rows. This is known as the property of reflection.

2. If all the elements of a row (or column) are zero, then the determinant is zero.

3. If the all elements of a row (or column) are proportional (identical) to the elements of some other row (or column), then the determinant is zero.

4. The interchange of any two rows (or columns) of the determinant changes its sign.

5. If all the elements of a determinant above or below the main diagonal consist of zeros, then the determinant is EQUAL to the product of diagonal elements