# Prove the following

\[\left| {\begin{array}{*{20}{c}}

{{{\left( {b + c} \right)}^2}}&{{a^2}}&{{a^2}} \\

{{b^2}}&{{{(c + a)}^2}}&{{b^2}} \\

{{c^2}}&{{c^2}}&{{{(a + b)}^2}}

\end{array}} \right| = 2abc{(a + b + c)^3}\]

Last updated date: 18th Mar 2023

•

Total views: 306.6k

•

Views today: 6.85k

Answer

Verified

306.6k+ views

Hint: Here we perform various operations on rows and columns of determinant to make it simple.

Taking the left-hand side of the questions and solving it further so that the left-hand side will become

equal to the right-hand side. So,

$ \Rightarrow $L. H. S =\[\left| {\begin{array}{*{20}{c}}

{{{\left( {b + c} \right)}^2}}&{{a^2}}&{{a^2}} \\

{{b^2}}&{{{(c + a)}^2}}&{{b^2}} \\

{{c^2}}&{{c^2}}&{{{(a + b)}^2}}

\end{array}} \right|\]

Now, to simplify the determinant we will do various operations so that the determinant becomes

easy and we can expand the determinant without any error. Expanding the determinant can be done

before simplification but, it will make the solution tedious and complicated. So, we will simplify the

determinant and then expand it. So, for simplification we will first subtract the column \[{C_3}\]from

the column ${C_1}$.

Applying ${C_1} \to {C_1} - {C_3}$

$ \Rightarrow $L. H. S = $\left| {\begin{array}{*{20}{c}}

{{{(b + c)}^2} - {a^2}}&{{a^2}}&{{a^2}} \\

{{b^2} - {b^2}}&{{{(c + a)}^2}}&{{b^2}} \\

{{c^2} - {{(a + b)}^2}}&{{c^2}}&{{{(a + b)}^2}}

\end{array}} \right|$

Also applying ${C_2} \to {C_2} - {C_3}$

$ \Rightarrow $L. H. S = $\left| {\begin{array}{*{20}{c}}

{{{(b + c)}^2} - {a^2}}&{{a^2} - {a^2}}&{{a^2}} \\

{{b^2} - {b^2}}&{{{(c + a)}^2} - {b^2}}&{{b^2}} \\

{{c^2} - {{(a + b)}^2}}&{{c^2} - {{(a + b)}^2}}&{{{(a + b)}^2}}

\end{array}} \right|$ ……………... (1)

Taking \[(a + b + c)\] common from column ${C_1}$and column \[{C_2}\]

$ \Rightarrow $L. H. S = ${(a + b + c)^2}\left| {\begin{array}{*{20}{c}}

{b + c - a}&0&{{a^2}} \\

0&{c + a - b}&{{b^2}} \\

{c - (a + b)}&{c - (a + b)}&{{{(a + b)}^2}}

\end{array}} \right|$

Also, applying ${R_3} \to {R_3} - {R_1} - {R_2}$

$ \Rightarrow $L. H. S = \[{(a + b + c)^2}\left| {\begin{array}{*{20}{c}}

{b + c - a}&0&{{a^2}} \\

0&{c + a - b}&{{b^2}} \\

{ - 2b}&{ - 2a}&{2ab}

\end{array}} \right|\]

Now, multiply and divide column ${C_1}$ by $a$ and column \[{C_2}\] by $b$.

$ \Rightarrow $L. H. S = \[\dfrac{{{{(a + b + c)}^2}}}{{ab}}\left| {\begin{array}{*{20}{c}}

{ab + ac - {a^2}}&0&{{a^2}} \\

0&{bc + ab - {b^2}}&{{b^2}} \\

{ - 2ab}&{ - 2ab}&{2ab}

\end{array}} \right|\]

Now, doing ${C_1} \to {C_1} + {C_3}$and ${C_2} \to {C_2} + {C_3}$

$ \Rightarrow $L. H. S = \[\dfrac{{{{(a + b + c)}^2}}}{{ab}}\left| {\begin{array}{*{20}{c}}

{ab + ac - {a^2} + {a^2}}&{{a^2}}&{{a^2}} \\

{{b^2}}&{bc + ab - {b^2} + {b^2}}&{{b^2}} \\

{ - 2ab + 2ab}&{ - 2ab + 2ab}&{2ab}

\end{array}} \right|\]

$ \Rightarrow $L. H. S = \[\dfrac{{{{(a + b + c)}^2}}}{{ab}}\left| {\begin{array}{*{20}{c}}

{ab + ac}&{{a^2}}&{{a^2}} \\

{{b^2}}&{bc + ab}&{{b^2}} \\

0&0&{2ab}

\end{array}} \right|\]

Now, our determinant has become simple. So, expanding determinant through row \[{R_3}\],

$ \Rightarrow $L. H. S = \[\dfrac{{{{(a + b + c)}^2}}}{{ab}}(2ab)\left| {\begin{array}{*{20}{c}}

{ab + ac}&{{a^2}} \\

{{b^2}}&{bc + ab}

\end{array}} \right|\]

$ \Rightarrow $L. H. S = \[\dfrac{{{{(a + b + c)}^2}}}{{ab}}(2ab)(ab)\{ (b + c)(c + a) - ab\} \]

$ \Rightarrow $L. H. S = \[\dfrac{{{{(a + b + c)}^2}}}{{ab}}(2ab)(ab)(bc + ab + {c^2} + ac - ab)\]

$ \Rightarrow $L. H. S = \[2abc{(a + b + c)^3}\]= R. H. S

Hence Proved.

Note: Such problems are easy but require a lot of concentration while doing. If there is lack of

concentration, then the problem may not be solved. Also, properties of determinant are important to

solve problems but without them the problem can be solved but the process is very complicated and

tedious as it includes many terms. Make the determinant as simple as possible to easily expand it.

Taking the left-hand side of the questions and solving it further so that the left-hand side will become

equal to the right-hand side. So,

$ \Rightarrow $L. H. S =\[\left| {\begin{array}{*{20}{c}}

{{{\left( {b + c} \right)}^2}}&{{a^2}}&{{a^2}} \\

{{b^2}}&{{{(c + a)}^2}}&{{b^2}} \\

{{c^2}}&{{c^2}}&{{{(a + b)}^2}}

\end{array}} \right|\]

Now, to simplify the determinant we will do various operations so that the determinant becomes

easy and we can expand the determinant without any error. Expanding the determinant can be done

before simplification but, it will make the solution tedious and complicated. So, we will simplify the

determinant and then expand it. So, for simplification we will first subtract the column \[{C_3}\]from

the column ${C_1}$.

Applying ${C_1} \to {C_1} - {C_3}$

$ \Rightarrow $L. H. S = $\left| {\begin{array}{*{20}{c}}

{{{(b + c)}^2} - {a^2}}&{{a^2}}&{{a^2}} \\

{{b^2} - {b^2}}&{{{(c + a)}^2}}&{{b^2}} \\

{{c^2} - {{(a + b)}^2}}&{{c^2}}&{{{(a + b)}^2}}

\end{array}} \right|$

Also applying ${C_2} \to {C_2} - {C_3}$

$ \Rightarrow $L. H. S = $\left| {\begin{array}{*{20}{c}}

{{{(b + c)}^2} - {a^2}}&{{a^2} - {a^2}}&{{a^2}} \\

{{b^2} - {b^2}}&{{{(c + a)}^2} - {b^2}}&{{b^2}} \\

{{c^2} - {{(a + b)}^2}}&{{c^2} - {{(a + b)}^2}}&{{{(a + b)}^2}}

\end{array}} \right|$ ……………... (1)

Taking \[(a + b + c)\] common from column ${C_1}$and column \[{C_2}\]

$ \Rightarrow $L. H. S = ${(a + b + c)^2}\left| {\begin{array}{*{20}{c}}

{b + c - a}&0&{{a^2}} \\

0&{c + a - b}&{{b^2}} \\

{c - (a + b)}&{c - (a + b)}&{{{(a + b)}^2}}

\end{array}} \right|$

Also, applying ${R_3} \to {R_3} - {R_1} - {R_2}$

$ \Rightarrow $L. H. S = \[{(a + b + c)^2}\left| {\begin{array}{*{20}{c}}

{b + c - a}&0&{{a^2}} \\

0&{c + a - b}&{{b^2}} \\

{ - 2b}&{ - 2a}&{2ab}

\end{array}} \right|\]

Now, multiply and divide column ${C_1}$ by $a$ and column \[{C_2}\] by $b$.

$ \Rightarrow $L. H. S = \[\dfrac{{{{(a + b + c)}^2}}}{{ab}}\left| {\begin{array}{*{20}{c}}

{ab + ac - {a^2}}&0&{{a^2}} \\

0&{bc + ab - {b^2}}&{{b^2}} \\

{ - 2ab}&{ - 2ab}&{2ab}

\end{array}} \right|\]

Now, doing ${C_1} \to {C_1} + {C_3}$and ${C_2} \to {C_2} + {C_3}$

$ \Rightarrow $L. H. S = \[\dfrac{{{{(a + b + c)}^2}}}{{ab}}\left| {\begin{array}{*{20}{c}}

{ab + ac - {a^2} + {a^2}}&{{a^2}}&{{a^2}} \\

{{b^2}}&{bc + ab - {b^2} + {b^2}}&{{b^2}} \\

{ - 2ab + 2ab}&{ - 2ab + 2ab}&{2ab}

\end{array}} \right|\]

$ \Rightarrow $L. H. S = \[\dfrac{{{{(a + b + c)}^2}}}{{ab}}\left| {\begin{array}{*{20}{c}}

{ab + ac}&{{a^2}}&{{a^2}} \\

{{b^2}}&{bc + ab}&{{b^2}} \\

0&0&{2ab}

\end{array}} \right|\]

Now, our determinant has become simple. So, expanding determinant through row \[{R_3}\],

$ \Rightarrow $L. H. S = \[\dfrac{{{{(a + b + c)}^2}}}{{ab}}(2ab)\left| {\begin{array}{*{20}{c}}

{ab + ac}&{{a^2}} \\

{{b^2}}&{bc + ab}

\end{array}} \right|\]

$ \Rightarrow $L. H. S = \[\dfrac{{{{(a + b + c)}^2}}}{{ab}}(2ab)(ab)\{ (b + c)(c + a) - ab\} \]

$ \Rightarrow $L. H. S = \[\dfrac{{{{(a + b + c)}^2}}}{{ab}}(2ab)(ab)(bc + ab + {c^2} + ac - ab)\]

$ \Rightarrow $L. H. S = \[2abc{(a + b + c)^3}\]= R. H. S

Hence Proved.

Note: Such problems are easy but require a lot of concentration while doing. If there is lack of

concentration, then the problem may not be solved. Also, properties of determinant are important to

solve problems but without them the problem can be solved but the process is very complicated and

tedious as it includes many terms. Make the determinant as simple as possible to easily expand it.

Recently Updated Pages

If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

KCN reacts readily to give a cyanide with A Ethyl alcohol class 12 chemistry JEE_Main

Trending doubts

What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?