# If $s = p + q + r$, then value of $\left| {\begin{array}{*{20}{c}}

{s + r}&p&q \\

r&{s + p}&q \\

r&p&{s + q}

\end{array}} \right|$ is ……………

$

A.{\text{ }}{s^3} \\

B.{\text{ 2}}{s^3} \\

C.{\text{ 2}}{s^2} \\

D.{\text{ 3}}{s^3} \\

$

Last updated date: 26th Mar 2023

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Answer

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308.7k+ views

Hint: In this question apply determinant rule without opening the determinant doing this we can easily solve the determinant so, first of all add the second and third column in the first column.

Given condition is

$s = p + q + r$

Then we have to find out the value of

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

{s + r}&p&q \\

r&{s + p}&q \\

r&p&{s + q}

\end{array}} \right|$

Now simplify the determinant using determinant rule and apply,

${c_1} \to {c_1} + {c_2} + {c_3}$, we have

Therefore above determinant becomes

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

{s + p + q + r}&p&q \\

{s + p + q + r}&{s + p}&q \\

{s + p + q + r}&p&{s + q}

\end{array}} \right|$

Now take $\left( {s + p + q + r} \right)$ outside the determinant from every row we have

$\left( {s + p + q + r} \right)\left| {\begin{array}{*{20}{c}}

1&p&q \\

1&{s + p}&q \\

1&p&{s + q}

\end{array}} \right|$

Now again simplify the determinant using determinant rule and apply,

${R_2} \to {R_2} - {R_1},{\text{ }}{R_3} \to {R_3} - {R_1}$, we have

$\left( {s + p + q + r} \right)\left| {\begin{array}{*{20}{c}}

1&p&q \\

0&s&0 \\

0&0&s

\end{array}} \right|$

Now open the determinant we have

$\left( {s + p + q + r} \right)\left[ {1\left| {\begin{array}{*{20}{c}}

s&0 \\

0&s

\end{array}} \right| - p\left| {\begin{array}{*{20}{c}}

0&0 \\

0&s

\end{array}} \right| + q\left| {\begin{array}{*{20}{c}}

0&s \\

0&0

\end{array}} \right|} \right]$

$

\left( {s + p + q + r} \right)\left[ {1\left( {{s^2} - 0} \right) - p\left( 0 \right) + q\left( 0 \right)} \right] \\

= \left( {s + p + q + r} \right)\left( {{s^2}} \right) \\

$

Now it is given that $s = p + q + r$, so above equation becomes,

$

\therefore \left( {s + s} \right)\left( {{s^2}} \right) \\

= 2s\left( {{s^2}} \right) = 2{s^3} \\

$

Hence, option (b) is correct.

Note: In such types of questions solve the determinant without opening the determinant if we direct open the determinant it will lead us to a very complex situation that will not help us so, first simplify the determinant using determinant rules as above, then expand the determinant as above after this substitute the given value, we will get the required answer.

Given condition is

$s = p + q + r$

Then we have to find out the value of

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

{s + r}&p&q \\

r&{s + p}&q \\

r&p&{s + q}

\end{array}} \right|$

Now simplify the determinant using determinant rule and apply,

${c_1} \to {c_1} + {c_2} + {c_3}$, we have

Therefore above determinant becomes

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

{s + p + q + r}&p&q \\

{s + p + q + r}&{s + p}&q \\

{s + p + q + r}&p&{s + q}

\end{array}} \right|$

Now take $\left( {s + p + q + r} \right)$ outside the determinant from every row we have

$\left( {s + p + q + r} \right)\left| {\begin{array}{*{20}{c}}

1&p&q \\

1&{s + p}&q \\

1&p&{s + q}

\end{array}} \right|$

Now again simplify the determinant using determinant rule and apply,

${R_2} \to {R_2} - {R_1},{\text{ }}{R_3} \to {R_3} - {R_1}$, we have

$\left( {s + p + q + r} \right)\left| {\begin{array}{*{20}{c}}

1&p&q \\

0&s&0 \\

0&0&s

\end{array}} \right|$

Now open the determinant we have

$\left( {s + p + q + r} \right)\left[ {1\left| {\begin{array}{*{20}{c}}

s&0 \\

0&s

\end{array}} \right| - p\left| {\begin{array}{*{20}{c}}

0&0 \\

0&s

\end{array}} \right| + q\left| {\begin{array}{*{20}{c}}

0&s \\

0&0

\end{array}} \right|} \right]$

$

\left( {s + p + q + r} \right)\left[ {1\left( {{s^2} - 0} \right) - p\left( 0 \right) + q\left( 0 \right)} \right] \\

= \left( {s + p + q + r} \right)\left( {{s^2}} \right) \\

$

Now it is given that $s = p + q + r$, so above equation becomes,

$

\therefore \left( {s + s} \right)\left( {{s^2}} \right) \\

= 2s\left( {{s^2}} \right) = 2{s^3} \\

$

Hence, option (b) is correct.

Note: In such types of questions solve the determinant without opening the determinant if we direct open the determinant it will lead us to a very complex situation that will not help us so, first simplify the determinant using determinant rules as above, then expand the determinant as above after this substitute the given value, we will get the required answer.

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