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Question: Consider the following matrix \(\left| \begin{matrix} l & m & n \\\ p & q & r \\\ s &...

Consider the following matrix ∣lmn pqr stu ∣=A\left| \begin{matrix} l & m & n \\\ p & q & r \\\ s & t & u \\\ \end{matrix} \right|=A, then ∣2l2m2n 2p2q2r 2s2t2u ∣=?\left| \begin{matrix} 2l & 2m & 2n \\\ 2p & 2q & 2r \\\ 2s & 2t & 2u \\\ \end{matrix} \right|=?

Explanation

Solution

We know that the determinant of a 3×33\times 3 matrix can be written as ∣abc def ghi ∣=a(ei−hf)−b(di−gf)+c(dh−ge)\left| \begin{matrix} a & b & c \\\ d & e & f \\\ g & h & i \\\ \end{matrix} \right|=a\left( ei-hf \right)-b\left( di-gf \right)+c\left( dh-ge \right). From this formula we will find the value of AA from given data. Now we have to find the value of ∣2l2m2n 2p2q2r 2s2t2u ∣\left| \begin{matrix} 2l & 2m & 2n \\\ 2p & 2q & 2r \\\ 2s & 2t & 2u \\\ \end{matrix} \right|, here also we will use the determinant formula and we will calculate the value of determinant and further we can simplify it and use the value of AA to get the result.

Complete step-by-step solution:
Given that, ∣lmn pqr stu ∣=A\left| \begin{matrix} l & m & n \\\ p & q & r \\\ s & t & u \\\ \end{matrix} \right|=A
We know that ∣abc def ghi ∣=a(ei−hf)−b(di−gf)+c(dh−ge)\left| \begin{matrix} a & b & c \\\ d & e & f \\\ g & h & i \\\ \end{matrix} \right|=a\left( ei-hf \right)-b\left( di-gf \right)+c\left( dh-ge \right), hence
∣lmn pqr stu ∣=A l(qu−tr)−m(pu−sr)+n(pt−sq)=A....(i) \begin{aligned} & \left| \begin{matrix} l & m & n \\\ p & q & r \\\ s & t & u \\\ \end{matrix} \right|=A \\\ & l\left( qu-tr \right)-m\left( pu-sr \right)+n\left( pt-sq \right)=A....\left( \text{i} \right) \\\ \end{aligned}
Now the value of ∣2l2m2n 2p2q2r 2s2t2u ∣\left| \begin{matrix} 2l & 2m & 2n \\\ 2p & 2q & 2r \\\ 2s & 2t & 2u \\\ \end{matrix} \right| is given by
∣2l2m2n 2p2q2r 2s2t2u ∣=2l[(2q)(2u)−(2t)(2r)]−2m[(2p)(2u)−(2s)(2r)]+2n[(2p)(2t)−(2q)(2s)]\left| \begin{matrix} 2l & 2m & 2n \\\ 2p & 2q & 2r \\\ 2s & 2t & 2u \\\ \end{matrix} \right|=2l\left[ \left( 2q \right)\left( 2u \right)-\left( 2t \right)\left( 2r \right) \right]-2m\left[ \left( 2p \right)\left( 2u \right)-\left( 2s \right)\left( 2r \right) \right]+2n\left[ \left( 2p \right)\left( 2t \right)-\left( 2q \right)\left( 2s \right) \right]
We know that (a)(b)=a×b\left( a \right)\left( b \right)=a\times b, then we will get
∣2l2m2n 2p2q2r 2s2t2u ∣=2l[4qu−4tr]−2m[4pu−4sr]+2n[4pt−4qs]\left| \begin{matrix} 2l & 2m & 2n \\\ 2p & 2q & 2r \\\ 2s & 2t & 2u \\\ \end{matrix} \right|=2l\left[ 4qu-4tr \right]-2m\left[ 4pu-4sr \right]+2n\left[ 4pt-4qs \right]
Taking common 44 from the terms [4qu−4tr]\left[ 4qu-4tr \right], [4pu−4sr]\left[ 4pu-4sr \right], [4pt−4qs]\left[ 4pt-4qs \right], then we will have
∣2l2m2n 2p2q2r 2s2t2u ∣=2l(4)[qu−tr]−2m(4)[pu−sr]+2n(4)[pt−qs]\left| \begin{matrix} 2l & 2m & 2n \\\ 2p & 2q & 2r \\\ 2s & 2t & 2u \\\ \end{matrix} \right|=2l\left( 4 \right)\left[ qu-tr \right]-2m\left( 4 \right)\left[ pu-sr \right]+2n\left( 4 \right)\left[ pt-qs \right]
Multiplying the terms 2l(4)2l\left( 4 \right), 2m(4)2m\left( 4 \right), 2n(4)2n\left( 4 \right), then we will have
∣2l2m2n 2p2q2r 2s2t2u ∣=8l[qu−tr]−8m[pu−sr]+8n[pt−qs]\left| \begin{matrix} 2l & 2m & 2n \\\ 2p & 2q & 2r \\\ 2s & 2t & 2u \\\ \end{matrix} \right|=8l\left[ qu-tr \right]-8m\left[ pu-sr \right]+8n\left[ pt-qs \right]
Taking common 88 from terms 8l[qu−tr]8l\left[ qu-tr \right], 8m[pu−sr]8m\left[ pu-sr \right], 8n[pt−qs]8n\left[ pt-qs \right], then we will get
∣2l2m2n 2p2q2r 2s2t2u ∣=8[l(qu−tr)−m(pu−sr)+n(pt−qs)]\left| \begin{matrix} 2l & 2m & 2n \\\ 2p & 2q & 2r \\\ 2s & 2t & 2u \\\ \end{matrix} \right|=8\left[ l\left( qu-tr \right)-m\left( pu-sr \right)+n\left( pt-qs \right) \right]
From equation (i)\left( \text{i} \right) we have the value l(qu−tr)−m(pu−sr)+n(pt−sq)=Al\left( qu-tr \right)-m\left( pu-sr \right)+n\left( pt-sq \right)=A, substituting this value in the above equation, then we will get
∣2l2m2n 2p2q2r 2s2t2u ∣=8A\left| \begin{matrix} 2l & 2m & 2n \\\ 2p & 2q & 2r \\\ 2s & 2t & 2u \\\ \end{matrix} \right|=8A
Hence the value of ∣2l2m2n 2p2q2r 2s2t2u ∣\left| \begin{matrix} 2l & 2m & 2n \\\ 2p & 2q & 2r \\\ 2s & 2t & 2u \\\ \end{matrix} \right| is 8A8A where A=∣lmn pqr stu ∣A=\left| \begin{matrix} l & m & n \\\ p & q & r \\\ s & t & u \\\ \end{matrix} \right|

Note: We can also solve the above problem with the rule of the determinate i.e. k∣abc def ghi ∣=∣kakbkc def ghi ∣k\left| \begin{matrix} a & b & c \\\ d & e & f \\\ g & h & i \\\ \end{matrix} \right|=\left| \begin{matrix} ka & kb & kc \\\ d & e & f \\\ g & h & i \\\ \end{matrix} \right| or k∣abc def ghi ∣=∣abc kdkekf ghi ∣k\left| \begin{matrix} a & b & c \\\ d & e & f \\\ g & h & i \\\ \end{matrix} \right|=\left| \begin{matrix} a & b & c \\\ kd & ke & kf \\\ g & h & i \\\ \end{matrix} \right| and also k∣abc def ghi ∣=∣abc def kgkhki ∣k\left| \begin{matrix} a & b & c \\\ d & e & f \\\ g & h & i \\\ \end{matrix} \right|=\left| \begin{matrix} a & b & c \\\ d & e & f \\\ kg & kh & ki \\\ \end{matrix} \right|or vice versa.
Here we have to find the value of ∣2l2m2n 2p2q2r 2s2t2u ∣\left| \begin{matrix} 2l & 2m & 2n \\\ 2p & 2q & 2r \\\ 2s & 2t & 2u \\\ \end{matrix} \right|
Taking 22 common from the first row then
∣2l2m2n 2p2q2r 2s2t2u ∣=2∣lmn 2p2q2r 2s2t2u ∣\left| \begin{matrix} 2l & 2m & 2n \\\ 2p & 2q & 2r \\\ 2s & 2t & 2u \\\ \end{matrix} \right|=2\left| \begin{matrix} l & m & n \\\ 2p & 2q & 2r \\\ 2s & 2t & 2u \\\ \end{matrix} \right| since we have k∣abc def ghi ∣=∣kakbkc def ghi ∣k\left| \begin{matrix} a & b & c \\\ d & e & f \\\ g & h & i \\\ \end{matrix} \right|=\left| \begin{matrix} ka & kb & kc \\\ d & e & f \\\ g & h & i \\\ \end{matrix} \right|
Taking 22 common from the second row then
∣2l2m2n 2p2q2r 2s2t2u ∣=2∣lmn 2p2q2r 2s2t2u ∣ =2×2∣lmn pqr 2s2t2u ∣ \begin{aligned} & \left| \begin{matrix} 2l & 2m & 2n \\\ 2p & 2q & 2r \\\ 2s & 2t & 2u \\\ \end{matrix} \right|=2\left| \begin{matrix} l & m & n \\\ 2p & 2q & 2r \\\ 2s & 2t & 2u \\\ \end{matrix} \right| \\\ & =2\times 2\left| \begin{matrix} l & m & n \\\ p & q & r \\\ 2s & 2t & 2u \\\ \end{matrix} \right| \\\ \end{aligned}
Now Taking 22 common from the third row then
∣2l2m2n 2p2q2r 2s2t2u ∣=2×2∣lmn pqr 2s2t2u ∣ =4×2∣lmn pqr stu ∣ =8∣lmn pqr stu ∣ \begin{aligned} & \left| \begin{matrix} 2l & 2m & 2n \\\ 2p & 2q & 2r \\\ 2s & 2t & 2u \\\ \end{matrix} \right|=2\times 2\left| \begin{matrix} l & m & n \\\ p & q & r \\\ 2s & 2t & 2u \\\ \end{matrix} \right| \\\ & =4\times 2\left| \begin{matrix} l & m & n \\\ p & q & r \\\ s & t & u \\\ \end{matrix} \right| \\\ & =8\left| \begin{matrix} l & m & n \\\ p & q & r \\\ s & t & u \\\ \end{matrix} \right| \\\ \end{aligned}
In the problem we have given that ∣lmn pqr stu ∣=A\left| \begin{matrix} l & m & n \\\ p & q & r \\\ s & t & u \\\ \end{matrix} \right|=A, substituting this value in above equation, then we will get
∣2l2m2n 2p2q2r 2s2t2u ∣=8∣lmn pqr stu ∣ =8A \begin{aligned} & \left| \begin{matrix} 2l & 2m & 2n \\\ 2p & 2q & 2r \\\ 2s & 2t & 2u \\\ \end{matrix} \right|=8\left| \begin{matrix} l & m & n \\\ p & q & r \\\ s & t & u \\\ \end{matrix} \right| \\\ & =8A \\\ \end{aligned}
Hence the value of ∣2l2m2n 2p2q2r 2s2t2u ∣\left| \begin{matrix} 2l & 2m & 2n \\\ 2p & 2q & 2r \\\ 2s & 2t & 2u \\\ \end{matrix} \right| is 8A8A where A=∣lmn pqr stu ∣A=\left| \begin{matrix} l & m & n \\\ p & q & r \\\ s & t & u \\\ \end{matrix} \right|
From both the methods we got the same result.