Question

Consider the following matrix $\left| \begin{matrix}
   l & m & n \\
   p & q & r \\
   s & t & u \\
\end{matrix} \right|=A$, then $\left| \begin{matrix}
   2l & 2m & 2n \\
   2p & 2q & 2r \\
   2s & 2t & 2u \\
\end{matrix} \right|=?$

Answer 1

Hint: We know that the determinant of a $3\times 3$ matrix can be written as $\left| \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right|=a\left( ei-hf \right)-b\left( di-gf \right)+c\left( dh-ge \right)$. From this formula we will find the value of $A$ from given data. Now we have to find the value of $\left| \begin{matrix}
   2l & 2m & 2n \\
   2p & 2q & 2r \\
   2s & 2t & 2u \\
\end{matrix} \right|$, 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 $A$ to get the result.

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

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