Question

The value of determinant $ \left| \begin{matrix}
   ka & {{k}^{2}}+{{a}^{2}} & 1 \\
   kb & {{k}^{2}}+{{b}^{2}} & 1 \\
   kc & {{k}^{2}}+{{c}^{2}} & 1 \\
\end{matrix} \right| $ is
\[\begin{align}
  & A.k\left( a+b \right)\left( b+a \right)\left( c+a \right) \\
 & B.kabc\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right) \\
 & C.k\left( a-b \right)\left( b-a \right)\left( c-a \right) \\
 & D.k\left( a+b-c \right)\left( b+c-a \right)\left( c+a-b \right) \\
\end{align}\]

Answer 1

Hint:
We have been asked to find the value of the given determinant. So, we will first split the second row and we will get that, given determinant will split in two parts. Now we will find the value of each determinant one by one. We will need that, when two rows (column) are identical then, the value of the determinant is zero. We will use the formula $ \det \left( kA \right)=k\cdot \det A $ to get the required determinant.

Complete step by step answer:
We are given $ 3\times 3 $ determinant as $ \left| \begin{matrix}
   ka & {{k}^{2}}+{{a}^{2}} & 1 \\
   kb & {{k}^{2}}+{{b}^{2}} & 1 \\
   kc & {{k}^{2}}+{{c}^{2}} & 1 \\
\end{matrix} \right| $ .
We have to find the value of the above determinant. As we look at it, we see the second column which we will consider as two terms. We know if we split the determinant, it is still the same, that is $ Det\left( A+B \right)=Det\left( A \right)+Det\left( B \right) $ .
So we will split the given determinant as,
 $ \left| \begin{matrix}
   ka & {{k}^{2}}+{{a}^{2}} & 1 \\
   kb & {{k}^{2}}+{{b}^{2}} & 1 \\
   kc & {{k}^{2}}+{{c}^{2}} & 1 \\
\end{matrix} \right|=\left| \begin{matrix}
   ka & {{k}^{2}} & 1 \\
   kb & {{k}^{2}} & 1 \\
   kc & {{k}^{2}} & 1 \\
\end{matrix} \right|+\left| \begin{matrix}
   ka & {{a}^{2}} & 1 \\
   kb & {{b}^{2}} & 1 \\
   kc & {{c}^{2}} & 1 \\
\end{matrix} \right|\cdots \cdots \cdots \left( 1 \right) $ .
Now we will simplify the given determinant easily.
Firstly in the determinant $ \left| \begin{matrix}
   ka & {{k}^{2}} & 1 \\
   kb & {{k}^{2}} & 1 \\
   kc & {{k}^{2}} & 1 \\
\end{matrix} \right| $ we have $ {{k}^{2}} $ common in column 2 so we take it out. So we get $ {{k}^{2}}\left| \begin{matrix}
   ka & 1 & 1 \\
   kb & 1 & 1 \\
   kc & 1 & 1 \\
\end{matrix} \right| $ .
Now as we know that, if two rows and columns are the same then the value of the determinant is zero. So, $ \left| \begin{matrix}
   ka & {{k}^{2}}+{{a}^{2}} & 1 \\
   kb & {{k}^{2}}+{{b}^{2}} & 1 \\
   kc & {{k}^{2}}+{{c}^{2}} & 1 \\
\end{matrix} \right|=0 $ .
Using this in (1) we get, $ \left| \begin{matrix}
   ka & {{k}^{2}}+{{a}^{2}} & 1 \\
   kb & {{k}^{2}}+{{b}^{2}} & 1 \\
   kc & {{k}^{2}}+{{c}^{2}} & 1 \\
\end{matrix} \right|=0+\left| \begin{matrix}
   ka & {{a}^{2}} & 1 \\
   kb & {{b}^{2}} & 1 \\
   kc & {{c}^{2}} & 1 \\
\end{matrix} \right| $ .
Taking k common from column one we get $ k\left| \begin{matrix}
   a & {{a}^{2}} & 1 \\
   b & {{b}^{2}} & 1 \\
   c & {{c}^{2}} & 1 \\
\end{matrix} \right| $ .
Now we will use row operation to simplify our determinant. We apply \[{{R}_{1}}\to {{R}_{1}}-{{R}_{2}}\text{ and }{{R}_{2}}\to {{R}_{2}}-{{R}_{3}}\] so we get, $ k\left| \begin{matrix}
   a-b & {{a}^{2}}-{{b}^{2}} & 0 \\
   b-c & {{b}^{2}}-{{c}^{2}} & 0 \\
   c & {{c}^{2}} & 1 \\
\end{matrix} \right| $ .
Now we take a-b common from $ {{R}_{1}} $ and b-c common from $ {{R}_{2}} $ . So we get,
 $ k\left( a-b \right)\left( b-c \right)\left| \begin{matrix}
   1 & a+b & 0 \\
   1 & b+c & 0 \\
   c & {{c}^{2}} & 1 \\
\end{matrix} \right| $ .
Now we expand along $ {{c}_{3}} $ we get $ k\left( a-b \right)\left( b-c \right)\left[ 0+0+1\left[ \left( b+c \right)-\left( a+b \right) \right] \right] $ .
Simplifying we get $ k\left( a-b \right)\left( b-c \right)\left( c-a \right) $ .
So we get that the value of the required determinant is $ k\left( a-b \right)\left( b-c \right)\left( c-a \right) $ .
So option C is the correct answer.

Note:
Always remember that, if we take common something from rows and columns, then that number will be multiplied by the remaining determinant i.e. $ \det \left( kA \right)=k\cdot \det A $ . Remember that \[\left( b+c \right)-\left( a+b \right)\ne b+c-a+b\] the negative sign will be multiplied by both a and b so we get, \[\left( b+c \right)-\left( a+b \right)=b+c-a-b=c-a\].