Question

If determinant \[f\left( x \right)=\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{3}} & 1 \\
\end{matrix} \\
\end{align} \right|\] then \[f'\left( x \right)=\lambda \left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{2}} & 1 \\
\end{matrix} \\
\end{align} \right|\] . Find the value of \[\lambda \] .

Answer 1

Hint: We have to differentiate \[f\left( x \right)\] . Write \[\Rightarrow \dfrac{df\left( x \right)}{dx}=\left| \begin{align}
  & \begin{matrix}
   \dfrac{d{{\left( x-a \right)}^{4}}}{dx} & {{\left( x-a \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   \dfrac{d{{\left( x-b \right)}^{4}}}{dx} & {{\left( x-b \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   \dfrac{d{{\left( x-c \right)}^{4}}}{dx} & {{\left( x-c \right)}^{3}} & 1 \\
\end{matrix} \\
\end{align} \right|\] + \[\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & \dfrac{d{{\left( x-a \right)}^{3}}}{dx} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & \dfrac{d{{\left( x-b \right)}^{3}}}{dx} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & \dfrac{d{{\left( x-c \right)}^{3}}}{dx} & 1 \\
\end{matrix} \\
\end{align} \right|\] + \[\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{3}} & \dfrac{d\left( 1 \right)}{dx} \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{3}} & \dfrac{d\left( 1 \right)}{dx} \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{3}} & \dfrac{d\left( 1 \right)}{dx} \\
\end{matrix} \\
\end{align} \right|\] . Now, simplify \[\dfrac{df\left( x \right)}{dx}\] using the formula, \[\dfrac{d{{\left( x-a \right)}^{n}}}{dx}=n{{\left( x-a \right)}^{n-1}}\] . We also know that the differentiation of the constant term is 0. Expand the determinant along the third column and get the determinant value of \[\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{3}} & 0 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{3}} & 0 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{3}} & 0 \\
\end{matrix} \\
\end{align} \right|\] . We know the property that if two rows or columns of a determinant are the same and identical then the determinant value of that determinant is equal to zero. So, the value of the determinant \[4\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{3}} & {{\left( x-a \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{3}} & {{\left( x-b \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{3}} & {{\left( x-c \right)}^{3}} & 1 \\
\end{matrix} \\
\end{align} \right|\] is equal to zero. Now, compare \[3\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{2}} & 1 \\
\end{matrix} \\
\end{align} \right|\] and \[\lambda \left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{2}} & 1 \\
\end{matrix} \\
\end{align} \right|\] , and get the value of \[\lambda \] ,

Complete step by step solution:
According to the question, we have a function which is given in the form of the determinant. We also have the derivative of the function \[f\left( x \right)\] in the form of the determinant.
\[f\left( x \right)=\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{3}} & 1 \\
\end{matrix} \\
\end{align} \right|\] …………………………….(1)
\[f'\left( x \right)=\lambda \left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{2}} & 1 \\
\end{matrix} \\
\end{align} \right|\] …………………………………(2)
We know that \[f'\left( x \right)\] is the derivative of \[f\left( x \right)\] . So,
\[f'\left( x \right)=\dfrac{df\left( x \right)}{dx}\] ……………………..(3)
From equation (1), we have \[f\left( x \right)\] .
Now, differentiating \[f\left( x \right)\] , we get
\[\dfrac{df\left( x \right)}{dx}=\dfrac{d\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{3}} & 1 \\
\end{matrix} \\
\end{align} \right|}{dx}\]

\[\Rightarrow \dfrac{df\left( x \right)}{dx}=\left| \begin{align}
  & \begin{matrix}
   \dfrac{d{{\left( x-a \right)}^{4}}}{dx} & {{\left( x-a \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   \dfrac{d{{\left( x-b \right)}^{4}}}{dx} & {{\left( x-b \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   \dfrac{d{{\left( x-c \right)}^{4}}}{dx} & {{\left( x-c \right)}^{3}} & 1 \\
\end{matrix} \\
\end{align} \right|\] + \[\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & \dfrac{d{{\left( x-a \right)}^{3}}}{dx} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & \dfrac{d{{\left( x-b \right)}^{3}}}{dx} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & \dfrac{d{{\left( x-c \right)}^{3}}}{dx} & 1 \\
\end{matrix} \\
\end{align} \right|\] + \[\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{3}} & \dfrac{d\left( 1 \right)}{dx} \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{3}} & \dfrac{d\left( 1 \right)}{dx} \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{3}} & \dfrac{d\left( 1 \right)}{dx} \\
\end{matrix} \\
\end{align} \right|\] …………………………………..(4)
We know the formula, \[\dfrac{d{{\left( x-a \right)}^{n}}}{dx}=n{{\left( x-a \right)}^{n-1}}\] ………………………………(5)
We also know that the differentiation of the constant term is 0 …………………………..(6)
Now, using equation (5) and equation (6), and simplifying equation (4), we get
\[\Rightarrow \dfrac{df\left( x \right)}{dx}=\left| \begin{align}
  & \begin{matrix}
   4{{\left( x-a \right)}^{4-1}} & {{\left( x-a \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   4{{\left( x-b \right)}^{4-1}} & {{\left( x-b \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   4{{\left( x-c \right)}^{4-1}} & {{\left( x-c \right)}^{3}} & 1 \\
\end{matrix} \\
\end{align} \right|\] + \[\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & 3{{\left( x-a \right)}^{3-1}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & 3{{\left( x-b \right)}^{3-1}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & 3{{\left( x-c \right)}^{3-1}} & 1 \\
\end{matrix} \\
\end{align} \right|\] + \[\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{3}} & 0 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{3}} & 0 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{3}} & 0 \\
\end{matrix} \\
\end{align} \right|\]
\[\Rightarrow \dfrac{df\left( x \right)}{dx}=\left| \begin{align}
  & \begin{matrix}
   4{{\left( x-a \right)}^{3}} & {{\left( x-a \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   4{{\left( x-b \right)}^{3}} & {{\left( x-b \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   4{{\left( x-c \right)}^{3}} & {{\left( x-c \right)}^{3}} & 1 \\
\end{matrix} \\
\end{align} \right|\] + \[\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & 3{{\left( x-a \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & 3{{\left( x-b \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & 3{{\left( x-c \right)}^{2}} & 1 \\
\end{matrix} \\
\end{align} \right|\] + \[\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{3}} & 0 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{3}} & 0 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{3}} & 0 \\
\end{matrix} \\
\end{align} \right|\] ………………………………..(7)
Expanding the determinant \[\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{3}} & 0 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{3}} & 0 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{3}} & 0 \\
\end{matrix} \\
\end{align} \right|\] along the column 3, we get the determinant value of the determinant \[\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{3}} & 0 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{3}} & 0 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{3}} & 0 \\
\end{matrix} \\
\end{align} \right|\] is equal to zero. So, \[\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{3}} & 0 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{3}} & 0 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{3}} & 0 \\
\end{matrix} \\
\end{align} \right|=0\] …………………………….(8)
Now, from equation (7) and equation (8), we get
\[\Rightarrow \dfrac{df\left( x \right)}{dx}=\left| \begin{align}
  & \begin{matrix}
   4{{\left( x-a \right)}^{3}} & {{\left( x-a \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   4{{\left( x-b \right)}^{3}} & {{\left( x-b \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   4{{\left( x-c \right)}^{3}} & {{\left( x-c \right)}^{3}} & 1 \\
\end{matrix} \\
\end{align} \right|\]+ \[\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & 3{{\left( x-a \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & 3{{\left( x-b \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & 3{{\left( x-c \right)}^{2}} & 1 \\
\end{matrix} \\
\end{align} \right|\] + 0
\[\Rightarrow \dfrac{df\left( x \right)}{dx}=4\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{3}} & {{\left( x-a \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{3}} & {{\left( x-b \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{3}} & {{\left( x-c \right)}^{3}} & 1 \\
\end{matrix} \\
\end{align} \right|\]+ \[3\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{2}} & 1 \\
\end{matrix} \\
\end{align} \right|\] ……………………………………(9)
We know the property that if two rows or columns of a determinant is same and identical then the determinant value of that determinant is equal to zero …………………………….(10)
In the determinant \[\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{3}} & {{\left( x-a \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{3}} & {{\left( x-b \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{3}} & {{\left( x-c \right)}^{3}} & 1 \\
\end{matrix} \\
\end{align} \right|\] , the elements of column 1 and column 2 are the same.
Using the property shown in equation (10), we can say that the determinant value of \[\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{3}} & {{\left( x-a \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{3}} & {{\left( x-b \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{3}} & {{\left( x-c \right)}^{3}} & 1 \\
\end{matrix} \\
\end{align} \right|\] is equal to zero …………………………..(11)
From equation (9) and equation (11), we get
\[\Rightarrow \dfrac{df\left( x \right)}{dx}=4\times 0\] + \[3\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{2}} & 1 \\
\end{matrix} \\
\end{align} \right|\]
\[\Rightarrow \dfrac{df\left( x \right)}{dx}=3\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{2}} & 1 \\
\end{matrix} \\
\end{align} \right|\] ………………………………….(12)
Using equation (3) and replacing \[\dfrac{df\left( x \right)}{dx}\] by \[f'\left( x \right)\] in equation (12), we get
\[\Rightarrow f'\left( x \right)=3\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{2}} & 1 \\
\end{matrix} \\
\end{align} \right|\] ………………………..(13)
From equation (2), we have the value of \[f'\left( x \right)\] .
Now, comparing equation (2) and equation (13), we get
\[\Rightarrow \lambda \left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{2}} & 1 \\
\end{matrix} \\
\end{align} \right|=\] \[3\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{2}} & 1 \\
\end{matrix} \\
\end{align} \right|\]
\[\Rightarrow \lambda =3\]
Therefore, the value of \[\lambda \] is equal to 3, \[\lambda =3\] .

Note: In this question, one might think to expand the determinants \[f\left( x \right)=\left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{3}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{3}} & 1 \\
\end{matrix} \\
\end{align} \right|\] and \[f'\left( x \right)=\lambda \left| \begin{align}
  & \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{2}} & 1 \\
\end{matrix} \\
\end{align} \right|\] , and then differentiate \[f\left( x \right)\] to make it equal to \[f'\left( x \right)\] . But if we do so, then our complexity increases and complexity would lead to the calculation mistakes.