Question

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

Answer 1

Hint: We have to differentiate $f\left( x \right)$ . Write $\Rightarrow \dfrac{df\left( x \right)}{dx}=\left| \begin{aligned} & \begin{matrix} \dfrac{d{{\left( x-a \right)}^{4}}}{dx} & {{\left( x-a \right)}^{3}} & 1 \\\ \end{matrix} \\\ & \begin{matrix} \dfrac{d{{\left( x-b \right)}^{4}}}{dx} & {{\left( x-b \right)}^{3}} & 1 \\\ \end{matrix} \\\ & \begin{matrix} \dfrac{d{{\left( x-c \right)}^{4}}}{dx} & {{\left( x-c \right)}^{3}} & 1 \\\ \end{matrix} \\\ \end{aligned} \right|$ + $\left| \begin{aligned} & \begin{matrix} {{\left( x-a \right)}^{4}} & \dfrac{d{{\left( x-a \right)}^{3}}}{dx} & 1 \\\ \end{matrix} \\\ & \begin{matrix} {{\left( x-b \right)}^{4}} & \dfrac{d{{\left( x-b \right)}^{3}}}{dx} & 1 \\\ \end{matrix} \\\ & \begin{matrix} {{\left( x-c \right)}^{4}} & \dfrac{d{{\left( x-c \right)}^{3}}}{dx} & 1 \\\ \end{matrix} \\\ \end{aligned} \right|$ + $\left| \begin{aligned} & \begin{matrix} {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{3}} & \dfrac{d\left( 1 \right)}{dx} \\\ \end{matrix} \\\ & \begin{matrix} {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{3}} & \dfrac{d\left( 1 \right)}{dx} \\\ \end{matrix} \\\ & \begin{matrix} {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{3}} & \dfrac{d\left( 1 \right)}{dx} \\\ \end{matrix} \\\ \end{aligned} \right|$ . 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 $\left| \begin{aligned} & \begin{matrix} {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{3}} & 0 \\\ \end{matrix} \\\ & \begin{matrix} {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{3}} & 0 \\\ \end{matrix} \\\ & \begin{matrix} {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{3}} & 0 \\\ \end{matrix} \\\ \end{aligned} \right|$ . 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 $4\left| \begin{aligned} & \begin{matrix} {{\left( x-a \right)}^{3}} & {{\left( x-a \right)}^{3}} & 1 \\\ \end{matrix} \\\ & \begin{matrix} {{\left( x-b \right)}^{3}} & {{\left( x-b \right)}^{3}} & 1 \\\ \end{matrix} \\\ & \begin{matrix} {{\left( x-c \right)}^{3}} & {{\left( x-c \right)}^{3}} & 1 \\\ \end{matrix} \\\ \end{aligned} \right|$ is equal to zero. Now, compare $3\left| \begin{aligned} & \begin{matrix} {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{2}} & 1 \\\ \end{matrix} \\\ & \begin{matrix} {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{2}} & 1 \\\ \end{matrix} \\\ & \begin{matrix} {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{2}} & 1 \\\ \end{matrix} \\\ \end{aligned} \right|$ and $\lambda \left| \begin{aligned} & \begin{matrix} {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{2}} & 1 \\\ \end{matrix} \\\ & \begin{matrix} {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{2}} & 1 \\\ \end{matrix} \\\ & \begin{matrix} {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{2}} & 1 \\\ \end{matrix} \\\ \end{aligned} \right|$ , 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.

& \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{aligned} \right|$$ …………………………….(1) $$f'\left( x \right)=\lambda \left| \begin{aligned} & \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{aligned} \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{aligned} & \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{aligned} \right|}{dx}$$ $$\Rightarrow \dfrac{df\left( x \right)}{dx}=\left| \begin{aligned} & \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{aligned} \right|$$ + $$\left| \begin{aligned} & \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{aligned} \right|$$ + $$\left| \begin{aligned} & \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{aligned} \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{aligned} & \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{aligned} \right|$$ + $$\left| \begin{aligned} & \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{aligned} \right|$$ + $$\left| \begin{aligned} & \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{aligned} \right|$$ $$\Rightarrow \dfrac{df\left( x \right)}{dx}=\left| \begin{aligned} & \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{aligned} \right|$$ + $$\left| \begin{aligned} & \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{aligned} \right|$$ + $$\left| \begin{aligned} & \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{aligned} \right|$$ ………………………………..(7) Expanding the determinant $$\left| \begin{aligned} & \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{aligned} \right|$$ along the column 3, we get the determinant value of the determinant $$\left| \begin{aligned} & \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{aligned} \right|$$ is equal to zero. So, $$\left| \begin{aligned} & \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{aligned} \right|=0$$ …………………………….(8) Now, from equation (7) and equation (8), we get $$\Rightarrow \dfrac{df\left( x \right)}{dx}=\left| \begin{aligned} & \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{aligned} \right|$$+ $$\left| \begin{aligned} & \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{aligned} \right|$$ + 0 $$\Rightarrow \dfrac{df\left( x \right)}{dx}=4\left| \begin{aligned} & \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{aligned} \right|$$+ $$3\left| \begin{aligned} & \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{aligned} \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{aligned} & \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{aligned} \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{aligned} & \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{aligned} \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{aligned} & \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{aligned} \right|$$ $$\Rightarrow \dfrac{df\left( x \right)}{dx}=3\left| \begin{aligned} & \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{aligned} \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{aligned} & \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{aligned} \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{aligned} & \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{aligned} \right|=$$ $$3\left| \begin{aligned} & \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{aligned} \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{aligned} & \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{aligned} \right|$$ and $$f'\left( x \right)=\lambda \left| \begin{aligned} & \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{aligned} \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.