Question

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

Answer 1

Take the matrix f(x) and find its determinant f’(x), i.e. differentiate ${{C}_{1}}$ while ${{C}_{2}}$ and ${{C}_{3}}$ are constant plus differentiate ${{C}_{2}}$ while ${{C}_{1}}$ and ${{C}_{3}}$ are constant. Thus, the columns become some value, apply property of determinants and solve it.

Complete step-by-step answer:
Given to us is that $f(x)=\left| \begin{matrix} {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{3}} & 1 \\\ {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{3}} & 1 \\\ {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{3}} & 1 \\\ \end{matrix} \right|$ and $f'(x)=\lambda \left| \begin{matrix} {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{2}} & 1 \\\ {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{2}} & 1 \\\ {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{2}} & 1 \\\ \end{matrix} \right|........(1)$
In the matrix of $f(x)=\left| \begin{matrix} {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{3}} & 1 \\\ {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{3}} & 1 \\\ {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{3}} & 1 \\\ \end{matrix} \right|$
Here, ${{C}_{1}}$, ${{C}_{2}}$ and ${{C}_{3}}$ are the three columns of the $3\times 3$ matrix.
Now let us differentiate column 1 and keep ${{C}_{2}}$ and ${{C}_{3}}$ as they are in the first. Then keep ${{C}_{1}}$ and ${{C}_{3}}$ constant and differentiate ${{C}_{2}}$. If we differentiate ${{C}_{3}}$, we will get the column as zero.

4{{\left( x-a \right)}^{3}} & {{\left( x-a \right)}^{3}} & 1 \\\ 4{{\left( x-b \right)}^{3}} & {{\left( x-b \right)}^{3}} & 1 \\\ 4{{\left( x-c \right)}^{3}} & {{\left( x-c \right)}^{3}} & 1 \\\ \end{matrix} \right|+\left| \begin{matrix} {{\left( x-a \right)}^{4}} & 3{{\left( x-a \right)}^{2}} & 1 \\\ {{\left( x-b \right)}^{4}} & 3{{\left( x-b \right)}^{2}} & 1 \\\ {{\left( x-c \right)}^{4}} & 3{{\left( x-c \right)}^{2}} & 1 \\\ \end{matrix} \right|$$ $$=4\left| \begin{matrix} {{\left( x-a \right)}^{3}} & {{\left( x-a \right)}^{3}} & 1 \\\ {{\left( x-b \right)}^{3}} & {{\left( x-b \right)}^{3}} & 1 \\\ {{\left( x-c \right)}^{3}} & {{\left( x-c \right)}^{3}} & 1 \\\ \end{matrix} \right|+3\left| \begin{matrix} {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{2}} & 1 \\\ {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{2}} & 1 \\\ {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{2}} & 1 \\\ \end{matrix} \right|$$ $${{C}_{1}}$$ and $${{C}_{2}}$$ in the first determinant is same, so the value becomes zero, i.e. $${{C}_{1}}={{C}_{2}}=0$$. i.e. the determinant of a matrix with two identical columns is zero. Thus, $$f'(x)=3\left| \begin{matrix} {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{2}} & 1 \\\ {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{2}} & 1 \\\ {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{2}} & 1 \\\ \end{matrix} \right|......(2)$$ Now let us compare equation (1) and (2). From this, we can say that $$\lambda =3$$. Thus, we got the value of $$\lambda =3$$. **Note:** It is one of the properties of determinants. If we have a matrix with two identical rows or columns then its determinant is equal to zero. In this question we had two columns same, i.e. $${{C}_{1}}={{C}_{2}}$$. Thus, finding its determinant comes out, to be zero at the end.