Question

If \[f(ax) = \left[ {\begin{array}{*{20}{c}}
  a&{ - 1}&0 \\
  {ax}&a&{ - 1} \\
  {a{x^2}}&{ax}&a
\end{array}} \right]\], using properties of determinants, find the value of \[f(2x) - f(x)\] .

Answer 1

Hint: You can easily understand, this is a numerical problem of determinants. Do you know what determinants are? Determinant is a scalar value that can be completed from the elements of a square matrix and includes certain properties of linear transformation described by the matrix is denoted as det(A) , det A or |A|. Determinants are mainly used as a theoretical tool.

Complete step by step solution:
Given data: \[f(ax) = \left[ {\begin{array}{*{20}{c}}
  a&{ - 1}&0 \\
  {ax}&a&{ - 1} \\
  {a{x^2}}&{ax}&a
\end{array}} \right]\]
We need to find out the value of \[f\left( {2x} \right) - f\left( x \right)\] .
We will simplify $f\left( {ax} \right)$ ,
\[f(ax) = \left[ {\begin{array}{*{20}{c}}
  a&{ - 1}&0 \\
  {ax}&a&{ - 1} \\
  {a{x^2}}&{ax}&a
\end{array}} \right]\]
Let us take $a$ common from column 1 (C1),
\[f\left( {ax} \right) = a\left| {\begin{array}{*{20}{c}}
  1&{ - 1}&0 \\
  {x }&a&{ - 1} \\
  {{x^2} }&{ax}&a
\end{array}} \right|\]
On applying the property of determinants \[{C_1} \to {C_2} + {C_1}\] , we get
\[f\left( {ax} \right) = a\left| {\begin{array}{*{20}{c}}
  0&{ - 1}&0 \\
  {x + a}&a&{ - 1} \\
  {{x^2} + ax}&{ax}&a
\end{array}} \right|\]
Since, $a$ is a constant and \[a \ne 0\] , we get
\[
  a \cdot f(ax) = a\left| {\begin{array}{*{20}{c}}
  0&{ - 1}&0 \\
  {x + a}&a&{ - 1} \\
  {{x^2} + ax}&{ax}&a
\end{array}} \right|\, \\
   \Rightarrow f(x) = \,\left| {\begin{array}{*{20}{c}}
  0&{ - 1}&0 \\
  {x + a}&a&{ - 1} \\
  {{x^2} + ax}&{ax}&a
\end{array}} \right| \\
 \]
Now, we will expand $f\left( x \right)$ to get the value of $f\left( x \right)$ :
\[
   \Rightarrow f\left( x \right) = 0 - \left( { - 1} \right)\left( {a\left( {x + a} \right) - \left( { - 1} \right)\left( {{x^2} + ax} \right)} \right) + 0 \\
   \Rightarrow f\left( x \right) = 1\left( {ax + {a^2} + 1\left( {{x^2} + ax} \right)} \right) \\
   \Rightarrow f\left( x \right) = {x^2} + {a^2} + 2ax \\
 \]
Using the basic formulas
\[ \Rightarrow f\left( x \right) = {\left( {x + a} \right)^2}\]
Hence, the value of \[f\left( x \right) = {\left( {x + a} \right)^2}\] .
Now, again, to find out the value of \[f\left( {2x} \right) - f\left( x \right)\]
Firstly, Lets replace $x$ with $2x$ in \[f\left( x \right) = {\left( {x + a} \right)^2}\]
\[
   \Rightarrow f\left( {2x} \right) = {\left( {2x + a} \right)^2} \\
   \Rightarrow f\left( {2x} \right) = 4{x^2} + {a^2} + 4ax \\
 \]
Secondly, \[f\left( x \right) = {\left( {x + a} \right)^2}\]
Now, \[
   \Rightarrow f\left( {2x} \right) - f\left( x \right) = \left( {4{x^2} + {a^2} + 4ax} \right) - {\left( {x + a} \right)^2} \\
   \Rightarrow f\left( {2x} \right) - f\left( x \right) = 4{x^2} + {a^2} + 4ax - \left( {{x^2} + {a^2} + 2ax} \right) \\
 \]
Open the bracket and solve it in the left hand side.
\[
   \Rightarrow f\left( {2x} \right) - f\left( x \right) = 4{x^2} + {a^2} + 4ax - {x^2} - {a^2} - 2ax \\
   \Rightarrow f\left( {2x} \right) - f\left( x \right) = 3{x^2} + 2ax \\
 \]
Take, $x$ common in the left hand side
\[ \Rightarrow f\left( {2x} \right) - f\left( x \right) = x\left( {3x + 2a} \right)\]
Hence, the required value of \[f\left( {2x} \right) - f\left( x \right)\] is \[x\left( {3x + 2a} \right)\].

Note: Students make mistakes in the properties of determinants. You should learn the properties nicely before doing the numerical of determinants. Also do not get confused between determinants and matrices. If in a matrix, any row or column has all elements equal to zero, then the determinant of that matrix is 0. This n-linear function is an alternating form: whenever two rows of a matrix are identical, its determinant is 0.