Question

Let for \[i = 1,2,3\], $P_i$(x) be a polynomial of degree 2 in x, $P_i$’(x) and $P_i$’’(x) be the first and second order derivatives of the $P_i$(x) respectively. Let,
$A(x) = \left( {\begin{array}{*{20}{c}}
  {{P_1}(x)}&{{P_1}'(x)}&{{P_1}''(x)} \\
  {{P_2}(x)}&{{P_2}'(x)}&{{P_2}''(x)} \\
  {{P_3}(x)}&{{P_3}'(x)}&{{P_3}''(x)}
\end{array}} \right)$
and $B(x) = {[A(x)]^T}A(x)$. Then determinant of B(x):
A. is a polynomial of degree 6 in x
B. is a polynomial of degree 3 in x
C. is a polynomial of degree 2 in x
D. does not depend on x

Answer 1

Hint: This is a determinant and matrix question. To solve this question first we will A(x) as a square matrix of 2 degree polynomial and their derivatives. Then we will find out the determinant of A(x) using the determinant formula. As the determinant of A(x) is equal to the determinant of transverse of A(x), by putting the determinant value of the both in the given equation we will get the determinant of B(x).

Complete step-by-step answer:
According to the question, Pi(x) be a polynomial of degree 2 in x, where \[i = 1,2,3\]
Given, $ A(x) = \left( {\begin{array}{*{20}{c}}
  {{P_1}(x)}&{{P_1}'(x)}&{{P_1}''(x)} \\
  {{P_2}(x)}&{{P_2}'(x)}&{{P_2}''(x)} \\
  {{P_3}(x)}&{{P_3}'(x)}&{{P_3}''(x)}
\end{array}} \right) $
As it is a polynomial of degree 2,
Let, $ A(x) = \left( {\begin{array}{*{20}{c}}
  {{a_1}{x^2} + {b_1}x + c}&{2{a_1}x + {b_1}}&{2{a_1}} \\
  {{a_2}{x^2} + {b_2}x + c}&{2{a_2}x + {b_2}}&{2{a_2}} \\
  {{a_3}{x^2} + {b_3}x + c}&{2{a_3}x + {b_3}}&{2{a_3}}
\end{array}} \right) $
Hence the determinant of A(x),
 $ \therefore \left| {A(x)} \right| = ({a_1}{x^2} + {b_1}x + c)\left( {\begin{array}{*{20}{c}}
  {2{a_2}x + {b_2}}&{2{a_2}} \\
  {2{a_3}x + {b_3}}&{2{a_3}}
\end{array}} \right) - 2{a_1}x + {b_1}\left( {\begin{array}{*{20}{c}}
  {{a_2}{x^2} + {b_2}x + c}&{2{a_2}} \\
  {{a_3}{x^2} + {b_3}x + c}&{2{a_3}}
\end{array}} \right) + 2{a_1}\left( {\begin{array}{*{20}{c}}
  {{a_2}{x^2} + {b_2}x + c}&{2{a_2}x + {b_2}} \\
  {{a_3}{x^2} + {b_3}x + c}&{2{a_3}x + {b_3}}
\end{array}} \right) $
Expanding the above equation we get,
 $ \therefore \left| {A(x)} \right| = ({a_1}{x^2} + {b_1}x + c)\{ (2{a_2}x + {b_2})2{a_3} - 2{a_2}(2{a_3}x + {b_3})\} - 2{a_1}x + {b_1}\{ 2{a_2}({a_3}{x^2} + {b_3}x + c) - 2{a_3}({a_2}{x^2} + {b_2}x + c)\} + 2{a_1}\{ ({a_2}{x^2} + {b_2}x + c)(2{a_3}x + {b_3}) - ({a_3}{x^2} + {b_3}x + c)(2{a_2}x + {b_2})\} $
Simplifying the above equation and separating them with respect to degree of x we get,
 $
  \left| {A(x)} \right| = {x^3}[(2{a_1}{a_2}{a_3} + 2{a_1}{a_2}{a_3} + 2{a_1}{a_2}{a_3}) - (2{a_1}{a_2}{a_3} + 2{a_1}{a_2}{a_3} + 2{a_1}{a_2}{a_3})] \\
   + {x^2}[({a_1}{b_2}{a_3} + 2{b_1}{a_2}{a_3} + {b_1}{a_2}{a_3} + 2{a_1}{a_2}{b_3} + {a_1}{a_2}{b_3} + 2{a_1}{b_2}{a_3}) - ({a_1}{b_2}{a_3} + 2{a_1}{a_2}{b_3} + {a_1}{a_2}{b_3} + 2{b_1}{a_2}{a_3} + {b_1}{a_2}{a_3} + 2{a_1}{b_2}{a_3})]............ \\
 $
From the above equation we get that the determinant of A(x) is a polynomial of degree 3 as its highest power is 3.
From the rules of determinant and matrices we know that the determinant of transpose of A(x) is equal to the determinant of A(x), i.e.
 $ \left| {A(x)} \right| = \left| {{{[A(x)]}^T}} \right| $
Hence, the determinant of transpose of A(x), $ \left| {{{[A(x)]}^T}} \right| $ is also a polynomial of degree 3.
According to the question we have,
 $ B(x) = {[A(x)]^T}A(x) $
Taking determinant of both of the side we get,
 $ \left| {B(x)} \right| = \left| {{{[A(x)]}^T}} \right|\left| {A(x)} \right| $
As a determinant of $ A(x) $ and the determinant of transverse of $ A(x) $ it has the highest power 3. So when we will multiply both of them, their highest power will be added according to formula, $ {a^3} \times {a^3} = {a^{3 + 3}} = {a^6} $
Hence the determinant of B, $ \left| {B(x)} \right| $ has the highest power = addition of the highest power of determinant of $ A(x) $ and the determinant of transverse of $ A(x) $ .
i.e. the determinant of B, $ \left| {B(x)} \right| $ has the highest power $ = 3 + 3 = 6 $
The determinant of B(x) is a polynomial of degree 6 in x.

So, the correct answer is “Option A”.

Note: You might have mistaken the answer to be a polynomial of degree 6 because the highest power of determinant of A(x) is 3. But upon simplifying that equation, all the terms containing x got cancelled leaving constant.
Matrix is a rectangular array of numbers, symbols or expressions arranged in rows and columns. A determinant is a scalar value that can be computed from the elements of a matrix and which encodes certain properties of the linear transformation described by the matrix.
Let A be a $ 3 \times 3 $ matrix, $ A = \left( {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right) $
Hence the determinant of A is given by,
 $ \left| A \right| = a\left| {\left( {\begin{array}{*{20}{c}}
  e&h \\
  f&i
\end{array}} \right)} \right| - b\left| {\left( {\begin{array}{*{20}{c}}
  d&f \\
  g&i
\end{array}} \right)} \right| + c\left| {\left( {\begin{array}{*{20}{c}}
  d&e \\
  g&h
\end{array}} \right)} \right| $
Or, $ \left| A \right| = aei + bfg + cdh - ceg - bdi – afh $
The transpose of a matrix is an operator which flips a matrix over its diagonal and it is denoted as $ {A^T} $ .
 $ {A^T} = \left( {\begin{array}{*{20}{c}}
  a&d&g \\
  b&e&h \\
  c&f&i
\end{array}} \right) $
You should remember all the formulas and properties of determinant and matrix.