Question

Let $\left| {\begin{array}{*{20}{c}}
  x&2&x \\
  {{x^2}}&x&6 \\
  x&x&6
\end{array}} \right| = {\text{A}}{x^4} + {\text{B}}{x^3} + {\text{C}}{x^2} + {\text{D}}x + {\text{E}}$. Then find the value of 5A+4B+3C+2D+E.
$
  {\text{A}}{\text{. 0}} \\
  {\text{B}}{\text{. }} - 16 \\
  {\text{C}}{\text{. 16}} \\
  {\text{D}}{\text{. }} - 11 \\
 $

Answer 1

Hint: Here, we will be proceeding by expanding the determinant for the given $3 \times 3$ order matrix in the LHS of the given equation and then we will compare the LHS and RHS of this equation to find the values of A, B, C, D and E.

Complete step-by-step answer:
As we know that by expanding the determinant of any $3 \times 3$ order matrix through first row, we have
 $\left| {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
  {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right| = {a_{11}}\left( {{a_{22}}{a_{33}} - {a_{23}}{a_{32}}} \right) - {a_{12}}\left( {{a_{21}}{a_{33}} - {a_{23}}{a_{31}}} \right) + {a_{13}}\left( {{a_{21}}{a_{32}} - {a_{22}}{a_{31}}} \right)$
The given determinant of a matrix of order $3 \times 3$ when expanded through first row, we have
$
  \left| {\begin{array}{*{20}{c}}
  x&2&x \\
  {{x^2}}&x&6 \\
  x&x&6
\end{array}} \right| = x\left( {6x - 6x} \right) - 2\left( {6{x^2} - 6x} \right) + x\left( {{x^3} - {x^2}} \right) = 0 - 12{x^2} + 12x + {x^4} - {x^3} \\
   \Rightarrow \left| {\begin{array}{*{20}{c}}
  x&2&x \\
  {{x^2}}&x&6 \\
  x&x&6
\end{array}} \right| = {x^4} - {x^3} - 12{x^2} + 12x + 0{\text{ }} \to {\text{(1)}} \\
 $
Since, it is given that $\left| {\begin{array}{*{20}{c}}
  x&2&x \\
  {{x^2}}&x&6 \\
  x&x&6
\end{array}} \right| = {\text{A}}{x^4} + {\text{B}}{x^3} + {\text{C}}{x^2} + {\text{D}}x + {\text{E }} \to {\text{(2)}}$
By comparing the RHS of equations (1) and (2), we get
A=1, B=-1, C=-12, D=12 and E=0
Therefore, the value of the expression 5A+4B+3C+2D+E can be obtained by putting the values of A, B, C, D and E obtained.
5A+4B+3C+2D+E$ = 5\left( 1 \right) + 4\left( { - 1} \right) + 3\left( { - 12} \right) + 2\left( {12} \right) + 0 = 5 - 4 - 36 + 24 = - 11$.
Hence, option D is correct.

Note: Here, we can also expand the determinant of the $3 \times 3$ order matrix given in the LHS of the given equation through any row or column but the results will always be the same no matter through which row or column the determinant is getting expanded.