# 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

Verified

363.3k+ views

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.

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.

Last updated date: 26th Sep 2023

•

Total views: 363.3k

•

Views today: 7.63k

Recently Updated Pages

What is the Full Form of DNA and RNA

What are the Difference Between Acute and Chronic Disease

Difference Between Communicable and Non-Communicable

What is Nutrition Explain Diff Type of Nutrition ?

What is the Function of Digestive Enzymes

What is the Full Form of 1.DPT 2.DDT 3.BCG

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

One cusec is equal to how many liters class 8 maths CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

How many millions make a billion class 6 maths CBSE

Which are the Top 10 Largest Countries of the World?

How many crores make 10 million class 7 maths CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE