Question

If ${{a}_{1}},{{a}_{2}},{{a}_{3}},5,4,{{a}_{6}},{{a}_{7}},{{a}_{8}},{{a}_{9}}$ are in H.P. and $D=\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\\ 5 & 4 & {{a}_{6}} \\\ {{a}_{7}} & {{a}_{8}} & {{a}_{9}} \\\ \end{matrix} \right|$ , then find the value of 21D.

Answer 1

Hint : Harmonic Progression (H.P.): The series of numbers where the reciprocals of the terms are in Arithmetic Progression, is called a Harmonic Progression.
i.e. if a, b and c are in H.P., then $\dfrac{1}{a}$ , $\dfrac{1}{b}$ and $\dfrac{1}{c}$ are in A.P., and vice versa.
Find the common difference of the A.P. using the values of ${{a}_{4}}=5$ and ${{a}_{5}}=4$ , and then the values of all the remaining terms.
Use these values to find the value of the given determinant D. Refer to the properties of determinants given in Note.

Complete step-by-step answer :
Let's say that the A.P. is ${{x}_{1}},{{x}_{2}},{{x}_{3}},\dfrac{1}{5},\dfrac{1}{4},{{x}_{6}},{{x}_{7}},{{x}_{8}},{{x}_{9}}$ , which are the reciprocals of the terms of the H.P. ${{a}_{1}},{{a}_{2}},{{a}_{3}},5,4,{{a}_{6}},{{a}_{7}},{{a}_{8}},{{a}_{9}}$ .
Equating the denominators of ${{x}_{4}}=\dfrac{1}{5}$ and ${{x}_{5}}=\dfrac{1}{4}$ , we get:
${{a}_{4}}=\dfrac{1}{5}=\dfrac{4}{20}$ and ${{a}_{5}}=\dfrac{1}{4}=\dfrac{5}{20}$ , with a common difference of $d={{a}_{5}}-{{a}_{4}}=\dfrac{1}{20}$ .
So, the 9 terms of the A.P. must be $\dfrac{1}{20},\dfrac{2}{20},\dfrac{3}{20},\dfrac{4}{20},\dfrac{5}{20},\dfrac{6}{20},\dfrac{7}{20},\dfrac{8}{20},\dfrac{9}{20}$ .
And the H.P. is: $\dfrac{20}{1},\dfrac{20}{2},\dfrac{20}{3},\dfrac{20}{4},\dfrac{20}{5},\dfrac{20}{6},\dfrac{20}{7},\dfrac{20}{8},\dfrac{20}{9}$ .
Substituting these values in $D=\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\\ 5 & 4 & {{a}_{6}} \\\ {{a}_{7}} & {{a}_{8}} & {{a}_{9}} \\\ \end{matrix} \right|$ , we get:
⇒ $D=\left| \begin{matrix} \dfrac{20}{1} & \dfrac{20}{2} & \dfrac{20}{3} \\\ \dfrac{20}{4} & \dfrac{20}{5} & \dfrac{20}{6} \\\ \dfrac{20}{7} & \dfrac{20}{8} & \dfrac{20}{9} \\\ \end{matrix} \right|$
Using the transformations ${{R}_{1}}\to \dfrac{{{R}_{1}}}{20},\ {{R}_{2}}\to \dfrac{{{R}_{2}}}{20},\ {{R}_{3}}\to \dfrac{{{R}_{3}}}{20}$ , we get:
⇒ $\dfrac{D}{{{20}^{3}}}=\left| \begin{matrix} 1 & \dfrac{1}{2} & \dfrac{1}{3} \\\ \dfrac{1}{4} & \dfrac{1}{5} & \dfrac{1}{6} \\\ \dfrac{1}{7} & \dfrac{1}{8} & \dfrac{1}{9} \\\ \end{matrix} \right|$
Using the transformations ${{C}_{2}}\to 2{{C}_{2}},\ {{C}_{3}}\to 3{{C}_{3}}$ , we get:
⇒ $\dfrac{D\times 2\times 3}{{{20}^{3}}}=\left| \begin{matrix} 1 & 1 & 1 \\\ \dfrac{1}{4} & \dfrac{2}{5} & \dfrac{1}{2} \\\ \dfrac{1}{7} & \dfrac{1}{4} & \dfrac{1}{3} \\\ \end{matrix} \right|$
Using the transformations ${{C}_{2}}\to {{C}_{2}}-{{C}_{1}},\ {{C}_{3}}\to {{C}_{3}}-{{C}_{1}}$ , we get:
⇒ $\dfrac{6D}{{{20}^{3}}}=\left| \begin{matrix} 1 & 0 & 0 \\\ \dfrac{1}{4} & \dfrac{3}{20} & \dfrac{1}{4} \\\ \dfrac{1}{7} & \dfrac{3}{28} & \dfrac{4}{21} \\\ \end{matrix} \right|$
Expanding along ${{R}_{1}}$ , we get:
⇒ $\dfrac{6D}{{{20}^{3}}}=1\left[ \left( \dfrac{3}{20} \right)\left( \dfrac{4}{21} \right)-\left( \dfrac{3}{28} \right)\left( \dfrac{1}{4} \right) \right]+0+0$
⇒ $\dfrac{6D}{{{20}^{3}}}=\dfrac{1}{5\times 7}-\dfrac{3}{16\times 7}$
Equating the denominators on the R.H.S., we get:
⇒ $\dfrac{6D}{{{20}^{3}}}=\dfrac{16}{5\times 16\times 7}-\dfrac{15}{5\times 16\times 7}$
⇒ $\dfrac{6D}{{{20}^{3}}}=\dfrac{1}{5\times 16\times 7}$
⇒ $D=\dfrac{{{20}^{3}}}{5\times 6\times 16\times 7}$
Multiplying both sides by 21, we get:
⇒ $21D=\dfrac{{{20}^{3}}\times 21}{5\times 6\times 16\times 7}$
Simplifying it by cancelling out the common factors, we get:
⇒ $21D=50$ , which is the required answer.
So, the correct answer is “50”.

Note : Arithmetic Progression (A.P.): The series of numbers where the difference of any two consecutive terms is the same, is called an Arithmetic Progression.
If a be the first term, d be the common difference and n be the number of terms of an AP, then the sequence can be written as follows:
$a,\ a+d,\ a+2d,\ ...,\ a+(n-1)d$
Properties of determinants:
The determinant of a diagonal matrix is the product of the diagonal entries.
If A and B are both $n\times n$ matrices, then $det(AB)=det(A)det(B)$ .
For a $n\times n$ matrix A, $det(kA)={{k}^{n}}det(A)$ .
The determinant of a square matrix is the same as the determinant of its transpose.