Question

If the following three linear equations have non-trivial solution, then
\[\begin{align}
  & x+4ay+az=0 \\
 & x+3by+bz=0 \\
 & x+2cy+cz=0 \\
\end{align}\]
A) a, b, c, are in AP
B) a, b, c, are in GP
C) a, b, c, are in HP
D) a + b + c = 0

Answer 1

Hint: To solve the question, we have to calculate the coefficient matrix and calculate the determinant of the obtained matrix which is equal to zero since by definition we know that for non-trivial solution of homogeneous equations the determinant of the coefficient matrix is equal to zero. For further solving we have to apply properties of determinants which state that the value of determinant is unchanged when the row and columns are subtracted or added. After simplifying the given determinant, apply the formula for the 3x3 matrix to calculate the answer.

Complete step-by-step answer:
The matrix representation of coefficient of the given equations is equal to \[\left[ \begin{matrix}
   1 & 4a & a \\
   1 & 3b & b \\
   1 & 2c & c \\
\end{matrix} \right]\]
We know that for non-trivial solutions of homogeneous equations the determinant of the coefficient matrix is equal to zero.
Thus, we get \[\left| \begin{matrix}
   1 & 4a & a \\
   1 & 3b & b \\
   1 & 2c & c \\
\end{matrix} \right|=0\]
We know that when the \[{{R}_{1}}={{R}_{1}}-{{R}_{2}}\], the determinant is unchanged. Thus, we get
\[\left| \begin{matrix}
   1-1 & 4a-3b & a-b \\
   1 & 3b & b \\
   1 & 2c & c \\
\end{matrix} \right|=0\]
\[\left| \begin{matrix}
   0 & 4a-3b & a-b \\
   1 & 3b & b \\
   1 & 2c & c \\
\end{matrix} \right|=0\]
We know that when the \[{{R}_{2}}={{R}_{2}}-{{R}_{3}}\], the determinant is unchanged. Thus, we get
\[\left| \begin{matrix}
   0 & 4a-3b & a-b \\
   1-1 & 3b-2c & b-c \\
   1 & 2c & c \\
\end{matrix} \right|=0\]
\[\left| \begin{matrix}
   0 & 4a-3b & a-b \\
   0 & 3b-2c & b-c \\
   1 & 2c & c \\
\end{matrix} \right|=0\]
We know that the formula for determinant of 3x3 matrix is given by
\[\left| \begin{matrix}
   {{a}_{1,1}} & {{a}_{1,2}} & {{a}_{1,3}} \\
   {{a}_{2,1}} & {{a}_{2,2}} & {{a}_{2,3}} \\
   {{a}_{3,1}} & {{a}_{3,2}} & {{a}_{3,3}} \\
\end{matrix} \right|={{a}_{1,3}}\left( {{a}_{2,2}}{{a}_{3,3}}-{{a}_{2,3}}{{a}_{3,2}} \right)-{{a}_{2,1}}\left( {{a}_{1,2}}{{a}_{3,3}}-{{a}_{1,3}}{{a}_{3,2}} \right)+{{a}_{3,1}}\left( {{a}_{2,3}}{{a}_{1,2}}-{{a}_{2,2}}{{a}_{1,3}} \right)\]
By comparing the given matrix with the general 4x4 matrix we get
\[{{a}_{1,3}}=0,{{a}_{2,1}}=0,{{a}_{3,1}}=1\]
By substituting the above values in the formula, we get
\[\begin{align}
  & 0+0+1\left( \left( 4a-3b \right)\left( b-c \right)-\left( 3b-2c \right)\left( a-b \right) \right)=0 \\
 & 4ab-4ac-3{{b}^{2}}+3bc-\left( 3ab-2ac-3{{b}^{2}}+2bc \right)=0 \\
 & ab-4ac-3{{b}^{2}}+3bc-3ab+2ac+3{{b}^{2}}-2bc=0 \\
 & ab-2ac+bc=0 \\
 & ab+bc=2ac \\
\end{align}\]
By dividing both the sides of equation with abc, we get
\[\begin{align}
  & \dfrac{ab+bc}{abc}=\dfrac{2ac}{abc} \\
 & \dfrac{ab}{abc}+\dfrac{bc}{abc}=\dfrac{2}{b} \\
 & \dfrac{1}{a}+\dfrac{1}{c}=\dfrac{2}{b} \\
\end{align}\]
We know the formula for harmonic mean of a, b, c is given by \[\dfrac{1}{a}+\dfrac{1}{c}=\dfrac{2}{b}\]
Thus, we get that a, b, c are in harmonic progression.
Hence, option (c) is the right choice.

Note: The possibility of mistake can be due to confusion of conditions of non-trivial solution for homogeneous equations, since the conditions of trivial, non-trivial solutions for homogeneous a nonhomogeneous equation are close in definition. The other possibility of mistake can be applying formula directly without rearranging the determinant which will lead to more calculation instead of easing the procedure of solving.