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  • If the system of linear equations  2x+2ay+az=0 2x+3by+bz=0, 2x+4cy+cz=0, where  are non-zero and distinct ; has a non-zero solution, the

If the system of linear equations  2x+2ay+az=0 2x+3by+bz=0, 2x+4cy+cz=0, where a,b,c\epsilon R are non-zero and distinct ; has a non-zero solution, then :
Option: 1 a+b+c=0
Option: 2 a,b,c are in AP
Option: 3 \frac{1}{a},\frac{1}{b},\frac{1}{c} are in A.P.
Option: 4a,b,c are in G.P.
 

Answers (1)

best_answer

 

 

Cramer’s law -

Cramer’s law for the system of equations in two variables :

\\\mathrm{Let \; a_1x +b_1y = c_1\; and \; a_2x + b_2y = c_2, where} \\\mathrm{\frac{a_1}{a_2}\neq\frac{b_1}{b_2}} \\\mathrm{on \; solving \; this \; equation \; by \; cross \; multiplication, we \; get} \\\mathrm{\frac{x}{b_1c_2-b_2c_1}=\frac{y}{a_1c_2-a_2c_1}=\frac{1}{a_1b_2-a_2b_1}} \\\mathrm{or \; \frac{x}{\begin{vmatrix} b_1 & c_1\\ b_2 & c_2 \end{vmatrix}}=\frac{y}{\begin{vmatrix} a_1 & c_1\\ a_2 & c_2 \end{vmatrix}}=\frac{1}{\begin{vmatrix} a_1 & b_1\\ a_2 & b_2 \end{vmatrix}}} \\\mathrm{or \; x=\frac{\begin{vmatrix} b_1 & c_1\\ b_2 & c_2 \end{vmatrix}}{\begin{vmatrix} a_1 & b_1\\ a_2 & b_2 \end{vmatrix}}, y=\frac{\begin{vmatrix} a_1 & c_1\\ a_2 & c_2 \end{vmatrix}}{\begin{vmatrix} a_1 & b_1\\ a_2 & b_2 \end{vmatrix}}}

We can observe that first row in the numerator of x is of constants and 2nd row in the numerator is of constants, and the denominator is of the coefficient of variables.

We can follow this analogy for the system of equations of 3 variable where third in the numerator of the value of z will be of constant and denominator will be formed by the value of coefficients of the variables.

 

 

\\\mathrm{so\; let \;us\; consider\; the \; system\; of \;equations} \\\mathrm{a_1x+b_1y +c_1z =d_1 \;\;...(i)} \\\mathrm{a_2x+b_2y +c_2z =d_2\;\;\;...(ii)} \\\mathrm{a_3x+b_3y +c_3z =d_3\;\;\;...(iii)} \\\mathrm{then \; \Delta,\; which\; will\; be \;determinant\; of\; coefficient\; of \;variables, will\; be } \\\mathrm{\Delta = \begin{vmatrix} a_1 & b_1 & c_1\\ a_2& b_2 & c_2\\ a_3 & b_3 & c_3 \end{vmatrix}, \Delta_1\; numerator\; of\; x \;is:} \\\mathrm{\Delta_1= \begin{vmatrix} d_1 & b_1 &c_1 \\ d_2 & b_2 & c_2\\ d_3 & b_3 & c_3 \end{vmatrix}, similarly\; \Delta_2 = \begin{vmatrix} a_1 & d_1 & c_1\\ a_2 & d_2 & c_2\\ a_3 & d_3 & c_3 \end{vmatrix}\; and \;} \\\mathrm{\Delta_3 = \begin{vmatrix} a_1 & b_1 & d_1\\ a_2 & b_2 & d_2\\ a_3 & b_3 & d_3 \end{vmatrix}, if \Delta \neq 0\; then\; \;x=\frac{\Delta_1}{\Delta}} \\\mathrm{y=\frac{\Delta_2}{\Delta}, z=\frac{\Delta_3}{\Delta}}

 

i) If ? ≠ 0, then the system of equations has a unique finite solution and so equations are consistent, and solutions are  \\\mathrm{x=\frac{\Delta_1}{\Delta}, y=\frac{\Delta_2}{\Delta}, z=\frac{\Delta_3}{\Delta}}

 

ii) If ? = 0, and any of \Delta_1\neq 0 \; or \;\Delta_2\neq 0 \; or \;\Delta_3\neq 0

Then the system of equations is inconsistent and hence no solution exists.

 

iii) If all \Delta =\Delta_1=\Delta_2=\Delta_3= 0 then

System of equations is consistent and dependent and it has an infinite number of solution.

-

 

 

For non zero solutions D = 0

\\D=\begin{vmatrix} 2 & 2a &a \\ 2& 3b &b \\ 2& 4c &c \end{vmatrix}=0\\C_2\rightarrow C_2-2C_3\\D=\begin{vmatrix} 1 &0 &a \\ 1& b &b \\ 1 & 2c & c \end{vmatrix}=0\\

on solving

\\-bc+2ac-ab=0\\\\\frac{1}{a}+\frac{1}{c}=\frac{2}{b}

Correct Option (3)

Posted by

Kuldeep Maurya

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