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The system of linear equations \lambda x+2y+2z=5 2\lambda x+3y+5z=8 4 x+\lambda y+6z=10 has :
Option: 1 no solution when \lambda =2 
Option: 2 infinitely many solutions when \lambda =2
Option: 3 no solution when \lambda =8
Option: 4 a unique solution when \lambda =-8
 

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.

-

 

 

 

D=\begin{vmatrix} \lambda &2 &2 \\ 2\lambda &3 & 5\\ 4 &\lambda &6 \end{vmatrix}

D = (\lambda + 8) ( 2 – \lambda) for \lambda = 2

D_1=\begin{vmatrix} 5 &2 &2 \\ 8 &3 & 5\\ 10 &2 &6 \end{vmatrix}

= 5[18 – 10] – 2 [48 – 50] + 2 (16 – 30]

= 40 + 4 – 28 \neq 0 No solutions for  \lambda= 2

Correct Option (1)

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vishal kumar

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