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  • If the system of linear equations, x+y+z=6 x+2y+3z=10 has more than two solutions, then <img alt="\mu-\lambda ^{2}" src

If the system of linear equations, x+y+z=6 x+2y+3z=10 3x+2y+\lambda z=\mu has more than two solutions, then \mu-\lambda ^{2} is equal to  _______.
Option: 1 13
Option: 2 17
Option: 3 21
Option: 4 25
 

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.

-

 

 

 

x + y + z = 6 …….. (1)

x + 2y + 3z = 10 …….. (2)

3x + 2y + \lambdaz = \mu…….. (3)

from (1) and (2)

if z = 0 \Rightarrow x + y = 6 and  x + 2y = 10

\Rightarrow y = 4, x = 2

(2, 4, 0)

if y = 0 \Rightarrow x + z = 6 and  x + 3z = 10

\Rightarrow z = 2 and x = 4

(4, 0, 2)

so, 3x + 2y + \lambdaz = \mu must pass through  (2, 4, 0) and  (4, 0, 2)

\mu = 14

and 12 + 2\lambda = \mu

\lambda = 1

(\mu-\lambda^2) = 13

Posted by

Ritika Jonwal

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