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This is one of our many quizzes on Solving systems of nonlinear equations. Click on this to access more such competitive quizzes on Algebra.

Solving systems of nonlinear equations Quiz 2

mathematics

 For what value of the $\beta $, the following system of the homogeneous equation having the non-trivial solution. $x + \beta y - z = 0 \\ 2x - y + \beta z = 0 \\ \beta x + y + 2z = 0$

 For a homogenous equation, the determinant is zero to get non-trivial solutions.  

$\left| {\begin{array}{*{20}{c}}  1&\beta &{ - 1} \\   2&{ - 1}&\beta  \\   \beta &1&2 \end{array}} \right| = 0 \\ (2 + \beta )({\beta ^2} - 2\beta  - 2) = 0 \\ \beta  =  - 2\,\,$Or$\,\beta  = 1 \pm \sqrt 3 $

 Choose the correct option if the nonlinear equations are given. $\left\{   2x + 3y = 3   \\  3x + 4y = 5   \\ \right.$

 Using the elimination method, we can eliminate one variable and then get the value of the variable after that and put the value of any of the equations to find another variable. 

 By Using elimination method  $2x + 3y = 3  \\ 3x + 4y = 5 \\ \overline{3 \times (2x + 3y = 3)} \\  2 \times (3x + 4y = 5)  \\ \overline{~~~~~~6x + 9y = 9} \\  ~~~~~~~6x + 8y = 10 \\  ~~~~~ -~-~~~~~~-   \\  \overline {~~~~~~~~0+ y=-1}\\ $ $y =  - 1$, $2x + 3y = 3$, $x = 3$, $y =  - 1$

 Suppose there are three variable systems having equations $x + 2y + z = 5 \\ x + y + 3z = 6 \\ x + y + ({p^2} - 6)z = p + 2 \\ $ Then which option is correct?  

 Since the two-equation is looked similar and by comparison of eq. 2 and 3, we get that value of the $\left| p \right| = 3$ system is inconsistent.

 $x + 2y + z = 5  ----(1)  \\ x + y + 3z = 6  ---(2) \\ x + y + ({p^2} - 6)z = p + 2  ----(3) $ Since the eq. $(2)$ and $(3)$ are identical except its RHS side, So $ \\  {p^2} - 6 = 3   \\  {p^2} = 9   \\  \left| p \right| = 3   \\ $ Inconsistent system for the $\left| p \right| = 3$ 

 Consider the following system of equation which is represent in matrix form choose the correct value of $k$ and $p$, $ \\ \left( {\begin{array}{*{20}{c}}  1&2&3 \\   3&4&5 \\   4&4&4 \end{array}} \right)\left( {\begin{array}{*{20}{c}}  l \\   m \\   n \end{array}} \right) = \left( {\begin{array}{*{20}{c}}  1 \\   k \\   p \end{array}} \right) \\ $  For which system has no solutions.

 For no solution, the determinant of the matrix must be zero.

$Det. = \left| {\begin{array}{*{20}{c}}  1&2&3 \\   3&4&5 \\   4&4&4 \end{array}} \right| \\ Det. = 0 \\ Det. = 1 \times (16 - 20) - 2(12 - 20) + 3 \times (12 - 16)=0 \\ Det. = 0 \\ $ For no solution system, any of ${D_l},\,\,{D_m},\,\,{D_n}$ must not be equal to $0$, So now $ \\ {D_l} = \left| {\begin{array}{*{20}{c}}  k&4&5 \\   1&2&3 \\   p&4&4 \end{array}} \right| \\ {D_m}\, = \left| {\begin{array}{*{20}{c}}  3&k&5 \\   1&1&3 \\   4&p&4 \end{array}} \right| \\ {D_n} = \left| {\begin{array}{*{20}{c}}  3&4&k \\   1&2&1 \\   4&4&p \end{array}} \right| \\ $ If the system has no solution then the $2k \ne p + 2$ $ \\ $ Hence the system has no solution for the pair $(k,p) = (3,4)$ 

 Given the system of the nonlinear equation which is shown below. $ \\ \alpha r + 2s + 2t = 5 \\ 2\alpha r + 3s + 5t = 8 \\ 4r + \alpha s + 6t = 10 \\ $ Select the value of $\alpha $ so that the system become inconsistent  

 To find the roots, set the determinants to zero and solve the quadratic equation.

$\left( {\begin{array}{*{20}{c}}  \alpha &2&2 \\   {2\alpha }&3&5 \\   4&\alpha &6 \end{array}} \right)\left( {\begin{array}{*{20}{c}}  r \\   s \\   t \end{array}} \right) = \left( {\begin{array}{*{20}{c}}  5 \\   8 \\   {10} \end{array}} \right)$ Determinant equivalent to zero and cross checking for $\alpha  = 2\,$ $ \\ Det. =  - {\alpha ^2} - 6\alpha  + 16 \\ Det. = 0 \\ {\alpha ^2} + 6\alpha  - 16 = 0 \\ $ $\alpha  = 2$ Or $\alpha  =  - 8$ $ \\   {D_r} = \left| {\begin{array}{*{20}{c}}  5&2&2 \\   8&3&5 \\   {10}&2&6 \end{array}} \right|   \\  {D_r} = 4 + 40 - 28 \ne 0   \\ $ Hence no solution for $\alpha  = 2\,$