For what value of a the pair of linear equations will have a unique solution
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Condition for the pair of system to have unique solution Show
a1/a2 ≠ b1/b2 Let the equations be, a1x + b1y + c1 = 0 a2x + b2y + c2 = 0 Since, x = – 1 and y = 3 is the unique solution of these two equations, then It must satisfy the equations – a1(-1) + b1(3) + c1 = 0 – a1 + 3b1 + c1 = 0 …(i) and a2(- 1) + b2(3) + c2 = 0 – a2 + 3b2 + c2 = 0 …(ii) Since for the different values of a1, b1, c1 and a2, b2, c2 satisfy the Eqs. (i) and (ii).  Hence, infinitely many pairs of linear equations are possible. A set of linear simultaneous equations may have (i) a unique solution, (ii) no solution, or (iii) infinitely many solutions. In a set of linear simultaneous equations, a unique solution exists if and only if, (a) the number of unknowns and the number of equations are equal, (b) all equations are consistent, and (c) there is no linear dependence between any two or more equations, that is, all equations are independent. In a system of linear simultaneous equations if one or more equations are inconsistent, the system does not have any solution. For example, if in a set of linear simultaneous equations with two equations and two unknowns, one equation is \(x+y=2\) and another equation is \(3x+3y=5\), these two equations are inconsistent within the given system. They are inconsistent because if \(x+y=2\), then \(3x+3y\) must be \(6\), not \(5\). One cannot solve the system of linear simultaneous equations \(x+y=2\) and \(3x+3y=5\) as they are inconsistent. Graphically, the solution of two linear simultaneous equations in two unknowns is equivalent to finding where the lines of the two equations cross. If these two equations are inconsistent, corresponding lines in the Cartesian plane are parallel and will never cross (See Practice Question 2). In a system of linear simultaneous equations if all equations are consistent, but (a) the number of independent equations is less than the number of unknowns, and/or (b) there exists a linear dependence between two or more equations in the system, there may exists infinitely many solutions that satisfy the system. Linear dependence, for example between two linear equations, refers to a situation when one equation in the system is a multiple of another equation. For example, equations \(y=x+2\) and \(2y=2x+4\) are linearly dependent as the later can be obtained by multiplying the former equation by \(2\). Consider a more general example. Suppose two linear simultaneous equations are: $$a_{11}x_{1} + a_{12}x_{2} = b_{1}$$ $$a_{21}x_{1} + a_{22}x_{2} = b_{2}$$ Where, \(a_{ij}\) is the coefficient, \(x_j\) is the variable, and \(b_i\) is the constant. In this system if \(a_{1j}=ka_{2j}\) and \(b_1=kb_2\), where \(k\) is a constant, the equations are linearly dependent. Graphically the lines representing the graphs of two equations coincide if the equations are linearly dependent, and every point on either line is a solution. [ See Practice Question 3] One interesting form of linear dependence may arise in a system of \(m×n\) linear simultaneous equations when one equation is the sum or difference of more than one equation in the system. For example, equations (i) \(x+y+z=10\),(ii) \(2x-2y-2z=4\), and (iii) \(3x-y-z=14\) have linear dependence. (Why?) To sum up, consider a system of linear simultaneous equations where all equations are consistent, however, due to the linear dependence between some equations the number of independent equations is less than the number of unknowns. Such a system has infinitely many solutions. It follows from the discussion in this section is that two linear simultaneous equations in two unknowns can have a unique solution, no solution or infinitely many solutions and this is true for every system of linear simultaneous equations with \(m\) equations and \(n\) unknowns. To read more about the existence of a unique solution, inconsistency, and linear dependence, please see the recommended books. The example shown previously in this module had a unique solution. The structure of the row reduced matrix was \[\begin{split}\begin{vmatrix} 1 & 1 & -1 & | & 5 \\ 0 & 1 & -5 & | & 8 \\ 0 & 0 & 1 & | & -1 \end{vmatrix}\end{split}\] and the solution was \[x = 1\] \[y = 3\] \[z = -1\] As you can see, each variable in the matrix can have only one possible value, and this is how you know that this matrix has one unique solution No solution¶Let’s suppose you have a system of linear equations that consist of: \[x + y + z = 2\] \[y - 3z = 1\] \[2x + y + 5z = 0\] The augmented matrix is \[\begin{split}\begin{vmatrix} 1 & 1 & 1 & | & 2 \\ 0 & 1 & -3 & | & 1 \\ 2 & 1 & 5 & | & 0 \end{vmatrix}\end{split}\] and the row reduced matrix is \[\begin{split}\begin{vmatrix} 1 & 0 & 4 & | & 1 \\ 0 & 1 & -3 & | & 1 \\ 0 & 0 & 0 & | & -3 \end{vmatrix}\end{split}\] As you can see, the final row states that \[0x + 0y + 0z = -3\] which impossible, 0 cannot equal -3. Therefore this system of linear equations has no solution. Let’s use python and see what answer we get. In [1]: import numpy as py from scipy.linalg import solve A = [[1, 1, 1], [0, 1, -3], [2, 1, 5]] b = [[2], [1], [0]] x = solve(A,b) x --------------------------------------------------------------------------- LinAlgError Traceback (most recent call last) As you can see the code gives us an error suggesting there is no solution to the matrix. Infinite Solutions¶Let’s suppose you have a system of linear equations that consist of: \[-3x - 5y + 36z = 10\] \[-x + 7z = 5\] \[x + y - 10z = -4\] The augmented matrix is \[\begin{split}\begin{vmatrix} -3 & -5 & 36 & | & 10 \\ -1 & 0 & 7 & | & 5 \\ 1 & 1 & -10 & | & -4 \end{vmatrix}\end{split}\] and the row reduced matrix is \[\begin{split}\begin{vmatrix} 1 & 0 & -7 & | & -5 \\ 0 & 2 & -3 & | & 1 \\ 0 & 0 & 0 & | & 0 \end{vmatrix}\end{split}\] As you can see, the final row of the row reduced matrix consists of 0. This means that for any value of Z, there will be a unique solution of x and y, therefore this system of linear equations has infinite solutions. Let’s use python and see what answer we get. In [2]: import numpy as py from scipy.linalg import solve A = [[-3, -5, 36], [-1, 0, 7], [1, 1, -10]] b = [[10], [5], [-4]] x = solve(A,b) x C:\Users\Said Zaid-Alkailani\Anaconda3\lib\site-packages\scipy\linalg\basic.py:223: RuntimeWarning: scipy.linalg.solve
Ill-conditioned matrix detected. Result is not guaranteed to be accurate.
Reciprocal condition number: 3.808655316038273e-19
' condition number: {}'.format(rcond), RuntimeWarning)
Out[2]: array([[-12.],
[ -2.],
[ -1.]])
As you can see we get a different type of error from this code. It states that the matrix is ill-conditioned and that there is a RuntimeWarning. This means that the computer took to long to find a unique solution so it spat out a random answer. When RuntimeWarings occur, the matrix is likely to have infinite solutions. Which pair of linear equations has a unique solution?∵aa=bb, the lines have unique solution.
For what value of p pair of linear equations will have unique solution?Therefore, the given system will have unique solution for all real values of p other than 4.
For what values of a the system of equations will have a unique solution?so for any value of k≠0 the given system of linear equation has a unique solution.
For what value of k will the pair of linear equations have unique solution kx 2y 5 3x y 1?∴ The system has a unique solution if k ≠ 6.
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