By the end of this section you should be able to:
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
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.
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) <ipython-input-1-afc47691740d> in <module>() 5 b = [[2], [1], [0]] 6 ----> 7 x = solve(A,b) 8 x C:\Users\Said Zaid-Alkailani\Anaconda3\lib\site-packages\scipy\linalg\basic.py in solve(a, b, sym_pos, lower, overwrite_a, overwrite_b, debug, check_finite, assume_a, transposed) 217 return x 218 elif 0 < info <= n: --> 219 raise LinAlgError('Matrix is singular.') 220 elif info > n: 221 warnings.warn('scipy.linalg.solve\nIll-conditioned matrix detected.' LinAlgError: Matrix is singular. As you can see the code gives us an error suggesting there is no solution to the matrix.
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.
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)
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.
Math Expert Joined: 02 Sep 2009 Posts: 87699
For what values of k will the pair of equations 3x + 4y = 12 and kx + [#permalink] 20 Jan 2022, 07:45
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Question Stats: 73% (01:03) correct 27% (01:37) wrong based on 26 sessionsHide Show timer StatisticsFor what values of k will the pair of equations \(3x + 4y = 12\) and \(kx + 12y = 30\) does not have a unique solution?A. 12B. 9C. 7.5D. 3 E. 2.5 _________________
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Re: For what values of k will the pair of equations 3x + 4y = 12 and kx + [#permalink] 23 Jan 2022, 20:26 For the equation to not have the unique solutions, the equations should be parrallel.If 3x+4y=12 than on multiplying this equation by 3 we get 9x+12y=36 which is parallel to kx+12y=30 but for k=9,the two equations have values9x+12y= 36 and 9x+12y=30.Hence for k=9, equation will not have the unique solutions. Posted from my mobile device
Re: For what values of k will the pair of equations 3x + 4y = 12 and kx + [#permalink] 23 Jan 2022, 23:11 Linear equations:1. ax + by = c2. ux + vy = wIf a/u = b/v = c/w -> Infinite solutionsIf a/u <>(not equal to) b/v -> Unique SolutionIf a/u = b/v <> c/w -> No Solution So, the above Questions asks for a value that isn't unique. So, we can plug in the values in the options and validate the equation.
Senior Manager Joined: 14 Jun 2014 Posts: 473
Re: For what values of k will the pair of equations 3x + 4y = 12 and kx + [#permalink] 24 Jan 2022, 07:05
Bunuel wrote: For what values of k will the pair of equations \(3x + 4y = 12\) and \(kx + 12y = 30\) does not have a unique solution?A. 12B. 9C. 7.5D. 3 E. 2.5 uniique solution when two lines are parallel.3/k = 4/12k =9
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Re: For what values of k will the pair of equations 3x + 4y = 12 and kx + [#permalink] 24 Jan 2022, 07:41
Bunuel wrote: For what values of k will the pair of equations \(3x + 4y = 12\) and \(kx + 12y = 30\) does not have a unique solution?
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Re: For what values of k will the pair of equations 3x + 4y = 12 and kx + [#permalink] 24 Jan 2022, 07:41 |