Learning Objectives¶By the end of this section you should be able to: Show
Unique Solution¶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. 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. 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. 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. What is the formula for unique solution?A system of linear equations a 1 x + b 1 y = 0 a 2 x + b 2 y = 0 has a unique solution, if a 1 a 2 ≠ b 1 b 2 .
When the system of equation has a unique solution the system is?Systems of equations can be classified by the number of solutions. If a system has at least one solution, it is said to be consistent . If a consistent system has exactly one solution, it is independent .
What is the condition for unique solution in determinant?The value of determinant (D) should not equal zero. So, this is the required condition of a unique solution of a pair of linear equations in two variables.
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