For what integer 0<n<1000, does the following system of n linear equations have no complex solution?
−10a1a1a1⋮a1+a2−10a2+a2+a2+a3+a3−10a3+a3+…+…+…+…+an+an+an−10an=9=99=999⋮=10n−1
Clarification: If the system always admits a solution, type your answer as 0. If there are multiple cases where the system doesn't admit a solution, type in any one of your answers.
To determine the integer value of n for which the given system of n linear equations has no complex solutions, we need to analyze the coefficients of the equations.
The equations can be rewritten as follows:
-10a1 + 1a2 + 0a3 + ... + 0an = 9
1a1 - 10a2 + 1a3 + ... + 0an = 9
0a1 + 1a2 - 10a3 + ... + 0an = 9
...
0a1 + 0a2 + 0a3 + ... + 10an = 9
Notice that all the linear equations have the same pattern of coefficients, except for the diagonal elements which are -10. The diagonal elements correspond to the coefficient of the term with the same variable in each equation.
For there to be no complex solutions, the determinant of the coefficient matrix must be non-zero. In this case, the coefficient matrix is a Toeplitz matrix, where each row is a cyclic shift of the previous row.
The determinant of a Toeplitz matrix can be calculated using the Levinson recursion, which is an efficient algorithm for computing determinants of Toeplitz matrices.
However, there is an easier method to determine the integer value of n. Since we are looking for the smallest value of n for which the determinant is non-zero, we can start by substituting values for n and check if the determinant is non-zero.
Let's start by substituting n = 0.
The coefficient matrix becomes:
[-10]
Since the determinant of a 1x1 matrix is simply the value of the element, the determinant of this matrix is -10, which is non-zero. Hence, the system admits a solution for n = 0.
Let's continue by substituting n = 1.
The coefficient matrix becomes:
[-10 1]
[ 1 -10]
Now, we can calculate the determinant of this 2x2 matrix.
Determinant = (-10)(-10) - (1)(1) = 100 - 1 = 99, which is non-zero. Therefore, the system admits a solution for n = 1 as well.
Continuing this process, we can substitute values for n up to n = 1000. However, it is highly likely that we will find a smaller value for n before reaching 1000.
Alternatively, since we want to find the smallest value of n for which the determinant is non-zero, we can save time by directly substituting larger values for n.
If we substitute n = 2, the coefficient matrix becomes:
[-10 1 0]
[ 1 -10 1]
[ 0 1 -10]
We can calculate the determinant of this 3x3 matrix using the Laplace expansion or other methods:
Determinant = (-10)((-10)(-10) - (1)(1)) - (1)((-10)(-10) - (0)(1)) + (0)((1)(1) - (-10)(0))
= (-10)(99) - (1)(99)
= -990 - 99
= -1089
Since the determinant is non-zero (-1089), we can conclude that the system admits a solution for n = 2.
Continuing this process, we can substitute larger values for n until we find the smallest one for which the determinant is non-zero.
However, since the complexity of evaluating determinants increases with the size of the matrix, it might be more efficient to use a computer program or a symbolic computation software to calculate determinants for larger values of n.
In conclusion, to determine the integer value of n for which the given system of n linear equations has no complex solutions, we need to substitute values for n and calculate the determinant of the corresponding coefficient matrix. By starting with smaller values for n and gradually increasing, we can determine the smallest value of n for which the determinant is non-zero.