answer:To prove that det(A) neq 0, we can consider the following strategies: 1. **Without finite fields:** Fix a matrix A with the given pattern. Define A^{(i,j)} to be the matrix obtained by flipping the sign of the i,j entry: A^{(i,j)}_{pq} = begin{cases} -A_{ij} & text{if } p = i text{ and } q = j A_{ij} & text{otherwise} end{cases} By applying Laplace's expansion to the determinant along the ith row, we get: det(A) - det(A^{(i,j)}) = 2C_{ij} where C_{ij} is the cofactor of the i,j entry. Since C_{ij} is an integer, flipping one sign won't change the parity of the determinant. It suffices to show that the determinant of a matrix with all off-diagonal entries equal to 1 is odd. This matrix can be represented as xx^T - I, where x is a column vector with all entries equal to 1. The determinant is then det(xx^T - I) = 1 - n, which is odd since n is even. 2. **Using finite fields (mathbb{F}_2):** Let B be an n times n matrix over mathbb{F}_2 with all entries equal to 1. Since n is even, B^2 = nB = 0. As B is nilpotent, its only eigenvalue is 0 (with multiplicity n). The matrix A modulo 2 is given by A = B + I. Since B has an eigenvalue of 0 with multiplicity n, A has an eigenvalue of 1 with the same multiplicity. Hence, det(A) equiv (1)^n equiv 1 mod 2, implying det(A) neq 0. 3. **Alternatively:** Note that B^2 = 0, so A = B + I is invertible because A^2 = B^2 + 2B + I = 0 + 0 + I = I. 4. **By manipulating the determinant:** Writing det(A) modulo 2 simplifies the problem, as pm 1 becomes 1. Consider the matrix: begin{vmatrix} 0 & 1 & 1 & dots & 1 1 & 0 & 1 & dots & 1 vdots & vdots & vdots & ddots & 1 1 & 1 & 1 & dots & 0 end{vmatrix} Sum every horizontal line to the first row to get an odd multiple of the determinant of a matrix with alternating -1 and 1 along the off-diagonal, which yields an odd determinant. Thus, det(A) is odd and cannot be 0. In all presented approaches, we have shown that det(A) is odd, hence non-zero.

answer:Standard deviation is a measure of how spread out a set of data is. The smaller the standard deviation, the more closely the data points are clustered around the mean. In this case, set (b) has the smallest standard deviation because the data points are all very close to each other. Set (a) has a wider range of values, with a difference of 95 points between the highest and lowest values. Set (c) also has a wider range of values, with a difference of 42 points between the highest and lowest values. Set (d) has a range of 37 points between the highest and lowest values. Therefore, set (b) has the smallest standard deviation. The set of scores with the smallest standard deviation is (b) 145, 143, 145, 147. This is because the values in this set are all very close to each other, with a range of only 4 points. The other sets have a wider range of values, which results in a higher standard deviation.

answer:Humanity shapes the future through its collective actions and decisions. By studying and understanding the past, we can learn from our mistakes and successes, and make more informed choices about the future. Additionally, by taking action in the present, we can directly influence the course of the future. For example, by investing in education and renewable energy, we can help to create a more sustainable and equitable future for all. By learning from the past and taking action in the present.

answer:For 2014, the IRS allowed a standard mileage rate of 0.56 per business mile. First, calculate the deduction for mileage: 20,000 business miles * 0.56 per mile = 11,200. Next, add non-mileage expenses that are deductible: - Parking: 500 - Tolls: 200 Total non-mileage deductions: 500 + 200 = 700. Now, combine the mileage and non-mileage deductions: 11,200 + 700 = 11,900. Therefore, Brittany can deduct 11,900 from her AGI.