answer:To find the eigenvectors, we first need to determine the eigenvalues. The characteristic equation is given by: left| begin{array}{cc} -frac{19}{3} - lambda & -frac{26}{3} -4 & -frac{23}{3} - lambda end{array} right| = 0 Solving for lambda gives us the eigenvalues: lambda = frac{1}{6} left(-1-sqrt{79}right), quad lambda = frac{1}{6} left(sqrt{79}-1right) For each eigenvalue, we solve the system of linear equations: left( begin{array}{cc} -frac{19}{3} - frac{1}{6} left(-1-sqrt{79}right) & -frac{26}{3} -4 & -frac{23}{3} - frac{1}{6} left(-1-sqrt{79}right) end{array} right) begin{bmatrix} x y end{bmatrix} = begin{bmatrix} 0 0 end{bmatrix} and left( begin{array}{cc} -frac{19}{3} - frac{1}{6} left(sqrt{79}-1right) & -frac{26}{3} -4 & -frac{23}{3} - frac{1}{6} left(sqrt{79}-1right) end{array} right) begin{bmatrix} x y end{bmatrix} = begin{bmatrix} 0 0 end{bmatrix} After solving these systems, we obtain the eigenvectors: v_1 = left{frac{1}{6} left(-1-sqrt{79}right),1right} and v_2 = left{frac{1}{6} left(sqrt{79}-1right),1right} The eigenvectors are normalized by dividing each by its largest component, ensuring they have a length of 1: v_1 = left{frac{1}{sqrt{left(-1-sqrt{79}right)^2 + 1^2}},1right} and v_2 = left{frac{1}{sqrt{left(sqrt{79}-1right)^2 + 1^2}},1right} Keep in mind that the normalization step might slightly vary the eigenvector components due to rounding errors.

answer:Let X_1, X_2, ..., X_n be a random sample from a normal distribution with mean mu and standard deviation sigma. Then the sample mean bar{X}_n = frac{1}{n}sum_{i=1}^n X_i is also normally distributed with mean mu and standard deviation frac{sigma}{sqrt{n}}. Therefore, a 95% confidence interval for mu is given by: bar{X}_n pm 1.96 frac{sigma}{sqrt{n}} If the population standard deviation sigma is unknown, it can be estimated using the sample standard deviation S_n = sqrt{frac{1}{n-1}sum_{i=1}^n (X_i - bar{X}_n)^2}. In this case, a 95% confidence interval for mu is given by: bar{X}_n pm t^* frac{S_n}{sqrt{n}} where t^* is the critical value from the t-distribution with n-1 degrees of freedom. To calculate a confidence interval for the sum of the random sample, we can use the fact that the sum of independent normal random variables is also normally distributed. Therefore, the sum T_n = sum_{i=1}^n X_i is normally distributed with mean nmu and standard deviation sqrt{n}sigma. Therefore, a 95% confidence interval for nmu is given by: T_n pm 1.96 sqrt{n}sigma If sigma is unknown, we can use the sample standard deviation S_n to estimate it. In this case, a 95% confidence interval for nmu is given by: T_n pm t^* sqrt{n}S_n where t^* is the critical value from the t-distribution with n-1 degrees of freedom.

answer:The (pseudo)inverse of a matrix A is denoted by A^+ and is defined as A^+ = (A^TA)^{-1}A^T. To compute the (pseudo)inverse of A, we first calculate its transpose and then its product with itself: A^T = left( begin{array}{ccc} -1 & 2 & 2 -1 & 2 & -4 -4 & 2 & -1 end{array} right), A^TA = left( begin{array}{ccc} 18 & 0 & 0 0 & 18 & 0 0 & 0 & 18 end{array} right). Next, we find the inverse of A^TA: (A^TA)^{-1} = frac{1}{18} left( begin{array}{ccc} 18 & 0 & 0 0 & 18 & 0 0 & 0 & 18 end{array} right) = left( begin{array}{ccc} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 end{array} right). Finally, we multiply the inverse of A^TA with A^T to obtain the (pseudo)inverse of A: A^+ = (A^TA)^{-1}A^T = left( begin{array}{ccc} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 end{array} right) left( begin{array}{ccc} -1 & 2 & 2 -1 & 2 & -4 -4 & 2 & -1 end{array} right) = left( begin{array}{ccc} frac{1}{6} & frac{5}{12} & frac{1}{6} frac{1}{6} & frac{1}{4} & -frac{1}{6} -frac{1}{3} & -frac{1}{6} & 0 end{array} right). The answer is A^+ = left( begin{array}{ccc} frac{1}{6} & frac{5}{12} & frac{1}{6} frac{1}{6} & frac{1}{4} & -frac{1}{6} -frac{1}{3} & -frac{1}{6} & 0 end{array} right).

answer:The least squares vector x is calculated as: x = (A^T A)^{-1} A^T b After performing the necessary matrix operations, the resulting least squares vector x is: left( begin{array}{c} 0.352 0.272 -0.136 -0.074 0.488 end{array} right)