Final answer:
The pivots of the matrix [1 3 3 2] are 1 and 3. The definiteness of the matrix cannot be determined without the full matrix provided. The matrix [1 3; 3 9] is semi-definite as one of the eigenvalues is zero. The approximation of f(x) = cos(x) at π/5 using the first four terms of the Taylor series can be found using the formula for the Taylor series expansion of cos(x).
Step-by-step explanation:
(1) The pivots of a matrix are the leading entries in each row of the matrix once it is in row-echelon form. In this case, the matrix [1 3 3 2] does not have any row operations performed on it, so the pivots are the non-zero entries in the first row, which are 1 and 3.
(2) To determine if the matrix is positive-definite, semi-definite, or indefinite, we need to find the eigenvalues of the matrix. The eigenvalues of a symmetric matrix can be found by solving the characteristic equation det(A-λI) = 0, where A is the matrix, λ is the eigenvalue, and I is the identity matrix. In this case, the matrix is not symmetric, so we cannot use this method to determine positive-definiteness. Instead, we can check the sign of the determinants of the upper left submatrices. If all the determinants are positive, the matrix is positive-definite. If some are positive and some are zero, it is semi-definite. If some are positive and some are negative, it is indefinite. However, without the full matrix provided, we cannot determine its definiteness.
(B) To show that the matrix [1 3; 3 9] is semi-definite using only the energy test for positive-definiteness, we need to determine if all the eigenvalues are positive or zero. You can find the eigenvalues of a matrix by solving the characteristic equation det(A-λI) = 0, where A is the matrix, λ is the eigenvalue, and I is the identity matrix. In this case, the characteristic equation is (1-λ)(9-λ)-3(3) = 0. Solving this equation yields the eigenvalues λ = 0 and λ = 10. Since one of the eigenvalues is zero, the matrix is semi-definite.
(C) To approximate f(x) = cos(x) at π/5 using the first four terms of the Taylor series, we can use the formula for the Taylor series expansion of cos(x), which is cos(x) ≈ 1 - (x^2)/2 + (x^4)/24. Plugging in π/5 for x into this formula and evaluating the expression gives an approximation for f(π/5).