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All are valid. Exercise 2. Only d represents a function. The range is the set of all nonpositive numbers. This would establish the uniqueness as required by the definition of a function. Being negative, this term serves to pull down the y values at the two extreme ends of the curve. If negative values can occur there will appear in quadrant III a curve which is the mirror image of the one in quadrant I.
The two curves would be parallel, with no equilibrium intersection point in Fig. The resulting equation is the same as the one in b above. Equation 3. No, because the equation system is nonlinear 4. No, not conformable. When u, v, and w all lie on a single straight line. Let the vector v have the elements a1 ,. The point of origin has the elements 0,. No, x0 Ax would then contain cross-product terms a12 x1 x2 and a21 x1 x2.
Unweighted sum of squares is used in the well-known method of least squares for fitting an equation to a set of data. Weighted sum of squares can be used, e. Exercise 4. The given diagonal matrix, when multiplied by itself, gives another diagonal matrix with the diagonal elements a , a ,.
Hence each aii must be either 1, or 0. Since each aii can thus have two possible values, and since there are altogether n of these aii , we are able to construct a total of 2n idempotent matrices of the diagonal type.
Two examples would be In and 0n. It is suggested that this particular problem could also be solved using a spreadsheet or other mathematical software. The student will be able to observe features of a Markov process more quickly without doing the repetitive calculations. We get the same results as in the preceding problem. In A1 keep row 1 as is, but add row 1 to row 2, to get A2. In A2 , divide row 2 by 5.
In B1 , divide row 1 by 6. In B2 , multiply row 2 by 2, then add the new row 2 to row 3. Matrix is singular. In C1 divide row 1 to 7. The question of nonsingularity is not relevant here because C is not square. But the interchange of rows 1 and 2 gives us simpler numbers to work with. Since the last two rows of D2 , are identical, linear dependence is obvious.
Again, the question nonsingularity is not relevant here. The link is provided by the third elementary row operation. This process involves the third elementary row operation. Exercise 5. Factoring out the k in each successive column or row —for a total of n columns or rows —will yield the indicated result.
In d , the rank is 3. In a , b and c , the rank is less than 3. The set in a can because when the three vectors are combined into a matrix, its determinant does not vanish. But the set in b cannot. They are ai3 Ci2 and a2j C4j , respectively. Yes, Matrices G and H in problem 4 are examples. Thus the matrix equation is 0. Element 0: Industry III does not use its own output as its input.
Element The open sector demands billion dollars of commodity II. Exercise 6. Yes, the limit is For example, 2. Yes; each function is not only continuous but also smooth. Refer to the following two graphs Exercise 7. Since the total-cost function shows zero fixed cost, the situation depicted is the long run. Exercise 7. The equations in 7. Hence, by 8. At least one of the partial derivatives in the vector of constants in 8. Exercise 8. Thus the implicit- function theorem is applicable.
A similar complication arises when the supply function is used to get the other comparative-static derivative. These are in the nature of relative extrema, thus a minimum can exceed a maximum.
A Exercise 9. It plots as an upward-sloping straight line emanating from the point of origin. Exercise 9. The critical values are 1 and 5. With its minimum at 3a zero output. The MC curve must be upward-sloping throughout. The value of marginal product must be equated to the wage rate.
Thus, by 9. Since the first nonzero derivative value is f 4 2. The value of t can be either positive, zero, or negative. Exercise The nonzero requirement serves to preclude this contingency. The graphs are of the same general shape as in Fig. Similarly to formula Thus, by Thus Both are the same as before.
Both roots being positive, u0 Du is positive definite. Both roots being negative, u0 Eu is negative definite. Since r1 is positive, but r2 is negative, u0 F u is indefinite.
Using r1 in The diagonal elements are all negative for problem 2, and all positive for problems 4 and 5. Thus the characteristic roots are all negative d2 z negative definite for problem 2, and all positive d2 z positive definite for problem 4. Thus the maximum in this problem is a unique absolute maximum. The highest 1 is cd1 ; the lowest is cd2.
Letting Pa0 alone vary i. In