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Systems of Linear Equations and Matrices Section 1. Note that both systems have infinitely many solutions. The solution is. The augmented matrix is 1 2 3 1. Add 2 times the first row to the second row, 1 times the first row to the third row, and 3 times the first row to the fourth row. Since the system has more unknowns 4 than equations 3 , it has nontrivial solutions. Since the system has more unknowns 3 than equations 2 , it has nontrivial solutions.

Multiply the second row by. The augmented matrix is 1 2 0 0. In Exercise 5, the following row echelon matrix occurred. The augmented matrix is 3 1 1 1 0. Section 1. The augmented matrix is. Multiply the second row by 1, then add 3 times the new second row to the third row and 1 times the new second row to the fourth row. Add 2 times the first row to the third row and 2 times the first row to the fourth row. Add 10 times the third row to the fourth row, 2 times the third row to the second row, and 3 times the third row to the first row.

The augmented matrix is 1 2 3 4 3 1 5 2. Add 3 times the first row to the second row and 4 times the first row to the third row.

Add 2 times the first row to the second row. Add 3 times the second row to the first row. For any other value of a, the solution of 1. Multiply the third row by 1 then add 3 times the new third row to the second row and 3 times the new third row to the first row.

Add 27 times the first row to the second row and 64 times the first row to the third row. Since the homogeneous system has only the trivial solution, using the same steps of GaussJordan elimination will reduce the augmented matrix of the nonhomogeneous system to the 1 0 0 d1 form 0 1 0 d 2. Chapter 1: Systems of Linear Equations and Matrices Since AB BA, the two terms cannot be combined. A similar statement can be made if A has a column of zeros.

The matrix is. The matrix is 0 1 0. Add 2 times the first row to the second and 4 times the first row to the third. Add 3 times the first row to the second and 2 times the first row to the third. Add 2 times the first row to both the second and third rows. Add 1 times the third row to both the first and second rows. Add 2 times the first row to the second.

Add 8 times the third row to the fourth. Add 12 times the third row to the first and 4 times the third row to the second. Add 5 times the fourth row to the third. Add the third row to the first and 2 times the third row to the second. Add 1 times and the first row to the second and third rows. Add 1 times the third row to the first and second rows. Use elementary matrices to reduce 0 4 3.

Use elementary matrices to reduce to 2 2 I2. E2 E1 A. Write and reduce the augmented matrix for both systems. Write and reduce the augmented matrix for the four systems.

The matrix is a diagonal matrix with nonzero entries on the diagonal, so it is invertible. The inverse is 2 0 15 3. The matrix is symmetric. The matrix is not invertible because it is upper triangular and has a zero on the main diagonal. For A to be invertible, the entries on the main diagonal must be nonzero. Thus x 1, 2, 4. Exercise Set 1. Label the network with nodes, flow rates, and flow directions as shown. Since I 2 is 5 5 5 negative, its direction is opposite to that indicated in the figure.

To keep the traffic flowing on all roads, each flow rate must be nonnegative. These equations form the The reduced row echelon form of this matrix is 1 0 0 31, 0 1 0 26, From this we conclude.

This shows that the economy is productive. If C has row sums less than 1, then CT has column sums less than 1. Thus, from Theorem 1. Reduce the augmented matrix. Thus, a 0, b 2 is required. Note that K must be a 2 2 matrix. The augmented matrix of the above system is 8 0 8 2 0 0 1 0 4 6. Note that all matrices must be square and of the same size. Consider the ith row of AB.

Exercise Set Add column five to column three and take advantage of another row of all but one zero to get. Substituting each of the points x1 , y1 ,. These together with the original equation form a homogeneous linear system with a non-trivial solution for c1, c2 , Thus the determinant of the coefficient matrix is zero, which is exactly Equation As in the previous problem, substitute the coordinates xi , yi , zi of each of the three.

Completing the squares in each variable yields the standard form. Substituting the coordinates xi , yi , zi of the four points into the equation. Thus the determinant of this system is zero, which is Equation Upon substitution of the coordinates of the three points x1 , y1 , x2 , y2 and x3 , y3 , we obtain the equations:.

This is a homogeneous system with a nontrivial solution for c1, c2 , c3 , c4 , so the determinant of the coefficient matrix is zero; that is, y.

In the figure the feasible region is shown and the extreme points are labeled. The values of the objective function are shown in the following table:. Expanding the determinant in Equation 9 by cofactors of the first row makes it apparent that.

If the three distinct points are collinear then two of the coordinates can be expressed in terms of the third. Without loss of generality, we can say that y and z can be expressed in terms of x, i.

The intersection of the five half-planes defined by the constraints is empty. Thus this problem has no feasible solutions. In the problems terms, if we use z there is 18 7 25 cup of milk and ounces of corn flakes, a 9 18 minimum cost of The feasible region for this problem, shown in the figure, is unbounded.

The feasible region and vertices for this problem are shown in the figure. The feasible region is shown in the figure. The feasible region and its extreme points are shown in the figure. Though the region is unbounded, x1 and x2 are always positive, so.

Section The number of oxen is 50 per herd, and there are 7 herds, so there are oxen. Let x1 be the number of pounds of ingredient A used and x2 the number of pounds of ingredient B. Note that this is, effectively, Gaussian elimination applied to the augmented matrix 1 1 10 1 1 7. The corresponding array is. Though the feasible region shown in the figure is unbounded, the objective function is always positive there and hence must have a minimum. The minimum value of z is 5 5 or It is sufficient to show that if a linear function has the same value at two points, then it has that value along the line connecting the two points.

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