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Since this covers all of the possibilities, there is never a unique solution. Exercise Set 1. Thus we have the same solution which we obtained in Problem 10 c. Any other value of a will yield a unique solution for z and hence also for y and x. This proof uses it twice. There are eight possibilities. The reduced row-echelon form of a matrix is unique, as stated in the remark in this section. The system can have a solution only if the 3 lines meet in at least one point which is common to all 3.
Let fij denote the entry in the ith row and jth column of C DE. In order to compute f23, we must calculate the elements in the second row of C and the third column of DE.
Thus [aij] has all zero elements below the diagonal. The second inequality says that the entry aij lies above the diagonal and also above the entries immediately above the diagonal ones. Each of these alternatives leads to a contradiction. A similar argument works for ATA, and since the sum of the squares of the entries of AT is the same as the sum of the squares of the entries of A, the result follows.
Every entry in the first row of AB is the matrix product of the first row of A with a column of B. Call the matrix A. By Theorem 1. Let A denote a matrix which has an entire row or an entire column of zeros. Then if B is any matrix, either AB has an entire row of zeros or BA has an entire column of zeros, respectively.
See Exercise 18, Section 1. We use Theorem 1. By Part d of Theorem 1. Thus, it is elementary. Thus it is elementary. Thus it is not an elementary matrix. Therefore, E1 must be the matrix obtained from I3 by interchanging Rows 1 and 3 of I3, i. Therefore, E3 must be the matrix obtained from I3 by replacing its third row by —2 times Row 1 plus Row 3, i. If A is an elementary matrix, then it can be obtained from the identity matrix I by a single elementary row operation.
Thus at least one entry in Row 3 must equal zero. From Theorem 1. Hence we have found, via elementary matrices, a sequence of elementary row operations which will put B in the same reduced row-echelon form as A.
Now suppose that A and B have the same reduced row-echelon form. Since the inverse of an elementary matrix is also an elementary matrix, we have that A and B are row equivalent. The matrix A, by hypothesis, can be reduced to the identity matrix via a sequence of elementary row operations.
Suppose we reduce A to its reduced row-echelon form via a sequence of elementary row operations. The resulting matrix must have at least one row of zeros, since otherwise we would obtain the identity matrix and A would be invertible.
Thus, if B were invertible, then A would also be invertible, contrary to hypothesis. For this system to have a unique solution, A — I must be invertible.
Let A and B be square matrices of the same size. If either A or B is singular, then AB is singular. Hence, there are 8 possible choices for x, y, and z, respectively, namely 4, 4, 4 , 4, 4, —1 , 4, —1, 4 , 4, —1, —1 , —1, 4, 4 , —1, 4, —1 , —1, —1, 4 , and —1, —1, —1. Therefore, the result does not hold. In general, suppose that A and B are commuting skew-symmetric matrices. To multiply two diagonal matrices, multiply their corresponding diagonal elements to obtain a new diagonal matrix.
Thus, if D1 and D2 are diagonal matrices with diagonal elements d1,. Continuing in this way, we can solve for successive values of xi by back substituting all of the previously found values x1, x2,. Supplementary Exercises 1 57 By virtue of Theorem 1. Note that all matrices must be square and of the same size. An argument similar to the one given above will serve, and we leave the details to you. Suppose that A is a square matrix whose entries are differentiable functions of x.
Suppose also that A has an inverse, A—1. Note that we never have to divide by a function which is identically zero. Then using Theorem 1. We prove this by induction. Exercise Set 2. This follows from Theorem 2. Let A be an upper not lower triangular matrix. Note that if we do so then A2,. Hence, X is upper triangular; the inverse of an invertible upper triangular matrix is itself upper triangular. Now apply Theorem 1. From I4 we see that such a matrix can have at least 12 zero entries i.
Expanding along that row shows that its determinant is necessarily zero. Since the given matrix is upper triangular, its determinant is the product of the diagonal elements. By Theorem 2. Take the transpose of the matrix. Therefore the matrix is not invertible. We work with the system from Part b. The solution is valid for all values of t.
Case 2: Let E be obtained by interchanging two rows of In. Case 3: Let E be obtained by adding a multiple of one row of In to another. If either A or B is singular, then either det A or det B is zero. Thus AB is also singular. If it could, then it would be invertible as the product of invertible matrices. The reduced row echelon form of A is the product of A and elementary matrices, all of which are invertible.
In general, reversing the order of the columns may change the sign of the determinant. There are 24 terms in this sum. Since the product of integers is always an integer, each elementary product is an integer. The result then follows from the fact that the sum of integers is always an integer. Now consider any elementary product a1j1a2j2 … anjn. Hence, a11a22 … ann is the only elementary product which is not guaranteed to be zero.
Since the column indices in this product are in natural order, the product appears with a plus sign. Thus, the determinant of U is the product of its diagonal elements. A similar argument works for lower triangular matrices. See Theorem 2.
We simply expand W. This will insure that the sum of the products of corresponding entries from the ith row of A and the ith column of A—1 will remain equal to 1. Call that matrix B. Now suppose that we add —c times the jth column of A—1 to the ith column of A—1.
Elementary Linear Algebra, Students Solutions Manual (e-only)
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.
The 3-tuples in b and c are not solutions of the system. The 3-tuples in a and c are not solutions of the system. The second equation is contradictory, so the original system has no solutions. The lines represented by the equations in that system have no points of intersection the lines are parallel and distinct.
Anton - Elementary Linear Algebra With Applications 10e Student Solutions Manual