MA 1B PRACTICAL - HOMEWORK SET 3 SOLUTIONS. Solution. (d) We have matrix form Ax = b and vector equation 4

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1 MA B PRACTICAL - HOMEWORK SET SOLUTIONS (Reading) ( pts)[ch, Problem (d), (e)] Solution (d) We have matrix form Ax = b and vector equation 4 i= x iv i = b, where v i is the ith column of A, and 4 A = 8 4, b = 4 5 We perform row reduction Using the first row to clear the first column of the other two rows, we get the augmented matrix Now the last two rows are multiples of each other, so we can eliminate the third row, factor a constant out of the second row, and then add the second row to the first row to arrive at 4 4 Here we take x and x 4 as free variables, whence x = 4+4x +x 4 and x = +x 4 Hence the solution set is, in vector form 4 4 x = + x + x 4 (e) We have matrix form Ax = b and vector equation 4 i= x iv i = b, where v i is the ith column of A, and A = 4 6, b = 5 We again perform row reduction, using the first row to clear the first entry in the second row: 9 6 Date: January, 7

2 MA B PRACTICAL - HOMEWORK SET SOLUTIONS Swapping the last two rows places the augmented matrix in row echelon form:, and a few more operations leave our augmented matrix in reduced row echelon form: Here x 4 is free, and x = +x 4, x = x 4, x = In vector form, the solution is x = + x 4 ( pts) Find a basis of each of the following vector spaces: (a) {(x, x,, x n ) T F n x + x + + x n = nx n } (b) {(x, x, x, x 4, x 5 ) T F 5 x = x, x = 7x 4, x + x + x 5 = } Solution (a) If n =, then our vector space is just F, and F is a single-element basis If n >, then for any (x,, x n ) in the space, we must have x n = (x + x + + x n )/(n ) Thus we can take x, x n as free variables So we have a basis {(,,,, (n ) ) T, (,,,, (n ) ) T,, (,,,, (n ) ) T } (b) We can rewrite the condition as 7 x = Row reduction leads to the following equivalent system: /4 /4 x = 7 We can thus take x 4 and x 5 as free variables, with x = x 5 /4, x = x 5 /4, and x = 7x 4 So we have the following basis for our vector space: {(,, 7,, ) T, ( /4, /4,,, ) T } 4 ( pts)[ch, Problem ] For what value of b does the system x 4 6 x = 4 x b have a solution Find the general solution of this system for this value of b

3 MA B PRACTICAL - HOMEWORK SET SOLUTIONS Solution To solve this, we will begin by determining the augmented matrix for our system of equations: b We now attempt to put this matrix into echelon form First, simply because it is easier to do it this way, we subtract half the second row from the third row to obtain b Finally, although not strictly necessary for this problem, we complete the reduction into row echelon form by dividing the second row by two and then subtracting the first row from it The result of this process is the augmented matrix b Recall now that a system is inconsistent if and only if there is a pivot in the last column of an echelon form of the augmented matrix Thus, in particular, we have that the system under consideration is inconsistent if and only if b Thus the unique value of b for which our system of equations has a solution is b = We now begin the process of finding a solution Take b = Then we have the system x x + x + x 4 6 x = x + 4x + 6x = 4 x x + x + x Note that the last and second to last conditions are saying the same thing, so in fact we have two linear equations in three unknowns, namely, x + x + x = and x + x + x = Subtracting the first condition from the second one, we immediately see that x = Now, this, in fact renders the two above equations into the same one, namely x + x =, or, rearranging, x = x + Thus, the general form of the solution to this equation is x x = x x x+ 5 ( pts)[ch Problem 4] Find the inverse of the matrices A = 7, B = 4 4 Show all steps

4 4 MA B PRACTICAL - HOMEWORK SET SOLUTIONS Solution We will find the inverses of these matrices via row reduction First we turn our attention to A We begin by forming (A I) We get 7 4 We begin by subtracting thrice the first row from the second row, since we note that this will yield ( ), which results in 4 We continue by subtracting twice the first row from the third, in order to clear out the left-most zero from the third column: We then clear out the second zero in the bottom row by adding row to row : 5 Now, we ensure that the leading coefficient in the bottom row is by dividing that row by to obtain 5 Now, we proceed to clear out the remainder of the top row We subtract row from row to obtain 7 5 Finally, we conclude by subtracting twice the second row from the first row to obtain Thus, reading off the inverse, we see that = 4 Now, we turn our attention to B We begin by forming (B I) We get 4 5

5 MA B PRACTICAL - HOMEWORK SET SOLUTIONS 5 We begin by clearing out the first position of the second row by subtracting row from row to obtain 4 4 Similarly, we subtract the first row from the third to obtain 4 We then clear out the second zero in the bottom row by subtracting row from row : 4 6 Now, we ensure that the leading coefficient in the bottom row is by dividing that row by 6 to obtain 4 We then add four times the final row to the second one to clear out the remaining nonzero term we wish to get rid of: We then divide the second row by to ensure that it is in the proper form: 6 Now, we begin our work on the top row We clear out the first lagging zero by subtracting off twice the last row to obtain 6 Finally, we add row to row to clear out the remaining lagging term: 6 Reading off the inverse, we obtain that = 6 = 4 6