This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Solutions to Midterm #2, MAT1341A Question # 1: (3 points) Let A = 1 1 3 and B = 2 4 1 1 2 3 1 3 5 . Compute AB T . Solution: AB T = 1 1 3 1 2 3 1 1 = 2 2 2 3 6 9 Question # 2: (2+2+2 = 6 points) The following augmented matrices are in RREF. Write down the general solution of the corresponding linear system. (a) 2 6 6 4 1 3 1 1 2 3 7 7 5 Solution: This system has a unique solution, since it is consistent and there is a leading one in every column. We read o the solution from the matrix; note that there are exactly 3 variables, one for each column, so the general solution is f (3 ; ; 2) g (b) 1 Solution: This question is similar to one given on HW5. It is a matrix in RREF, so we proceed as usual. There are 4 variables but only x 2 corresponds to a leading 1; it gives equation x 2 = 0. The rest are nonleading variables, so correspond to parameters. We get x 1 = r; x 2 = 0 ; x 3 = s; x 4 = t r; s; t 2 R so the general solution is f ( r; ; s; t ) j r; s; t 2 R g . (c) 2 4 1 1 3 1 1 2 3 3 5 Solution: This system is inconsistent, because the last line corresponds to a degenerate equation. Therefore the general solution, which is by de nition the set of all solutions to the system, is the empty set (denoted ; or fg ). Question # 3: (5 points) Reduce the following matrix to RREF, using Gaussian elimination. 2 6 6 4 1 2 2 1 1 2 4 1 1 1 3 3 3 1 2 1 4 3 7 7 5 1 2 Solution: 2 6 6 4 1 2 2 1 1 2 4 1 1 1 3 3 3 1 2 1 4 3 7 7 5 R 2 ! R 2 +2 R 1 ! R 4 ! R 4 R 1 2 6 6 4 1 2 2 1 1 3 3 3 3 3 3 3 1 3 3 7 7 5 ! R 2 ! 1 3 R 2 2 6 6 4 1 2 2 1 1 1 1 1 3 3 3 3 1 3 3 7 7 5 R 3 ! R 3 +3 R 2 ! R 4 ! R 4 +3 R 2 2 6 6 4 1 2 2 1 1 1 1 1 2 6 3 7 7 5 R 3 R 4 ! R 3 ! 1 2 R 3 2 6 6 4 1 2 2 1 1 1 1 1 1 3 3 7 7 5 R 2 ! R 2 R 3 ! R 1 ! R 1 R 3 2 6 6 4 1 2 2 2 1 2 1 3 3 7 7 5 ! R 1 ! R 1 2 R 2 2 6 6 4 1 2 2 1 2 1 3 3 7 7 5 Recall that RREF means reduced row echelon form  the best form for writing down the solution to a linear system; it is the optimal form in the sense of having the most zeros. Do notice that you should follow a systematic procedure to get to RREF (although shortcuts are ne once you know what you're doing!). Putting zeros \at random" in the matrix will not achieve your goal....
View Full
Document
 Winter '10
 DAIGLE
 Matrices

Click to edit the document details