1. Math 3C Practice Midterm Solutions Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

2. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 1) Solve the following system of linear equations: Augmented matrix form Row reduction: The 3 rd row represents the equation 0x+0y+0z = -5. This equation has no solution, so the original system of equations is inconsistent (not solvable).

3. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 2) Consider the following matrix: a)Find the determinant of this matrix. b)Is this matrix invertible? If so, find the inverse. c)What is the determinant of the inverse of this matrix? d)Using this information, solve the following system of linear equations: a) Since A is a triangular matrix, the determinant is the product of the diagonal elements – det(A) = -312 b) Since det(A) is not zero, A is invertible, and we can use row reduction to find the inverse.

4. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 2) Consider the following matrix: a)Find the determinant of this matrix. b)Is this matrix invertible? If so, find the inverse. c)What is the determinant of the inverse of this matrix? d)Using this information, solve the following system of linear equations: c) Since A -1 is a triangular matrix, the determinant is the product of the diagonal elements: det(A -1 ) = -1/312 d) Notice that we have the inverse for the coefficient matrix, so we just need to multiply:

5. 3) Consider the following set of vectors in R 3 a)How many linearly independent vectors are in this set? b)Is the span of this set of vectors R 3 ? c)Is this a basis for R 3 ? d)Is each of the following vectors within the span of this set? If so, express them as a linear combination of the vectors in the set. a) Put the column vectors in a matrix and perform elementary row operations: This has 2 pivot columns, so there are 2 linearly independent vectors in the set. In other words, the span of this set is 2-dimensional. b)No, the span is 2-dimensional, and R 3 is 3-dimensional. c)No, this is not a basis because it does not span R 3, and there are 4 vectors, not 3. d) We have several options for this part. One way is to make a good guess. If you notice that v 2 is just 2 times a 3 then we have our answer for that one: For v 1 there is no obvious answer, so we need to try to solve for it. Since we know that our span is only 2-dimensional, we only need two basis vectors. Choose any 2 independent vectors from our set, say a 1 and a 2, and try to write v 1 as a linear combination of them. We get an inconsistent set of equations, so v 1 is not in the span. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

6. 4) Consider the following set of vectors in P 2 : a)How many linearly independent vectors are in this set? b)Does this set span P 2 ? c)Is this set a basis for P 2 ? d)Express each of the standard basis vectors {1, x, x 2 } as a linear combination of the vectors in the set. a) These are 3 independent vectors. We can see this if we try to form a linear combination of them, and find that there is no way to cancel them out: b) P 2 is a 3-dimensional space, and we have 3 independent vectors in P 2 – yes they span P 2. c) P 2 is a 3-dimensional space, and we have 3 independent vectors in P 2 – yes they form a basis for P 2. d) Form a linear combination of a 1, a 2 and a 3 : Similar analysis yields the other 2 combinations. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB