Linear Algebra Review Solutions for Exam 2022

Review Solutions for Exam 1 Definitions and Basic Theorems 1. Finish the definition: (a) A linear combination of vectors {v1 , . . . , vn } is: any vector of the form c1 v1 + c2 v2 + · · · + cn vn (b) A set of vectors {v1 , . . . , vn } are said to be linearly independent if: the only solution to c1 v1 + c2 v2 + · · · + cn vn = 0 is the trivial solution, c1 = c2 = · · · = cn = 0. (c) The span of a set of vectors {v1 , . . . , vn } is: the set of all linear combinations of the vectors. (d) A system of equations is inconsistent if: there is no solution that satisfies all of the equations. (e) A system of equations is homogeneous if: the equations are set to zero. (f) Two matrices are row equivalent if: you can change one into the other by a sequence of row operations. (g) A transformation T : X → Y is said to be linear if: T (x1 + x2 ) = T (x1 ) + T (x2 ), for all x1 , x2 in X, and T (cx) = cT (x) for all x ∈ X and all scalars c. (h) A transformation T : X → Y is said to be one to one if the preimage of every y ∈ Y is either empty or a single value of x ∈ X. Alternative: If T (x1 ) = T (x2 ), then x1 = x2 . Altnerative: If x1 6= x2 , then T (x1 ) 6= T (x2 ). (i) The definition of Ax: This is a linear combination of the columns of A using the weights in x. If A is m × n, then: Ax = x1 a1 + x2 a2 + · · · + xn an 2. What information about T : IRn → IRm do we need to know in order to compute the standard matrix for the transformation? SOLUTION: The best information to know is where T sends the standard basis vectors. That is, T (e1 ), . . . , T (en ). That’s because these are the columns of the matrix. It is possible to use any linearly independent set of n vectors, but we would have some work to do to find the matrix. 3. Fill in the blanks for the Existence and Uniqueness Theorem. Your answers should refer to pivot columns: 1 • A linear system is consistent if and only if: the rightmost column of the augmented matrix is not a pivot column. • Furthermore, if the system is consistent, the solution is unique if: every column of the coefficient matrix is a pivot column. 4. Give three statements that are logically equivalent to saying that A has a pivot in every row. I’ll give some hints so you can fill in the blanks: • • • • For each b ∈ IRm , the equation Ax = b has at least one solution for every b. The columns of A span IRm Each b is a linear combination of the columns of A The mapping x → Ax will be ONTO. 5. Similar to the last problem, give two statements that are logically equivalent to saying that: A has a pivot in every column. SOLUTION: Here are some options: (a) (b) (c) (d) The mapping x → Ax is 1-1. For every b ∈ IRm , the equation Ax = b has AT MOST 1 solution. The homogeneous equation has only the trivial solution. The columns of A are linearly independent. 6. Suppose A is m × n with m > n. Is it possible that the mapping x → Ax is 1-1? Onto? (Explain). SOLUTION: If m > n and we have the maximum number of pivots (n of them), then every column is a pivot column. This means that the equation Ax = b has at most one solution for every b. That means the mapping x → Ax could be 1 − 1 (it is possible, but not necessary). On the other hand, if we have the maximum number of pivots, we would still not have enough for all the rows of A, thus we must have at least one row of A that is not a pivot row. That means that we can find b so that Ax = b has no solution, and so the mapping can never be “onto”. 7. Same question, but let A be m × n with m < n. SOLUTION: With the matrix “wide” instead of “tall”, we might be able to get a pivot in every row, but we will never get a pivot in every column. Thus, while it is possible that the equation Ax = b is consistent for every b (meaning it is possible that the mapping x → Ax is “onto”), the solution will never be unique (for any b). Therefore, the mapping Ax = b cannot be 1 − 1. 2 Computational Questions 1. Find the row reduced echelon form of the matrix A given below. Be sure to show all your work: SOLUTION:     1 −7 0 6 5 1 −7 0 6 5 0 1 −2 −3  →  0 0 1 −2 −3  A= 0 −1 7 −4 2 7 0 0 0 0 0 2. If the matrix given above was actually an augmented matrix, use your row reduced echelon form to give the solution to the system. SOLUTION: If this had been an augmented matrix, then the RREF would be:         x1 = 5 +7x2 −6x4 5 7 −6 5 1 −7 0 6  0      x = x2 +x2  1 +x4  0   0 0 1 −2 −3  → 2 →x=  −3   0   2  x3 = −3 +2x4 0 0 0 0 0 x4 = x4 0 0 1 3. Do the three lines x1 − 4x2 = 1, 2x1 − x2 = −3 and −x1 − 3x2 = 4 have a common point of intersection? Explain. SOLUTION: If there is a common point of intersection, it would be found by solving the given system of equations (and there would be a unique solution):     1 1 −4 1 1 −4  2 −1 −3  →  0 7 −5  −1 −3 0 0 4 0 Every column is a pivot column, and the system is consistent, so the solution is unique, and the three lines do indeed have a point of intersection. 4. Let a3 = 2a1 − 3a2 . Let A = [a1 a2 a3 ]. If A is 3 × 3 with 2 pivots, write the solution to Ax = 0 in parametric form. SOLUTION: If we have two pivots, then there is one free variable. Therefore, we know the solution set is a line through the origin. Also, since 2a1 − 3a2 − a3 = 0 then the solution set must be: s(2, −3, −1) (recall that this is scalar s times a column vector in our textbook notation), where s is a free value (parameter). 5. Write the equation of a plane that spans (1, 2, 3, 4), (1, −1, −1, 1) and has been translated by (3, 0, 0, 1) 3 SOLUTION:    3 1  0   2   + s  0   3 1 4   1     + t  −1    −1  1  6. Find the general solution (in parametric vector form) to the system: x1 + 3x2 + x3 + x4 = −1 −2x1 − 6x2 − x3 = 5 x1 + 3x2 + 2x3 + 3x4 = 2 SOLUTION: Row reduction gets you the matrix to the left; solving gives you the parametric equation on the right:          −4 −3 1 1 3 0 −1 −4 1 3 1 1 −1  0   1   0      −2 −6 −1 0 5 ∼ 0 0 1 2 3 ⇒x=  3 +r  0 +s  −2 2 0 1 3 2 3 0 0 0 0 0 0 1 7. Suppose the solution set of a certain system of linear equations is given by x1 = 5+4x4 , x2 = −2 + 7x4 with x3 = 2 + x4 and x4 is a free variable. (a) Use vectors to describe the solution set as a (parametric) line in IR4 . SOLUTION: x = (5, −2, 2, 0) + x4 (4, 7, 1, 1) (b) Was the original system homogeneous? If not, give the solution to the homogeneous system of equations, if you have enough information. SOLUTION: No, the system was not homogeneous since 0 is not a solution. However, we know all solutions are of the form xp + xh , where the parameters are in xh . Therefore, we can think of (5, −2, 2, 0) as the particular part of the solution to Ax = b, and x4 (4, 7, 1, 1) is the homogeneous part of the solution.     x1 4x1 − 2x2 8. Show that the mapping T is not linear: T = x2 3x2 + 1 Quick method: T (~0) = (0, 1) 6= (0, 0), therefore the mapping is not linear. 9. Determine if the mapping is linear: T (x1 , x2 ) = x21 + 3x2 . SOLUTION: We check if T (x1 + y1 , x2 + y2 ) = T (x1 , x2 ) + T (y1 , y2 ): T (x1 + y1 , x2 + y2 ) = (x1 + y1 )2 + 3(x2 + y2 ) And T (x1 , x2 ) + T (y1 , y2 ) = (x21 + y12 ) + 3(x2 + y2 ) 4     These two quantities are not the same- The first quantity has 2x1 y1 term that the second does not. Therefore, T is not linear. Alternative: Check that T (cx1 , cx2 ) = cT (x1 , x2 ) T (cx1 , cx2 ) = (cx1 )2 + 3cx2 = c2 x21 + c3x1 = c(cx21 + 3x2 ) 6= cT (x1 , x2 ) 10. Given the matrix A below, explain whether or not the system Ax = b has a solution in terms of h. If there are restrictions on b, give them.   1 −3 A= 2 −h SOLUTION: Row reduce:     b1 1 −3 1 −3 b1 ∼ 2 −h b2 0 6 − h b2 − 2b1 Therefore, if h 6= 6, we have a unique solution for all vectors b. If h = 6, the system is consistent only when b2 = 2b1 , and in that case we have an infinite number of solutions to the system. Otherwise, if b2 6= 2b1 , the system is inconsistent. 11. Let A be a 3 × 4 matrix, let y 1 , y 2 be vectors, and let w = y 1 + y 2 . Suppose that y 1 = Ax1 and y 2 = Ax2 for some vectors x1 and x2 . (a) What size must y 1 , y 2 , x1 , x2 be? SOLUTION: The vectors x1 , x2 are in IR4 and y1 , y2 are in IR3 (b) Does Ax = w have a solution? Why or why not? SOLUTION: Yesw = y1 + y2 = Ax1 + Ax2 = A(x1 + x2 ) so the solution to Ax = w is x + x1 + x2 . 12. Suppose that:  T 1 2   = −3 0   , and T −2 1   = 1 2  Find a matrix A so that T (x) = Ax. SOLUTION: In this case, we don’t know T (e1 , T (e2 ), so we’ll try something different. Let the matrix A be given below, so that we need to find a, b, c, d:  a b c d  1 2   = −3 0   and 5 a b c d  −2 1   = 1 2  These lead us to the system of equations: a + 2b = −3 −2a + b = 1 c + 2d = 0 ⇒ −2c + d = 2  A= −1 −1 −4/5 2/5  13. Determine the matrix for the linear transformation T given below: T (x1 , x2 , x3 , x4 ) = 3x1 − 4x2 + 8x4 SOLUTION: You could think about T (ei ), or just reason it out. Note that T : IR4 → IR, so A is 1 × 4, and A = [3, −4, 0, 8] 14. Let T : IR3 → IR2 , where T (e1 ) = (1, 4), T (e2 ) = (−2, 9), and T (e3 ) = (3, −8). Find a matrix A so that T (x) = Ax. SOLUTION: First, note the size of A should be 2 × 3, and  we alreadyhave the action 1 −2 3 of T on the standard basis vectors. The matrix is A = 4 9 −8 15. Let T : IR2 → IR3 so that T (x) = Ax, where   1 1 A =  −2 −1  −1 −3 Is T 1-1? Explain. Is T onto? Explain. SOLUTION: Examining the matrix, we see that there are two pivots. Therefore, if the system is consistent (and the last row of the rref form of A will be all zeros), there is exactly one solution. This corresponds to the mapping being 1 − 1 but not onto- That is, the solution to Ax = b, if it exists, is unique (but it may not exist). 16. Suppose we want to determine a quadratic function f (x) = a0 + a1 x + a2 x2 that interpolates the data (1, 1), (−1, 2), and (2, 3). Write down the system of equations (and the corresponding matrix equation) we need to solve (do not actually solve the system) to find the polynomial. SOLUTION: Put each point in turn into the given polynomial. That gives the system of equations.   a0 +a1 (1) +a2 (1)2 = 1 1 1 1 1 a0 +a1 (−1) +a2 (−1)2 = 2 →  1 −1 1 2  1 2 4 3 a0 +a1 (2) +a2 (2)2 = 3 6 Discussion Questions 1. Suppose A is 3 × 3 an y is a vector in IR3 such that the equation Ax = y does not have a solution. Does there exist a vector z in IR3 such that Ax = z has a unique solution? SOLUTION: The setup is telling us that the RREF of the coefficient matrix A has a row of zeros. Therefore, there are at most two pivots, and we have three columns. Therefore, at least one column of A is not a pivot column, so if the equation is consistent, we must have an infinite number of solutions (we have at least one free variable). 2. Let A be n × n. If the equation Ax = 0 has only the trivial solution, do the columns of A span IRn ? Why or why not? Is your answer different if A is n × m? SOLUTION: Yes. If the equation Ax = 0 has only the trivial solution, there is a pivot in every column. Since there are n columns, we have n pivots, so there is a pivot in every row, and therefore, Ax = b has a solution for every b, and therefore the columns of A span IRn . The answer is false if A is not square. Such a matrix would necessarily be tall, and the columns would not span the appropriate space. 3. Let T be a linear transformation. Show that if {v 1 , v 2 , v 3 } are linearly dependent vectors, then {T (v 1 ) , T (v 2 ) , T (v 3 )} are linearly dependent vectors. SOLUTION: If the set of vectors is linearly dependent, there is a non-trivial solution to c1 v 1 + c2 v 2 + c3 v 3 = 0 Apply T to both sides and use the linearity of T : T (c1 v 1 + c2 v 2 + c3 v 3 ) = T (0) ⇒ c1 T (v 1 ) + c2 T (v 2 ) + c3 T (v 3 ) = 0 This equation says that there is a non-trivial solution, so the vectors {T (v 1 ) , T (v 2 ) , T (v 3 )} are linearly dependent vectors. 4. If H is 7 × 7 matrix and Hx = v is consistent for every v in IR7 , then is it possible for Hx = v to have more than one solution for some v ∈ IR7 ? Why or why not? SOLUTION: If the given equation has a solution for every v, then there is a pivot in every row of H. Since there are the same number of rows as columns, H also must have a pivot in every column, and so there can not be more than one solution for any v. 5. Suppose that the third column of B is the sum of the first two columns, which are not linear combinations of each other. If B is 4 × 3, give the matrix which should be the RREF of B. 7 SOLUTION: The first two columns are basic variables, the third is not. Also, from the information given, we can find the values of the last column to be:   1 0 1  0 1 1     0 0 0  0 0 0 6. Suppose that the full solution to Ax = b is given by       4 2 5      x = 0 + c1 1 + c2 0  0 0 1 (a) Give the RREF of A, if A is 3 × 3. SOLUTION: We note from the position of the 1’s in the last two vectors, that c1 is actually x2 and c2 is actually x3 . Therefore:   x1 = 4 +2x2 +5x3 1 −2 −5 4 0 0 0  x2 = x2 ⇒  0 0 0 0 0 x3 = x3 (b) If the following row operations were applied to A (in order): 3R1 + R2 → R2 − 5R1 + R3 → R3 Find the matrix A and the vector b. SOLUTION: To “invert” the operations, take the last one first, and perform the following to the RREF: 5R1 + R3 → R3 and − 3R1 + R2 → R2 Resulting in the final matrix (the original [A|b]:   1 −2 −5 4  −3 6 15 −12  −5 −10 −25 20 True or False (and explain)? 1. If A and B are row equivalent, then they have the same row reduced echelon form. TRUE- If A and B are row equivalent, then some (invertible) sequence of elementary row operations will take you from matrix A to B. Similarly, there is an (invertible) sequence of elementary row operations that will take A to its unique RREF. 8 2. If vectors v1 and v2 are linearly dependent, it is still possible that {v1 , v2 , v3 } to be linearly independent for some v3 . FALSE. If v1 and v2 are linearly dependent, then there is a nontrivial solution to the vector equation: c1 v1 + c2 v2 = ~0 This has the same (nontrivial) solution as: c1 v1 + c2 v2 + 0 · v3 = ~0 which, by definition, means that the full set of three vectors is linearly dependent. 3. A mapping T : IRn → IRm is one-to-one if each vector in IRn maps onto a unique vector in IRm . FALSE, this is a statement of what it means to be a function. To be 1-1, each y ∈ IRm must have come from a unique x ∈ IRn (or have no solution). 4. If A is 5 × 5, and the columns of A do not span IR5 , it is possible that A is invertible. FALSE: There are several ways of describing what’s happening here- For example, If the columns do not span IR5 , the mapping is not onto. Because A is square, the mapping is also not 1-1, and so the mapping is not invertible. 5. A linear transformation preserves the operations of vector addition and scalar multiplication. TRUE: This is another way to state the definition of linearity. 6. If Ax = b has more than 1 solution, so does Ax = 0. TRUE: True. Theorem 6 in Section 1.5 essentially says that when Ax = b is consistent, the solution sets of the non-homogeneous equation and the homogeneous equation are translates of each other. In this case, the two equations have the same number of solutions. 7. In some cases, it is possible for four vectors to span IR5 . FALSE: Construct the matrix A from the four vectors, so that A is 5 × 4. Then to say the vectors span IR5 means that Ac = b has a solution for every b. But we can have at most 4 pivots (the number of columns), so the last row in the rref of A must be a row of zeros, so the last column could be a pivot column for some choices of b. 8. If A, B are row equivalent m × n matrices, and if the columns of A span IRm , then so do the columns of B. 9 True. If the columns of A span IRm , then the reduced echelon form of A is a matrix U with a pivot in each row (For more info, see Theorem 4 in Section 1.4). Since B is row equivalent to A, B can be transformed by row operations first into A and then further transformed into U . Since U has a pivot in each row, so does B. Therefore, the columns of B span IRm . 9. The span{u, v} is always visualized as a plane through the origin. FALSE. If the two vectors are both zero, the span is a point. If only one vector is non-zero, or if the two vectors are scalar multiples of each other, then the span is a line. If the two vectors are non-zero and are not multiples of each other, then the span can be visualized as a plane. 10