Linear Algebra Math 221

Transcription

1 Linear Algebra Math Open Book Exam Open Notes 8 Oct, 004 Calculators Permitted Show all work (except #4). (0 pts) Let A = 3 a) (0 pts) Compute det(a) by Gaussian Elimination. 3 3 swap(i)&(ii) (iii) (iii)+( )(i) A = 3 A = A = (iii) (iii)+()(ii) A = A3 = As A3 is upper triangular det(a3) is the product of the diagonal elements: det(a3)=()()(-) =- Of the steps in the reduction, each but the first is a simple replacement: (a) (a)+(n)(b), which doesn t affects the determinant: - = det(a3) = det(a) = det(a). The first step in the reduction was a row swap, which multiplies the determinant by (-), so det(a) = (-)det(a) = (-)(-) = Answer: det(a) = b) (0 pts) Compute det(a) by Cofactor Reduction. (all determinants, even smaller than 3x3 must be computed by cofactor reduction) We choose to expand on the first row because it has a low density (only one non-zero element): A = 3 = ( ) + (0) 3 + ( )+ () 3 + ( )+3 (0) = ( ) 3 Now we look at the remaining x determinant. No row or column gives us an advantage as they all have density two, so we arbitrarily choose to expand over the first row. (Note that the notation 3 and are x determinants, not absolute values.) A = 3 = ( ) 3 = ( ) ( ( )+ () + ( ) + (3)) = ( )( + ( 3)) = Answer: det(a) = Check: Compare the answers in parts (a) and (b)

2 . (0 pts) Compute the inverse of A = by Gaussian Elimination. We can compute A - by augmenting A by I and bringing A to reduced echelon form. ( A I) = 0 (ii) (ii)+( )( i) 0 (i) (i)+( )(ii) (ii) ( )(ii) Answer: A = Check: Multiply and check AA - =I. 3. (0 pts) If A = 4 4 a) (0 pts) Compute an LU decomposition (without pivoting) of A We write A as I times A and then step through the Gaussian Reduction of A to its echelon form U, while converting I to L by applying the corresponding steps. (The pivots of A and U are emphasized by putting them in parentheses.): 0 0 ( ) 0 0 ( ) 0 0 ( ) IA = ( ) Answer: Thus L = 0 and U = Check: Multiply and confirm that LU = A. (In fact, at any step of the reduction we can multiply the two matrices to get a product of A.) b) (0 pts) Solve Ax v = 0 for the vector x using the LU decomposition of A. 6 We recall that Ax=b can also be written as (LU)x=b. We define y=ux and first solve Ly=b for y: 0 0 ( L b) = From the first row we get the equation y =. ( )

3 Second row: y + y = 0, so y = 0 y = 0 () = -. Third row: ()y + (-)y + y 3 = 6, so y 3 = 6 - y + y = 6 () + (-) = 0. Check: Multiply and confirm that Ly=b. Now we use this value of y to solve Ux=y for x: ( U y) = Note first that the system is consistent, so there is a solution. We now note that the second column is non-pivot, so x is a free variable. From the third row we read the equation 0 = 0, which gives us no information. Second row: ()x 3 = -, so x 3 = -/ = -. First row: ()x + ()x + ()x 3 =, so ()x = - ()x - ()x 3 and we solve to get x = ( - x - ()(-))/ = - x / x / v Solution: x = x = 0 + x 0 Check: Multiply and confirm that Ax=b Note: The same solution set can be expressed in a number of equivalent ways. By adding two copies of the second vector to the first we get: v x = + ( x ) = 0 + s (where s= (x -)/(-)) 0 4. (0 pts) If A =. a) (0 pts) Find all matrices B such that AB=I (B is a right inverse of A). We can set this up as a system of equations and directly solve it. As A has two rows the product AB must have two rows and I must be the x identity matrix. a b Thus B must have three rows and two columns, so it has form B =. The e f matrix equation I=AB= a b = 0 gives us the system of four 0 e f equations in six unknowns: a + c + e = 0 a + c + e = 0 which gives us the augmented matrix 0 0. b + d + f = 0 b + d + f = 0 We reduce this system as

4 We read off the solutions as: e=free, f=free, a=-/3 e/3, b=/3-f/3, c=/3-e/3, d=-/3-f/3 Easier approach: Note that, just like inverting a matrix, we can more efficiently solve this as follows: 0 0 red ech form The first augment column gives us a, c and e, so e is free and a=-/3 e/3, b=/3-f/3, c=/3-e/3 as before. The second augment column gives us the relations among b, d and f, so f is free and b=/3-f/3, d=-/3-f/3, as before. a b e f Answer: Thus B = = e f e f e f Check: Multiply and confirm that CA = I b) (0 pts) Find all matrices C such that CA=I (C is a left inverse of A). Answer: There is no matrix C such that CA=I The matrix A is equivalent to a linear map from R 3 to R. Thus, any matrix C such that CA=I must be an onto map from R to R 3, not a possible condition. A brute force approach to this problem is to solve the system and demonstrate that it is inconsistent. As A has three columns the product CA=I must have three columns, and it must be the 3x3 identity matrix. C must have three rows a b and two columns, so it has form C =. The matrix equation e f a b 0 0 I=CA= = gives us the system of nine equations in e f 0 0 six unknowns:

5 a + b = a + b = a + b = c + d = c + d = which gives us the augmented matrix c + d = e + f = e + f = e + f = After the first few steps of Gaussian Elimination the first three rows become: and the last row gives us an inconsistent equation 0 = -/3. Easier approach: As we are solving the equation CA=I for C we cannot use the usual matrix vector notation. We can, however, rewrite this as A T C T =(CA) T =I. Now, we can efficiently solve this as follows: red ech form We immediately see that the system is inconsistent as the last row is interpreted as the three equations 0a+0b=, 0c+0d= and 0e+0f=-3, each of them inconsistent. 5. (0 pts) Multiple Choice: ALWAYS/SOMETIMES/NEVER (5 pts per question) a) If the 4x3 matrix A has three pivots then it defines a linear transformation T: R 3 R 4 which is onto. NEVER A linear transformation from R 3 R 4 cannot be onto. A linear transformation which is onto corresponds to a matrix which has a pivot in every row this cannot be the case for A which has four rows and three columns, hence no more than three pivots. b) If the matrices A and B are both invertible then A+B is invertible. SOMETIMES There are cases where this is true (an example is A=B=I, so A+B=I) but there are also cases where this is false (an example is B=-A, so A+B=0). c) If the matrices A and B are both invertible then AB is invertible. ALWAYS A and B are invertible iff det(a) and det(b) are non-zero. Thus det(ab)=det(a)det(b) is also non-zero and AB is invertible.

6 d) I A is a 4x4 matrix, then det(3a) = 8 det(a). ALWAYS The series of elementary row operations which take the matrix A to the matrix 3A is scaling each of the four rows by 3. Each of these operations multiplies the determinant by 3, so the total change is to multiply the determinant by 3 4 =8.