Linear Algebra Practice Exam 2 PDF

MA 2150 AA Linear Algebra: Practice Exam 2 Key Definitions, Concepts and Formulas Note the operations of matrix multiplication, finding the trace and the transpose will play an important role on this exam. Be sure to review them. The ( i, j )-minor of an n × n matrix, M i j , is the d...

MA 2150 AA Linear Algebra: Practice Exam 2 Key Definitions, Concepts and Formulas Note the operations of matrix multiplication, finding the trace and the transpose will play an important role on this exam. Be sure to review them. The ( i, j )-minor of an n × n matrix, M i j , is the determinant of the submatrix formed by deleting row i and column j. The ( i, j )-cofactor of an n × n matrix, C i j is defined as C i j = (−1) i+ j M i j. The determinant of an n × n matrix can be found by using a cofactor expansion using any row or column (due to Laplace’s Cofactor Expansion Theorem). det( A ) = a i1 C i1 + a i2 C i2 + · · · + a in C in (row i ) = a 1 j C 1 j + a 2 j C 2 j + · · · + a n j C n j (column j ) Cofactor expansions are extremely inefficient. The number multiplications needed to evaluate an n × n determinant using cofactor expansion is given by the following recursive formula m n = n · m n−1 + n, m1 = 0 There are many useful properties that make the process more efficient. Namely, if A is upper triangular, det( A ) = a 11 a 22 · · · · · a nn , which means the determinant of a triangular matrix can be found by simply multiplying the entries on the main diagonal. Be careful not to confuse this operation with the trace. We may use our elementary row operations to convert any determinant into a triangular determinant, as long as we keep the following in mind: – If matrix B is produced by swapping a pair of rows of matrix A , det(B) = − det( A ). – If matrix B is produced by multiplying any row of matrix A by k ∈ R, k ̸= 0, det(B) = k det( A ). – If matrix B is produced by adding a multiple of one row to another row of matrix A , det(B) = det( A ). Other useful “tricks” to evaluate a determinant by inspection include recognizing that – an entire row or column is filled with zeros means that the determinant must be equal to zero. – one row (or column) is a scalar multiple of another row (or column) means that the determine must be equal to zero. Even more properties of the determinant are as follows. If A and B are n × n matrices, and k ∈ R, – det( kA ) = k n det( A ). – det( AB) = det( A ) det(B). – det( A T ) = det( A ). 1 – det( A −1 ) = , provided A is invertible. det( A ) An n × n matrix A is invertible if, and only if, there exist an n × n matrix, B, such that AB = BA = I n and we write B = A −1. If the matrix does not have an inverse, we say the matrix is non-invertible or singular. There are different methods for finding the inverse of an invertible matrix, A. – Using Gauss-Jordan elimination setup up the£ augmented matrix [ A | I n ] and reduce this augmented matrix to reduced row-echelon form. This will lead to I n | A −1. ¤ – Using the adjoint of A , find the adjoint of A , which is adj( A ) = [C i j ]T , find the determinant of A and use the formula 1 A −1 = adj( A ) det( A ) – If the matrix is 2 × 2, we can use the following “special” formula for its inverse. ¸−1 ¸−1 1 · · a b d −b = , provided ad − bc ̸= 0 c d ad − bc − c a  Cramer’s rule is a method that may be used to solve an n timesn linear system that has a unique solution. If A is n × n, such that det( A ) ̸= 0 and b is n × 1, then the system Ax = b has solution det( A i ) xi = , det( A ) where A i is formed by replacing the i th column of A with the column b. An elementary matrix is one that is produced by producing exactly one row operation on the identity matrix. Alterna- tively, an elementary matrix is one that can be reduced to the identity matrix using exactly one row operation. Each type of elementary matrix is invertible. This is due to the fact that each row operation is reversible. For k ̸= 0, Elementary Operation Its Inverse ri ↔ r j ri ↔ r j 1 kr i k ri k · ri + r j −k · r i + r j If A is invertible, A and its inverse can be expressed as a product of elementary matrices. In general, we have A −1 = E k E k−1 · · · E 2 E 1 and A = E −1 −1 −1 −1 1 E 2 · · · E k−1 E k , where each E i is the elementary matrix that corresponds to the i th operation used to reduce A to I n. Questions 1. For the matrices given below, find AB. −1 2     3 −1 2 5  0 1 A = 1 −1 0 4 , B= −3  1 1 2 2 0 2 −1 · ¸ · ¸ −3 4 1 −5 2. Suppose A = and B =. Find: 6 7 3 4 (a) AB (b) A T B (c) AB T (d) tr( AB) (e) tr( A T B) (f) tr( AB T ) 3. Consider the following information. Matrix Dimension A 2×4 B 3×3 C 2×3 D 1×2 Which of the following products is well-defined? Give the dimension of the product. If a product is not well-defined, give a reason why. (a) A 2 (b) BC (c) CB (d) D A (e) B2 (f) A T C 4. Let f ( x) = −4 x2 + 7 x − 8. Find f ( A ) for each of the following. · ¸ −4 10 (a) A = 1 2   2 1 −1 (b) A = 0 2 4 5 3 0 5. Find, if possible, the inverse of each matrix.    · ¸ · ¸ 1 1 1 1 2 −1 −4 8 1 2 (a) A = (b) B = (c) C = 3 5 4 (d) D = 3 7 −10 1 −2 −3 7 3 6 5 7 16 −21 · ¸ 4 x 6. Find the value of x such that the matrix is singular. −2 −3 7. Use the appropriate inverse to solve the following for X.     1 −6 −4 −5  2 −10 −9 X = −4 −1 6 5 3 8. Is each matrix elementary? Explain. · ¸ 10 0 (a) A = 0 −1 0 0 0 1   0 1 0 0 (b) B =  0 0 1 0  1 0 0 0 9. Suppose E is an elementary matrix formed by performing kr 20 + r 17 , for k ∈ R, on the corresponding identity matrix, in which entry will there be k? 10. Find an elementary matrix, E , such that E A = B, where · ¸ · ¸ 3 −4 3 −4 A= and B= 2 5 −4 13 11. What is the inverse of the following elementary matrix? 1 0 0 0 0   0 10 0 0 0   0 E= 0 1 0 0  0 0 0 1 0 0 0 0 0 1 · ¸ 2 −4 12. Write the inverse of A = as a product of elementary matrices. −2 3   1 −1 0 13. Write the matrix A = −1 2 1 as a product of elementary matrices. 2 1 0 14. Using any method discussed, find the determinant of each matrix. 5 0 0 0 0   2 0 0 4     · ¸ 1 −4 2 −4 2 0 0 0 −4 7 0 0 −2 3   (a) A = (b) B = 0 −3 1 (c) C =    10 (d) D =  6 −1 0 0 5 0. 5 4 1 1 0  2 5 8  4 4 5 6 0 2 1 2 5 −1 −4 −3 1 −1 15. Find the value(s) of k such that ¯ ¯ ¯k + 4 40 1 ¯¯ ¯ ¯ 0 −2 k + 1 −9 ¯¯ = 0. ¯ ¯ 0 0 k − 3¯ · ¸ · ¸ 3 5 6 −2 16. Let A = and B =. Show that det( A + B) ̸= det( A ) + det(B). 7 −11 4 −1 17. If A and B are both 2 × 2 matrices such that | A | = 4 and |B| = −3, compute the following using appropriate properties of the determinant. (a) | AB| (b) | A −1 | (c) |5 A | (d) |4 A −1 B| (e) | A T B T | 18. Solve the linear system using Cramer’s rule. 2 x1 + 3 x2 − x3 = 2 3 x1 − 2 x2 + x3 = −1 −5 x1 − 4 x2 + 2 x3 = 3 19. Find the adjoint of the matrix A. Use the adjoint to find the inverse of A.   2 4 −3 A = −3 1 0 5 6 −2 20. Consider the following matrix.   10 0 0 0 0 0  0 −2 0 0 0 0    0 0 1 0 0 0   A=  0 0 0 1 0 0     0 0 0 0 3 0 0 0 0 0 0 4 (a) What type of special matrix is A ?  (b) What is A 3 ? (c) Is A invertible? Explain. If so, find A −1. (d) What is the determinant of A ? (e) What is the trace of A ? 21. Determine, with reason, if each statement is true or false. (a) The transpose of an upper triangular matrix is a lower triangular. (b) The (3, 4)-minor of some square matrix is equal to the (3, 4)-cofactor of the same matrix. (c) If det( A ) = 4, then the equation A X = B must have a unique solution. (d) If A is an n × n matrix, and B is obtained by swapping 3 pairs of rows in A , then det(B) = det( A ). (e) Every square matrix can be written as a product of elementary matrices. (f) The product of two elementary matrices is never another elementary matrix. (g) If A and B are invertible n × n matrices, then AB is also an invertible n × n matrix. (h) If tr( A ) = 0, then A must be singular. Answers 1.  −1 2        3 −1 2 5  3 · (−1) + (−1) · 0 + 2 · (−3) + 5 · 2 3 · 2 + (−1) · 1 + 2 · 1 + 5 · (−1) 1 2 1 0 1  −1 0 4 ·  −3  = 1 · (−1) + (−1) · 0 + 0 · (−3) + 4 · 2 1 · 2 + (−1) · 1 + 0 · 1 + 4 · (−1) =  7 −3 1  1 2 2 0 1 · (−1) + 2 · 0 + 2 · (−3) + 0 · 2 1 · 2 + 2 · 1 + 2 · 1 + 0 · (−1) −7 6 2 −1 · ¸· ¸ · ¸ · ¸ −3 4 1 −5 −3 + 12 15 + 16 9 31 2. (a) AB = = = 6 7 3 4 6 + 21 −30 + 28 27 −2 · ¸· ¸ · ¸ · ¸ −3 6 1 −5 −3 + 18 15 + 24 15 39 (b) A T B = = = 4 7 3 4 4 + 21 −20 + 28 25 8 · ¸· ¸ · ¸ · ¸ −3 4 1 3 −3 − 20 −9 + 16 −23 7 (c) AB T = = = 6 7 −5 4 6 − 35 18 + 28 −29 46 (d) tr( AB) = 9 − 2 = 7 (e) tr( A T B) = 15 + 8 = 23 (f) tr( AB T ) = −23 + 46 = 23 3. (a) A 2 = A · A is undefined because A has four columns, but only two rows. (b) BC is undefined because B has three columns, but C has two rows. (c) CB is well-defined, and dim(CB) = 2 × 3. (d) D A is well-defined, and dim(D A ) = 1 × 4. (e) B2 is well-defined, and dim(B2 ) = 3 × 3. (f) A T C is well-defined, and dim( A T C ) = 4 × 3. 4. (a) So that all the operations are defined for matrices, we have to interpret the “−8” as “−8 I 2.” So, · ¸2 · ¸ · ¸ −4 10 −4 10 1 0 f ( A ) = −4 +7 −8 1 2 1 2 0 1 · ¸· ¸ · ¸ · ¸ −4 10 −4 10 −4 10 1 0 = −4 +7 −8 1 2 1 2 1 2 0 1 · ¸ · ¸ · ¸ 16 + 10 −40 + 20 −4 10 1 0 = −4 +7 −8 −4 + 2 10 + 4 1 2 0 1 · ¸ · ¸ · ¸ −104 80 −28 70 8 0 = + − 8 −56 7 14 0 8 · ¸ −140 150 = 15 −50 (b) Similar to the process in part (a), for a well-defined function evaluation, we have to interpret the “−8” as “−8 I 3.” Note the slight adjustment due to the fact that we are now evaluating f at a 3 × 3 matrix. So,  2     2 1 −1 2 1 −1 1 0 0 f ( A ) = −4 0 2 4 + 7 0 2 4  − 8 0 1 0  5 3 0 5 3 0 0 0 1        2 1 −1 2 1 −1 2 1 −1 1 0 0 = −4 0 2 4  0 2 4  + 7 0 2 4 − 8 0 1 0 5 3 0 5 3 0 5 3 0 0 0 1       4+0−5 2+2−3 −2 + 4 + 0 2 1 −1 1 0 0 = −4 0 + 0 + 20 0 + 4 + 12 0 + 8 + 0 + 7 0 2 4  − 8 0 1 0 10 + 0 + 0 5 + 6 + 0 −5 + 12 + 0 5 3 0 0 0 1       4 −4 −8 14 7 −7 8 0 0 = −80 −64 −32 +  0 14 28 − 0 8 0 −40 −44 −28 35 21 0 0 0 8   10 3 −15 = −80 −58 −4 −5 −23 −36 5. (a) A is not invertible, because det( A ) = (−4) · (−2) − 8 · 1 = 0. (b) Since det(B) = 7 + 6 = 13 ̸= 0, B is invertible. Using the “special” formula for the inverse of a 2 × 2 matrix, we get 1 7 · ¸ −1 −2 B = 13 3 1 "7 2 # 13 − 13 = 3 1 13 13 (c) Let’s use Gauss-Jordan elimination to attempt to find the inverse of C. First, we augment C with I 3 , which gives.   1 1 1 1 0 0 3 5 4 0 1 0 3 6 5 0 0 1 Using −3 r 1 + r 2 and −3 r 1 + r 3 produces   1 1 1 1 0 0 0 2 1 −3 1 0 0 3 2 −3 0 1 Using 12 r 2 produces   1 1 1 1 0 0 1 0  1 2 − 32 1 2 0 0 3 2 −3 0 1 Using −3 r 2 + r 3 and − r 2 + r 1 produces 1 5 − 12   1 0 2 2 0 1 0 1 − 32 1 0   2 2 1 3 0 0 2 2 − 32 1 Using 2 r 3 produces 1 5 − 12   1 0 2 2 0 1 0 1 − 32 1 0   2 2 0 0 1 3 −3 2 Finally, using − 21 r 3 + r 2 and − 12 r 3 + r 1 produces   1 0 0 1 1 −1 0 1 0 −3 2 −1 0 0 1 3 −3 2   1 1 −1 Therefore, C −1 = −3 2 −1 3 −3 2 (d) Using the same process as in part (c), we start by augmenting D with I 3 and apply the Gauss-Jordan process.   1 2 −1 1 0 0 3 7 −10 0 1 0 7 16 −21 0 0 1 Using −3 r 1 + r 2 and −7 r 1 + r 3 produces   1 2 −1 1 0 0  0 1 −7 −3 1 0  0 2 −14 −7 0 1 Using −2 r 2 + r 3 produces   1 2 −1 1 0 0  0 1 −7 −3 1 0  0 0 0 −1 −2 1 D is not invertible, because it cannot be reduced to the 3 × 3 identity.  6. The matrix is singular if 4 · (−3) + 2 x = 0. Solving this equation for x, gives x = 6. 7. First up, we need to find the inverse of the coefficient. This sounds like a job for Gauss-Jordan elimination or the adjoint matrix. Let’s use Gauss-Jordan elimination. First, we augment the augment the coefficient matrix with I 3 :   1 −6 −4 1 0 0  2 −10 −9 0 1 0  −1 6 5 0 0 1 Using −2 r 1 + r 2 and r 1 + r 3 produces   1 −6 −4 1 0 0  0 2 −1 −2 1 0  0 0 1 1 0 1 Using 21 r 2 produces 1 −6 −4 1 0 0    0 1 − 12 −1 1 2 0  0 0 1 1 0 1 1 Using 2 r3 + r2 and 4 r 3 + r 1 produces 1 −6 0 5 0 4    0 1 0 − 21 1 2 1 2  0 0 1 1 0 1 Finally, using 6 r 2 + r 1 produces 1 0 0 2 3 7    0 1 0 − 12 1 2 1 2  0 0 1 1 0 1 2 3 7   This means A −1 = − 21 1 1 2 . Once the coefficient matrix of a matrix equation has an inverse, the equation must  2 1 0 1 have a unique solution and that unique solution can be found using the formula X = A −1 · B. Therefore, X = A −1 B 2 3 7 −5    1 1 1 = − 2 2 2 −4 1 0 1 3 −10 − 12 + 21   =  25 − 2 + 32  −5 + 0 + 3   −1 =  2 −2 8. (a) A is not an elementary matrix since it would require two row operations, performed on I 2 , to create A. Alterna- tively, it requires two row operations to reduce A to I 2 , hence, A is not elementary. For an elementary matrix, only one row operation is allowed to create or reduce it! (b) B is elementary since performing one operation, r 1 ↔ r 4 , on I 4 will create B. Alternative, B is elementary since it would take exactly one operation, r 1 ↔ r 4 , performed on B to create I 4. 9. k will be in row 17, column 20. · ¸ 1 0 10. Note that apply −2 r 1 + r 2 to A will produce B. Therefore, E = −2 1 1 11. Since E is produced by performing 10 r 2 ( I 5 ), the inverse of E will be produced by performing 10 r 2 ( I 5 ). Therefore,   1 0 0 0 0 0 1  10 0 0 0  E −1 = 0 0 1 0 0   0 0 0 1 0   0 0 0 0 1 12. This type of matrix factorization is only possible for invertible matrices. To identify the elementary matrix factors, we keep of each elementary matrix that corresponds to each row operation used to reduce A to I 2. First, using 21 r 1 , we get "1 # 0 · ¸ 1 −2 2 =⇒ E1 = −2 3 0 1 Next, we perform 2 r 1 + r 2 to get · ¸ · ¸ 1 −2 1 0 =⇒ E2 = 0 −1 2 1 Next, we perform − r 2 to get · ¸ · ¸ 1 −2 1 0 =⇒ E3 = 0 1 0 −1 Finally, we perform 2 r 2 + r 1 to get · ¸ · ¸ 1 0 1 2 =⇒ E4 = 0 1 0 1 Let’s summarize what we have done using matrix multiplication: E4 E3 E2 E1 A = I2 So, by definition, A −1 must be equal to E 4 E 3 E 2 E 1. Therefore, ¸"1 # 0 · ¸· ¸· −1 1 2 1 0 1 0 2 A = 0 1 0 −1 2 1 0 1 Note, the sequence of elementary depends the order in which the row operations are performed. So, there is not a unique factorization. 13. We will use the exact same process in part (a). First, using r 1 + r 2 , we get     1 −1 0 1 0 0 0 1 1 =⇒ E 1 = 1 1 0 2 1 0 0 0 1 Next, we perform −2 r 1 + r 3 to get     1 −1 0 1 0 0 0 1 1 =⇒ E2 =  0 1 0 0 3 0 −2 0 1 Next, we perform −3 r 2 + r 3 to get     1 −1 0 1 0 0 0 1 1 =⇒ E 3 = 0 1 0 0 0 −3 0 −3 1 Next, we perform − 13 r 3 to get 1 0 0     1 −1 0 0 1 1 =⇒ E 4 = 0 1 0 0 0 1 0 0 − 31 Next, we perform − r 3 + r 2 to get     1 −1 0 1 0 0 0 1 0 =⇒ E 5 = 0 1 −1 0 0 1 0 0 1 Finally, we perform r 2 + r 1 to get     1 0 0 1 1 0 0 1 0 =⇒ E 6 = 0 1 0 0 0 1 0 0 1 This means, E6 E5 E4 E3 E2 E1 A = I3 and A −1 = E 6 E 5 E 4 E 3 E 2 E 1.  However, the questions wants a factorization of A , not A −1. Since ( A −1 )−1 = A and ( AB)−1 = B−1 A −1 , A = (E 6 E 5 E 4 E 3 E 2 E 1 )−1 = E −1 −1 −1 −1 −1 −1 1 E2 E3 E4 E5 E6 Applying the properties relating an elementary matrix and its inverse,     1 0 0 1 0 0 1  E 1 = 1 1 0 =⇒ E − 1 = −1 1 0 , 0 0 1 0 0 1     1 0 0 1 0 0 1  E 2 =  0 1 0 =⇒ E − 2 = 0 1 0 ,  −2 0 1 2 0 1     1 0 0 1 0 0 − 1 E 3 = 0 1 0 =⇒ E 3 = 0 1 0 , 0 −3 1 0 3 1 1 0 0     1 0 0 1  E 4 = 0 1 0 =⇒ E −4 = 0 1 0 , 1 0 0 −3 0 0 −3     1 0 0 1 0 0 −1 E 5 = 0 1 −1 =⇒ E 5 = 0 1 1 , and 0 0 1 0 0 1     1 1 0 1 −1 0 1  E 6 = 0 1 0 =⇒ E − 6 = 0 1 0 0 0 1 0 0 1 Putting this all together we have the following factorization of A !        1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 −1 0 A = −1 1 0 0 1 0 0 1 0 0 1 0 0 1 1 0 1 0 0 0 1 2 0 1 0 3 1 0 0 −3 0 0 1 0 0 1 Note, the sequence of elementary factors depends the order in which the row operations are performed. So, there is not a unique factorization. You could, in theory, obtain a different factorization. 14. (a) Using the formula for the determinant of a 2 × 2 matrix, | A | = −2 − 35 = −37. (b) B has a zero in row 2, column 1. So, using either of these to do a cofactor expansion would be a good choice. However, if we introduced another zero in column 1, the cofator expansion would be even easier. By a property of determinants, performing −2 r 1 + r 3 on ¯ ¯ ¯1 −4 2¯ ¯ ¯ ¯0 −3 1¯ ¯ ¯ ¯2 5 8¯ will produce an “easier” determinant (of equal value) to compute. This easier determinant is ¯ ¯ ¯1 −4 2¯ ¯ ¯ ¯0 −3 1¯ ¯ ¯ ¯0 13 4¯ Now, using a cofactor expansion with column 1, we get ¯ ¯ ¯ ¯ ¯1 −4 2¯ ¯1 −4 2¯¯ ¯ ¯ ¯ ¯0 −3 1¯ = ¯0 −3 1¯¯ ¯ ¯ ¯ ¯2 5 8¯ ¯0 13 4¯ = a 11 C 11 + a 21 C 21 + a 31 C 31 ¯ ¯ 2 ¯−3 1¯ ¯ ¯ = 1 · (−1) ¯ +0+0 13 4¯ = 1 · (−12 − 13) = −25 Therefore, |B| = −25.  (c) For a 4 × 4 determinant, we know obvious patterns (apart from a few sporadically placed zeros), let’s use our row operations to convert the determinant to a triangular form. Note that for the purposes of evaluating determi- nants, it is not terribly important to have pivots of 1 (certainly not as important as when solving a linear system). So, let’s start with ¯2 0 0 4¯¯ ¯ ¯ ¯ ¯0 0 −2 3¯ ¯ ¯ ¯4 1 1 0¯¯ ¯ ¯2 1 2 5¯ and perform −2 r 1 + r 3 and − r 1 + r 3 to get ¯2 0 0 4¯¯ ¯ ¯ ¯ ¯0 0 −2 3¯¯ ¯ ¯0 1 1 −8¯¯ ¯ ¯0 1 2 1¯ To get a pivot in row 2, let’s perform the row swap r 2 ↔ r 3 , which changes the sign of the determinant: ¯2 0 0 4¯¯ ¯ ¯ ¯ ¯0 1 1 −8¯¯ − ¯¯ ¯0 0 −2 3¯¯ ¯0 1 2 1¯ Next, we eliminate the 1 in row 4, column 2 using − r 2 + r 4 to get: ¯2 0 0 4¯¯ ¯ ¯ ¯ ¯0 1 1 −8¯¯ − ¯¯ ¯0 0 −2 3¯¯ ¯0 0 1 9¯ Lastly, we can obtain a triangular determinant by performing 21 r 3 + r 4 : ¯2 0 0 4¯ ¯ ¯ ¯ ¯ ¯0 1 1 −8¯¯ − ¯¯ ¯0 0 −2 3¯¯ ¯0 21 ¯ 0 0 2 With the negative sign in front of the determinant, we get ¡ 21 ¢ det(C ) = −2 · 1 · (−2) · 2 = 42 (d) This is a triangular matrix, hence, the determinant of D may be found by multiplying the entries on the main diagonal. So. det(D ) = 5 · 2 · (−1) · 6 · (−1) = 60. 15. Since the matrix is upper triangular, its determinant is the product of the entries of the main diagonal, which is ( k + 4)(−2 k + 1)( k − 3). Setting the determinant equal to zero gives the solution: k = −4, k = 1/2, or k = 3. So, there are three possible values of k that will make this determinant equal to zero. · ¸ 9 3 16. det( A )+det(B) = (−33−35)+(−6+8) = −66. A +B = and det( A +B) = −108−33 = −141. Therefore, det( A +B) ̸= 11 −12 det( A ) + det(B). 17. (a) | AB| = | A ||B| = 4 · (−3) = −12 1 1 (b) | A −1 | = = | A| 4 (c) Since A is 2 × 2, |5 A | = 52 | A | = 25 · 4 = 100 1 1 (d) |4 A −1 B| = 42 · · |B| = 16 · · (−3) = −12 | A| 4 (e) | A T B T | = | A T | · |B T | = | A | · |B| = 4 · (−3) = −12 18. Here are the four matrices needed to apply Cramer’s rule to solve a system with 3 equations, 3 variables. Recall, A i is     2 3 −1 2 3 −1 formed by replacing the i th column of A with the column of constants, b. A =  3 −2 1, A 1 = −1 −2 1, −5 −4 2 −3 −4 2     2 2 −1 2 3 2 A 2 =  3 −1 1, and A 3 =  3 −2 −1. Let’s find the determinant of each using cofactor expansion, all of −5 3 2 −5 −4 3 them using the first row. ¯ ¯ ¯ 2 3 −1¯¯ ¯ det( A ) = ¯¯ 3 −2 1¯¯ ¯−5 −4 2¯ = 2(−4 + 4) − 3(6 + 5) − 1(−12 − 10) = 0 − 33 + 22 = −11 ¯ ¯ ¯ 2 3 −1¯¯ ¯ det( A 1 ) = ¯¯−1 −2 1¯¯ ¯ 3 −4 2¯ = 2(−4 + 4) − 3(−2 − 3) − 1(4 + 6) = 0 + 15 − 10 =5 ¯ ¯ ¯ 2 2 −1¯¯ ¯ det( A 2 ) = ¯¯ 3 −1 1¯¯ ¯−5 3 2¯ = 2(−2 − 3) − 2(6 + 5) − 1(9 − 5) = −10 − 22 − 4 = −36 ¯ ¯ ¯ 2 3 2¯¯ ¯ det( A 3 ) = ¯¯ 3 −2 −1¯¯ ¯−5 −4 3¯ = 2(−6 − 4) − 3(9 − 5) + 2(−12 − 10) = −20 − 12 − 44 = −76 Also, recall that det( A i ) xi = , det( A ) So, x1 = −5/11, x2 = 36/11, and x3 = 76/11. 19. To find the adjoint of a matrix, we first need all the cofactors of the matrix. C 11 = (−1)2 (−2 − 0) = −2, C 12 = (−1)3 (6 − 0) = −6, C 13 = (−1)4 (−18 − 5) = −23, C 21 = (−1)3 (−8 + 18) = −10, C 22 = (−1)4 (−4 + 15) = 11, C 23 = (−1)5 (12 − 20) = 8, C 31 = (−1)4 (0 + 3) = 3, C 32 = (−1)5 (0 − 9) = 9, and C 33 = (−1)6 (2 + 12) = 14. So, the adjoint of A is  T   −2 −6 −23 −2 −10 3 adj( A ) = −10 11 8 =  −6 11 9 3 9 14 −23 8 14 Next, we need to find the determinant of A is 41. Using the first row and a cofactor expansion, we get det( A ) = a 11 C 11 + a 12 C 12 + a 13 C 13 = 2 · (−2) + 4 · (−6) + (−3) · (−23) = −4 − 24 + 69 = 41  So, the inverse of A is  2 10 3    − 41 − 41 41 −2 −10 3 1 1  A −1 =  6 11 9   adj( A ) = −6 11 9 = − 41   41 41  det( A ) 41 −23 8 14 − 23 41 8 41 14 41 20. (a) A is a diagonal matrix. (b) To raise a diagonal matrix to a positive integer power, raise each of the entries on the main diagonal to that power. So,  3    10 0 0 0 0 0 1000 0 0 0 0 0  0  (−2)3 0 0 0 0  0  −8 0 0 0 0 3  0 0 13 0 0 0  0 0 1 0 0 0     A = =  0 0 0 13 0 0  0 0 0 1 0 0  0 33 0   0      0 0 0 0 0 0 27 0 0 0 0 0 0 43 0 0 0 0 0 64 (c) A diagonal matrix is invertible if, and only if, there are no zeros on the main diagonal. Since this is the case, A is invertible. The inverse of A is found by taking the reciprocal of each of the entries on the main diagonal. 1  10 0 0 0 0 0  0 − 1 0 0 0 0  2   0 0 1 0 0 0   −1 A =  0 0 0 1 0 0  0 0 0 13 0    0 0 0 0 0 0 14 (d) The determinant of a diagonal matrix (in fact, any triangular matrix) is found by multiplying the entries of the main diagonal. Therefore, det( A ) = 10 · (−2) · 1 · 1 · 3 · 4 = −240 (e) The trace of any square matrix is the sum of the entries on the main diagonal. Therefore, tr( A ) = 10 + (−2) + 1 + 1 + 3 + 4 = 17 21. (a) True. If A is upper triangular, then a i, j = 0 for i > j. If B = A T , then b i, j = a j,i = 0 for j < i. Therefore, B is lower triangular. (b) False. C 3,4 = (−1)7 M3,4 = − M3,4. In general, C 3,4 and M3,4 will have opposite signs. Note, the only time they will be equal is if M3,4 = 0. (c) True. Since det( A ) ̸= 0, A −1 exists, and the unique solution is given by X = A −1 B. (d) False. By performing 3 row swaps, det(B) = (−1)3 det( A ) = − det( A ). (e) False. Only invertible matrices may be expressed as a product of elementary matrices.       2 0 0 4 0 0 8 0 0 (f) False. Let E 1 = 0 1 0 and E 2 = 0 1 0, then E 1 E 2 = 0 1 0, which is another elementary matrix. 0 0 1 0 0 1 0 0 1 (g) True. First, note that AB is well-defined and dim( AB) = n × n. Since A and B are both invertible, det( A ) ̸= 0 and det(B) ̸= 0, Since det( AB) = det( A ) det(B) ̸= 0, AB must also be invertible. (h) This is definitely false. Here is a nice 4 × 4 counterexample. 10 0 0 0    0 −10 0 0 A=  0  0 −5 0 0 0 0 5 The trace of A is 10 + (−10) + (−5) + 5 = 0, however, A is a diagonal matrix (we love diagonal matrices!) with no zeros on the main diagonal. Hence, A must be invertible. Moreover, here is the inverse of A : 1  10 0 0 0  0 − 101 0 0   A −1 =    0  0 − 51 0  0 0 0 15