Matrix transformations #3 - Questions and Answers

Question 1 A and B are n × n × n matrices. Mark each statement True or False. Justify your answer. a. A row replacement operation does not affect the determinant of a matrix. b. The determinant of A is the product of the pivots in any echelon form U of A, multiplied by (−1)^r, where r is the number of row interchanges made during row reduction from A to U. c. If the columns of A are linearly dependent, then det A = 0. d. det(A + B) = det A + det B. Answer (a) The statement is true. Adding a multiple of one row to another row does not change the value of the determinant according to Theorem 3. (a) True. (b) The statement is false. This statement is only true for invertible matrices since their determinants are not equal to zero. Since pivots are defined as the first nonzero entries in each of the rows of the row echelon form of a matrix, it follows that their product is also a nonzero number. Thus, determinant of a singular matrix cannot be the product of its pivots multiplied by (−1)^r, where r is the number of row interchanges made during its transformation to its row echelon form. (b) False. (c) The statement is true. There are multiple ways to prove this. Since the columns of A are linearly dependent, it follows that the rows of 𝐴 𝑇 A T are linearly dependent as well. Therefore, its row echelon form contains a zero row. Keep in mind that the row echelon form of a matrix is always an upper triangular matrix. Since the determinant of the row echelon form of 𝐴 𝑇 A T has the same determinant as the matrix 𝐴 𝑇 A T , and the determinant of a triangular matrix is equal to the product of its diagonal entries, it follows that det 𝐴 = 0 detA=0 (at least one of the rows of its row echelon form is a zero row). Now, according to Theorem 5, it follows that det ( 𝐴 ) = 0 det(A)=0. (c) True. (d) The statement is false. Consider the following example of 2 × 2 2×2 matrices. det ( 𝐴 ) = ∣ 1 1 1 1 ∣ = 0 ,  det ( 𝐵 ) = ∣ 1 0 0 1 ∣ = 1 ,  det ( 𝐴 + 𝐵 ) = ∣ 2 1 1 2 ∣ = 3. det(A)= ​ 1 1 ​ 1 1 ​ ​ =0, det(B)= ​ 1 0 ​ 0 1 ​ ​ =1, det(A+B)= ​ 2 1 ​ 1 2 ​ ​ =3. Therefore, it follows that generally det ( 𝐴 ) + det ( 𝐵 ) ≠ det ( 𝐴 + 𝐵 ) det(A)+det(B)  =det(A+B). (d) False. Question 2 Use the given inverse of the coefficient matrix to solve the following system 5 𝑥 1 x 1 ​ + 3 𝑥 2 x 2 ​ = 6 −6 𝑥 1 x 1 ​ − 3 𝑥 2 x 2 ​ = −2 𝐴 − 1 = [ − 1 − 1 2 3 5 ] A −1 =[ −1 2 ​ −1 5 3 ​ ​ ] Answer 1 𝐴 𝑥 = 𝐵 Ax=B 𝑥 = 𝐴 − 1 𝐵 x=A −1 B 𝑥 = [ − 1 − 1 2 3 5 ] [ 6 − 2 ] . x=[ −1 2 ​ −1 5 3 ​ ​ ][ 6 −2 ​ ]. Step-by-step calculations yield: 𝑥 1 = − 4 ,   𝑥 2 = 26 5 . x 1 ​ =−4, x 2 ​ = 5 26 ​ . Answer 2 Same process as above confirms: 𝑥 1 = − 4 x 1 ​ =−4, 𝑥 2 = 26 5 . x 2 ​ = 5 26 ​ . Question 3 What is the area of the parallelogram whose vertices are listed? (0,0), (5,2), (6,4), (11,6) Answer 1 Using the determinant formula: Area = det ∣ 5 2 6 4 ∣ . Area=det ​ 5 6 ​ 2 4 ​ ​ . Result: 8. Answer 2 Using cross products of vectors: 𝐴 𝐵 = ( 5 , 2 ) ,   𝐴 𝐷 = ( 11 , 6 ) . AB=(5,2), AD=(11,6). Cross product magnitude confirms: 8 square units. Answer 3 Using determinants and vectors similarly confirms: Area = 8 square units. Answer 4 A miscalculation leads to a conclusion of 0 square units, inconsistent with previous results. Question 4 Show that if 𝐴 2 A 2 is the zero matrix, then the only eigenvalue of A is 0. Answer Eigenvalue proof: 𝐴 2 𝑣 = 0    ⟹    𝜆 2 𝑣 = 0. A 2 v=0⟹λ 2 v=0. Since 𝑣 ≠ 0 v  =0, 𝜆 = 0 λ=0. Question 5 Find an orthogonal basis for the column space of each matrix. Answer Using the Gram-Schmidt process, the orthogonal basis for 𝐴 A is: { [ 3 1 − 1 ] , [ 9 / 2 3 / 2 − 3 / 2 ] , [ 3 1 − 3 ] } . ⎩ ⎨ ⎧ ​ ​ 3 1 −1 ​ ​ , ​ 9/2 3/2 −3/2 ​ ​ , ​ 3 1 −3 ​ ​ ⎭ ⎬ ⎫ ​ .