Questions and Answers #6 Matrices

Questions and Answers Sheet 6 Matrices Question #1 Construct a 3 × 3 matrix, not in echelon form, whose columns span 𝑅3 . Show that the matrix you construct has the desired property. Answer: 2 1 1   Let A = 2 2 0 be matrix is not in echelon form    2 0 3 Reduce matrix to reduced echelon form: 2 1 1  2 2 0  , multiply first row with -1 and add to second and third    2 0 3 2 1 1  ~ 0 1 −1 , add second row to third 0 −1 2  2 1 1  ~ 0 1 −1 , divide first row with 2  0 0 1  1 1  1 2 2    ~ 0 1 −1 0 0 1      1 add third row to second and multiply third row with − and add lo first 2  1  ~ 0 0   1 ~ 0 0 1  0 2  1 0  , multiply second row with − 12 and add to first 0 1   0 0 1 0  0 1  Since there are 3 pivot columns they span 𝑅3 Answer 2 1 1 A =  2 2 0   2 0 3 Question #2 Compute the determinants using a cofactor expansion across the first row. Also compute the determinant by a cofactor expansion down the second column. 0 4 1 5 −3 0    2 3 1  Answer: First we will calculate determinant by cafactor expansion across the first row: 0 4 1  5 −3 0 = 0   −3 0 − 4  5 0 + 1 5 −3  3 1 2 1 2 3           2 3 1  = 0 − 4 ⋅ (5 ⋅ 1 − 0 ⋅ 2) + (5 ⋅ 3 − (−3) ⋅ 2) = −4 ⋅ 5 + 21 = 1 First we will calculate determinant by cafactor expansion across the second column: 0 4 1 5 −3 0 = −4  5 0 + (−3)  0 1 − 3  0 1  2 1  2 1 5 0           2 3 1  = −4 ⋅ (5 ⋅ 1 − 0 ⋅ 2) − 3 ⋅ (0 ⋅ 1 − 1 ⋅ 2) − 3 ⋅ (0 ⋅ 0 − 1 ⋅ 5) = −4 ⋅ 5 − 3 ⋅ (−2) − 3 ⋅ (−5) = 1 Question #3 List five vectors in Span {𝐯𝟏 , 𝐯𝟐 } For each vector, show the weights on 𝐯𝟏 and 𝐯𝟐 used to generate the vector and list the three entries of the vector. Do not make a sketch. Answer: Step1 Remember that Span{𝐯𝟏 , 𝐯𝟐 }is the set of all linear combinations of the vectors 𝐯𝟏 and 𝐯𝟐 . That means that it is the collection of all vectors that can be written as 𝑎𝐯𝟏 + 𝑏𝐯𝟐 for some numbers a and b. The numbers a and b are called the weights of the vector relative to {𝐯𝟏 , 𝐯𝟐 } Step2 So to get any vector in the span, we just pick any two numbers a and b form the linear combination with those weights. Step 3 Vector1: Here is the vector in the span with weights 1 and 2. 7  −5  7   −10  −3   1  1  + 2  3  =  1  +  6  =  7   −6  0   −6  0   −6 Vector 2: Weights: 0 and -2. 7  −5 10    0  1  − 2  3  =  −6  −6  0   0  Vector 3: Weights: 0 and 0. 7  −5 0   0  1  − 2  3  = 0  −6  0  0 Vector 4: Weights: 5 and 7. 7  −5  0    5  1  + 7  3  =  26   −6  0   −30 Vector 5: Weights: -3 and 1  7   −5  −26 −3  1  + 1  3  =  0 ].  −6  0   18  Result:  −4 10  0  0   −26  7  ,  −6 , 0 ,  26  ,  0             −6  0  0  −30  18  Question #4 Determine if b is a linear combination of the vectors formed from the columns of the matrix A.  1 −4 2  3   A =  0 3 5  , b =  −7   −2 8 −4  −3 Answer: We need to determine if 3 b =  −7   −3 is a linear combination the vectors formed from the columns of the matrix A  1 −4 2 [ A | b] =  0 3 5  −2 8 −4  R1right  R2  −2  ~0 1 R3 + R1  R3  2 0  | 3   1 −4 2 | 3  | −7   0 3 5 | −7  | −3  −2 8 −4 | −3   8 −4 | −3  3 5 | −7  3 0 0 |  2 3 The third equation is 0 = , therefore we conclude that the system has no solution. 2 Hence b, is not a linear combination of the vectors formed from the columns of the matrix A Result: b is not a linear combination of the vectors formed from the columns of the matrix A Question #5 Determine if the system has a nontrivial solution. Try to use as few row operations as possible. 2𝑥1 − 5𝑥2 + 8𝑥3 = 0 −2𝑥1 − 7𝑥2 + 𝑥3 = 0 4𝑥1 + 2𝑥2 + 7𝑥3 = 0 Answer: The given system is 2𝑥1 − 5𝑥2 + 8𝑥3 = 0 −2𝑥1 − 7𝑥2 + 𝑥3 = 0 4𝑥1 + 2𝑥2 + 7𝑥3 = 0 The augmented matrix is  2 −5 8 0   2 −5 8 0   −2 −7 1 0   −2 −7 1 0      4 2 7 0   4 2 7 0  R1right  R3 R2 + 1/ 2 R1  R2 R3 − 1/ 2 R1  R3    4 2 7 0 4 2 7 0     0 -6 9 0   0 -6 9 0     2 2     9 9  0 -6 0   0 -6 0  2  2  R 3 -R 2  R 3 -1/6R 2  R 2 4 2 7 0 4 2 7 0    0 1 - 3 0 0 1 - 3 0 4 4    0 0 0 0 0 0 0 0    R 1 -2R 2  R 11/4R 1  R 1 17   1 0 8 0    0 1 - 3 0    4 0 0 0 0      Thus we get 17 17 𝑥1 + 𝑥3 = 0 ⇒ 𝑥1 = − 𝑥3 8 3 3 4 4 8 𝑥2 − 𝑥3 = 0 ⇒ 𝑥2 = 𝑥3 Hence, the given system has a nontrivial solution because variable 𝑥3 is a free. Result The system has a nontrivial solution because variable 𝑥3 is a free Question #6 Find a basis for the subspace spanned by the given vectors. What is the dimension of the subspace?  1   −3  2   −4  −3  9   −1  5   , , ,   2   −6  4   −3          −4 12   2   7  Answer: We need to find a basis for the subspace spanned by the given vectors and the dimension of the subspace. The augmented matrix is  1 −3 2 −4   −3 9 −1 5     2 −6 4 −3    −4 12 2 7  Reduce matrix to row echelon form  1 −3 2 −4   −3 9 −1 5     2 −6 4 −3    −4 12 2 7  R2 + 3R1  R2 ~ R3 − 2 R2  R3  1 −3 2 −4   1 −3  0 0 5 7  0 0    0 0 0 5  0 0    −4 12 2 7   −4 12 R4 − 4 R1  R4 ~ R4 − 2 R2  R4 1 −3 0 0  0 0  0 0 2 −4  1 5 7  0 0 5  0  0 5  0 1 0 R4 − R3  R4 ~  0  0 2 −4  5 7  0 5  2 7 −3 0 0 0 2 −4  5 7  0 5  0 5 −3 2 −4  0 5 7  0 0 5  0 0 5 Thus we get the matrix has pivot columns in first, third and fourth. According to this basis for the subspace spanned by given vector are  1   2   −4   −3  −1  5   , ,   2   4   −3        −4   2   7  and dim 𝐻 = 3 Result  1   2   −4   −3  −1  5    ,   ,   , dim H = 3  2   4   −3        −4   2   7  Question #7 Let 3 y=  1  and 8  u=  6  Compute the distance from y to the line through u and the origin. Answer: Formula for orthogonal projection: 𝑦⋅𝑢 𝑦̂ = 𝑢 𝑢⋅𝑢 Formula for distance: ||𝑦 − 𝑦̂|| 𝑦 ⋅ 𝑢 = (3,1) ⋅ (8,6) = 24 + 6 = 30 𝑢 ⋅ 𝑢 = (8,6) ⋅ (8,6) = 64 + 36 = 100 yˆ = 30 3 8 24 /10 u=  = 100 10 6 18 /10  3  24 /10  3 / 5  y − yˆ =   −  =  1 18 /10   −4 / 5 ||𝑦 − 𝑦̂|| − √ 9 16 + = √1 = 1 25 25 Question #8 Let  3 −6   −1 2    Construct a 2 × 2 matrix B such that AB is the zero matrix. Use two different nonzero columns for B. Answer:  3 −6    −1 2  Let  We need to construct a 2 × 2 matrix B such that AB is the zero matrix that is AB=0 Let a b  B=  c d  Find the product of AB  3 −6 a b  3a + (−6)c 3b + (−6)d  AB =   =  −1 2   c d  (−1)a + 2c (−1)b + 2d  Since AB=0, then  3a − 6c 3b − 6d  0 0 = =  −a + 2c −b + 2d  0 0 Two matrices are equal if their corresponding inputs are equal, so we get a system of equations 3a-6c=0 3b-6d=0 -a+2c=0 -b+2d=0 From the first equation we get a=2c Replace this in the third, we get −2𝑐 + 2𝑐 = 0 ⇒ 0 = 0 Thus, this is true. From the second we get b=2d, replace this in the fourth equation to get −2𝑑 + 2𝑑 = 0 ⇒ 0 = 0 Therefore, we get a=2c and b=2c We choose some arbitrary values for a and b and get in the first case 𝑎=2⇒𝑐=1 𝑏=4⇒𝑑=2 Thus, we get 1 2 B=   2 4 Find the product of AB  3 −6   2 4  AB =     −1 2  1 2  3  2 + (−6) 1 3  4 + (−6)  2  =   (−1)  2 + 2 1 (−1)  4 + 2  2  0 0 =  0 0 In the second case we get 𝑎=4⇒𝑐=2 𝑏=8⇒𝑑=4 Find the product of AB  3 −6 4 8  3  4 + (−6)  2 3  8 + (−6)  4 0 0 AB =   = =  −1 2  2 4 (−1)  4 + 2  2 (−1)  8 + 2  4  0 0 Hence,  2 4  4 8  1 2   2 4     Question #9 For the matrices, list the eigenvalues, repeated according to their multiplicities.  4 −7 0 2   0 3 −4 6     0 0 3 −8   0 0 0 1  Answer: By theorem, the eigenvalues of a triangular matrix are the entries on its main diagonal.  4 −7 0 2   0 3 −4 6     0 0 3 −8   0 0 0 1  The matrix as given is triangular. These are the entries on the main diagonal of the matrix, and therefore its eigenvalues. 1,4,3,3 Result: 1,4,3,3 Question #10 Construct a 3 × 3 matrix A, with nonzero entries, and a vector b in 𝑅𝟛 such that b is not in the set spanned by the columns of A. Answer: For the matrix A in the result, any vector in the span of the column vectors is of the form on the left. 1 1 1  x1 + x2 + x3  x1 1 + x2 1 + x3 1 =  x1 + x2 + x3  1 1 1  x1 + x2 + x3  After multiply the scalars and adding the vectors, we get a vector of this form. Notice that all the entries vector are equal. Since the entries in vector b are not all same, 1,2,3,b cannot be in the span of the column vectors. Result: 1 1 1 A = 1 1 1 1 1 1 and 1  b =  2   3  Question #11 Determine if the columns of the matrix span 𝑅𝟜  7 2 −5 8   −5 −3 4 −9     6 10 −2 7     −7 9 2 15  Answer: We will first enter the matrix A using the comand A = 7, 2 − 5,8; −5, −, 4, −9; −7,9, 2,15 Use rref(A) reduce matrix to reduced row echelon, we get 1 0 −0.6364 0.5455  0 1 −0.2727 2.0909    0 0  0 0 From Theorem the columns of A span 𝑅𝑚 if A has a pivot columns of A span 𝑅𝑚 . Here, A has not a pivot of position in ebery row. Hence the columns of A not span 𝑅𝑚 Result: Not span 𝑅4 Question #12 Find the cross product 𝑎 × 𝑏 and verify that it is orthogonal to both a and b. 𝑎 = ⟨6,0, −2⟩, 𝑏 = ⟨0,8,0⟩ Answer: The given in the problem, Given: 𝑎 =< 6,0, −2 >, 𝑏 =< 0,8,0 > 𝑛 =𝑎×𝑏 i j k  0 −2 a  b = 6 0 −2 = +i  − 8 0   0 8 0  6 −2 6 0  j +k   0 0  0 8  Solved as follows: = ((𝑎 × 𝑑) − (𝑏 × 𝑐)) +𝑖((0 × 0) − (−2 × 8)) − 𝑗((−2 × 0)) + 𝑘((6 × 8) − (0 × 0)) = 16𝑖 + 0𝑗 + 48𝑘 if the dot product equals 0 then it's orthogonal 𝑛. 𝑎 = ((𝑎 × 𝑏). 𝑎) = (16,0,48) ⋅ (6,0, −2) = (966 + 0 − 96) = 0 if the product equals 0 then it's orthogonal 𝑛. 𝑎 = ((𝑎 × 𝑏) ⋅ 𝑏) = (16,0,48) ⋅ (0,8,0) = 0 Question #13 Find the size of the matrix. Determine if it is a square, column, or row matrix. 6 1 0 2 0 6 Answer: 6 1 0 2 0 6 Clearly it has 3 columns and 2 rows, so it’s a 2 × 3 matrix It isn’t column matrix (𝑛 × 1), neither row (1 × 𝑛) not square (𝑛 × 𝑛) matrix. Question #14 Find the determinant of the matrix: [[2,1,4], [−3,1,5], [4,0,0]] Answer: [[2,1,4], [−3,1,5], [4,0,0]] = 2(0 − 0)– 1(0 − 20) + 4(0– 4) = 0 + 20– 16 = 4 𝐷 = 4 Question #15 Let 𝐴 = [[4,0,5], [−1,3,2]] 𝐵 = [[1,1,1], [3,5,7]] 𝐶 = [[2, , −3], [0, ,1]] Find 3𝐴 – 𝐵; 𝐶 ∗ 𝐵 + 𝐴. Answer: 𝐴 = [[4,0,5], [−1,3,2]] 𝐵 = [[1,1,1], [3,5,7]] 𝐶 = [[2, , −3], [0, ,1]] 1) (3𝐴– 𝐵) = 3 × [[4,0,5], [−1,3,2]] − [[1,1,1], [3,5,7]] = = [[12,0,15], [−3,9,6]] − [[1,1,1], [3,5,7]] = = [[11, −1,14], [−6,4, −1]] 2) (𝐶 ∗) = [[2, , −3], [0, ,1]] ∗= [[2, ,0], [−3, ,1]] (𝐶 ∗ 𝐵) = [[2, ,0], [−3, ,1]] × [[1,1,1], [3,5,7]] = = [[2,2,2], [0,2,4]] (𝐶 ∗ 𝐵 + 𝐴) = [[2,2,2], [0,2,4]] + [[4,0,5], [−1,3,2]] = = [[6,2,7], [−1,3,6]] Question #16 Write the matrix equation as a system of equations. [[3, −1], [2,4]][[𝑥], [𝑦]] = [[−2], [4]] Answer: [[3, −1], [2,4]][[𝑥], [𝑦]] = [[−2], [4]] (3𝑥, −, 𝑦, =, −1), (2𝑥, +,4𝑦, = ,3) Question #17 Write the matrix equation as a system of equations. [[4, −2], [2,3]][[𝑥], [𝑦]] = [[−2], [4]] Answer: [[4, −2], [2,3]][[𝑥], [𝑦]] = [[−2], [4]] (4𝑥, −,2𝑦, =, −2), (2𝑥, +,3𝑦, = ,4) Question #18 Tell which of the following subsets are subspaces of the vector space of nxn matrices, 𝑀(𝑛𝑥𝑛) a) The set of nxn matrices with a zero determinant. b) The invertible matrices. c) The diagonal matrices. d) Upper triangular matrices. e) The nxn matrices, all of 𝑀(𝑛𝑥𝑛) . Answer: Subspace of the vector space of nxn matrices, 𝑀(𝑛𝑥𝑛) are the subsets of the diagonal matrices and upper triangular matrices. Question #19 Tell, please, what are the ≺ and ≻ operators for when you use it with matrices? Answer: Most commonly, the relation 𝐴 ≻ 𝐵 is used to indicate ''A-B'' is positive definite. 𝐴 ≻ 7.3 means 𝐴 ≻ 7.3𝐼 ; that is, ''A is symmetric, and its eigenvalues are greater than 7.3.'' Question #20 What are the relationships between det(𝐴 + 𝐵) and A+B? Answer: There are formulas for det(𝐴 + 𝐵) in terms of A and B, but they are not nicely self-contained, as one needs to refer to compositions that can't be expressed in terms of basic matrix operations. det(𝐴 + 𝐵) = det(𝐴) + det(𝐵) + 𝑡𝑟(𝐴′ 𝐵) , where A' is the adjugate of A. The series acquires more terms as the size of the matrices increases.