Question

# Questions 10~14Further problems2. Suppose A =2 33 4.a) What is det A ?b) What is det(5A) ?c)

If B is an arbitrary 2 × 2 matrix, what is the ratio det(5B)/ det(B) ?Think how the area of a parallelogram scales if all its edges are muiplied by a factor of 5.d) If B is an arbitrary 3 × 3 matrix, what is the ratio det(5B)/ det(B)?Think how the volume of a parallelepiped scales if all its edges are muiplied by 5.3. What are all the possible values for the determinant of an n × n orthogonal matrix Q?You can approach this question geometrically, recalling that every orthogonal matrix isa composition of rotations and reflections. You can also approach the question algebraically,starting from the formula QTQ = I and using the algebraic properties of determinants.3.3. If A is a 5 × 5 matrix and det A = 3, finda) det(−A) b) det(2A) c) det(A−1) d) det(ATA) e) the rank of A3.7. If P is a projection onto a plane in R3, what is det P ?4. Find the LU decomposition of A =a bc d. What are the pivots? Muiply the pivotsto recover the formula det A = ad − bc.5. Let A =1 −1 23 −4 52 1 1.a) Use elimination to calculate det A. Namely: bring A into upper-triangular form ‘U’,as you would in LU decomposition; then muiply the pivots to get det A. (You don’t need tokeep track of the individual elimination matrices or L, since their determinants are always 1.)b) Use the Big Formula to calculate det A.6. Let A =1 −1 23 −3 52 1 1.a) Use elimination to calculate det A. Careful: you will need a permutation at somepoint, which can change the sign of the determinant.b) Use the Big Formula to calculate det A.7. Find the determinant of A =1 2 1 31 1 0 12 −1 −1 23 2 1 −1. You can use either elimination or theBig Formula; one might be significantly easier than the other.98. Let A =0 3 −51 25 40 0 0 4 101 −200 0 −1 10 0 2 3 40 0 0 0 1. Use row exchanges to bring A to triangular form.Then calculate det A by using properties P3’ and P8 to getdet A = (−1)# row exchanges(product of pivots).9. There have been lots of problems in this class asking whether three vectors u, v, w in R3are dependent or independent. (Or asking whether u, v, w span all of R3 or just a plane,etc.) These can be answered with determinants, sincedetu v w= 0 ⇔ u, v, w are linearly dependent.This statement is equivalent to property P5 above.a) Let u =12−1, v =01−2, and w =214. Compute detu v wany way youlike (including via MATLAB) to determine whether u, v, w are independent or not.b) Let u =1913−25, v =5971−47, w =3123−39. What sort of subspace of R3 do thesevectors span: a point, a line, a plane, or all R3? Answer by computing a determinant, viaMATLAB or any other way.10. In a similar way, determinants can be used to ascertain whether or not linear equationshave solutions. For example, consider the equationA1 2−1 22 3xx1x2=bb1b2b3 .This equation has a solution if and only if b is a linear combination of the columns of A.In other words, there is a solution if and only if the vectors1−12,223, andb1b2b3 arelinearly dependent.a) Use the Big Formula to computedet1 2 b1−1 2 b22 3 b3b) Use the resu of (a) to determine the relation that b1, b2, b3 must satisfy in order forAx = b to have a solution.In the past, you have gotten this by using elimination/RREF form. The methods arecompletely equivalent.1011. Compute the determinant of A =1 −1 0−1 0 12 −4 2 (any way you like, including MATLAB)to determine whether the equation A x = 0 can have a nonzero solution.Hint: Remember that A x is a linear combination of the columns of A. So a nonzerosolution to A x = 0 exists if and only if the columns are linearly dependent.12. In the final week of the class, we will discuss eigenvalues. From a computationalperspective, the eigenvalues λ of a matrix A are solutions to the equation det(A − λI) = 0.For example, if A =1 20 3, thenA − λI =1 20 3− λ1 00 1=1 − λ 20 3 − λ,and det(A − λI) = (1 − λ)(3 − λ). The solutions to det(A − λI) = 0 are λ = 1 and λ = 3.a) Let A =17 00 5. Find the solutions to det(A − λI) = 0.b) Let A =17 890 5 . Find the solutions to det(A − λI) = 0.c) Let A =1 11 1. Find the solutions to det(A − λI) = 0.d) Let A =0 11 0. Find the solutions to det(A − λI) = 0.e) Let A =0 −11 0 . Find the solutions to det(A − λI) = 0.The last two problems involve the cross product. They are optional, and need not be turned in.13. The cross-product in R3 has a formal linear-algebra definition that may look a bitawkward at first, but starts to make more and more sense after years of use… Here it is.If u and v are vectors in R3, their cross-product w (denoted w = u × v) is the uniquevector in R3 with the property thatw · t = dett u vfor all t in R3.In other words, w is defined indirectly by saying that its dot product with any t has to bethe same as the determinant of the matrix with columns t, u, and v.a) By choosing t = u in the above relation, show that w must be perpendicular to u.b) By choosing t = v in the above relation, show that w must be perpendicular to v.Therefore, w must be perpendicular to the entire plane spanned by u and v.c) By choosing t =w||w|| in the above relation, deduce that the norm ||w|| must be equao the area of the parallelogram with edges u and v. (This is a parallelogram sitting in theplane spanned by u and v.)These are the fundamental properties of a cross-product. Note that they follow (relatively)easily from the abstract definition, in contrast to the usual cross-product formula.11Continuing the notation of Problem 13, let i =100, j =010, and k =001 denotethe standard basis vectors in R3. Let’s also write u =u1u2u3, v =v1v2v3, and w =w1w2w3.We want to find the components of w, given u and v. Observe thatw1 = w · i , w2 = w · j, w3 = w · k .14. a) Set t = i in the definition of the cross-product from Problem 13 to find w1 in termsof the entries of v and w.b) Similarly, set t = j and t = k to find w2 and w3.The resu of Problem 14 leads to the usual cross-product formula, written in the intuitive(but slightly illegal) wayw = deti u1 v1j u2 v2k u3 v3 or w = deti j ku1 u2 u3v1 v2 v3 .Final remark: the cross-product in R3 has a natural generalization to any Rn. In Rn, thecross-product takes n − 1 vectors u1, u2, …, un−1 as inputs, and outputs a single vector wwith the property thatw · t = dett u1 u2 · · · un−1for all t in Rn.In analogy with Problem 13, one can deduce that w is perpendicular to the (n − 1)-dimensional hyperplane spanned by u1, u2, …, un−1; and that the norm ||w|| equals the volumeof the (n−1)-dimensional parallelepiped with edges u1, u2, …, un−1. If we let i1, i2, …, indenote the standard basis vectors in Rn, then a concrete (slightly illegal) formula for w isw = deti1i2…u1 u2 · · · un−1in.12

10a) 1 2 b1 det −1 2 b2 = (1)(2)(b3 ) − (1)(b2 )(3) + (2)(b2 )(2) − (2)(−1)(b3 ) + (b1 )(−1)(3) − (b1 )(2)(2) 2 3 b3
= −7b1 + b2 + 4b3 10b) A needs to be singular for the columns to…
Math