21 Questions on Linear Algebra - Final Practice Examination | MATH 3013, Exams of Linear Algebra

Download 21 Questions on Linear Algebra - Final Practice Examination | MATH 3013 and more Exams Linear Algebra in PDF only on Docsity! Prackcce | Raya 1. Find all eigenvalues and associated eigenspaces for the following matrices: 2. Determine if A = 1 is an eigenvalue of a) we No AO ADH 3. The matrix has eigenvectors 1 1 2 v=] -l |ve=]0],v3=] 0 1 1 1 What are the eigenvalues of A associated with these eigenvectors? 4. Use expansion by minors to find the determinant of the following: 3 00-2 4 020 0 0 (a) | 0-10 5 -3 4 01 0 6 0-10 3 2 010 00 000 08 (bs) |}200 40 005 00 100-10 5. Let A bea 4 x 4 matrix with det A= —3. Compute the following determinants: (a) det A? (b) det Aq? (c) det As (d) det(2A) (e) det(B.Ay’ 6. Suppose A, B and C are all 3 x 3 matrices such that det A = 2, det B = 5 and det C = 7. Find (a) det(3AB) (b) det(A?BO™!) 7. Let 1 3 -1 A= QO «a 6 -2 -2 2 (a) Find det A. (b) For what values of z is A invertible? c) For those « such that A~ exists find det A7’. 9. For each matrix, determine if it is diagonalizable. If it is, find a diagonal matrix D and an invertible matrix P so that P~'AP = D. -404 wa] 1 ‘ 20 2 20 0 (b) A= | —5 17 —20 —5 15 -18 2-1 2 () A=] 5-3 3 -1 0 -2 10. Let A be a 4 x 4 matrix with eigenvalues 5,2 and —2. If the eigenspace for 4 = 2 is two-dimensional, can we conclude that A is diagonalizable? Explain. Directions: Answer the following questions on a separate piece of paper. You may use a calculator. If you do not show your work, you will receive no credit. CIRCLE YOUR FINAL ANSWER. IF YOU DO NOT FOLLOW DIRECTIONS YOU WILL BE PENALIZED! Note: This practice exam is longer than the actual exam. It is meant to give an idea of types of questions that will be asked. 1, Find all eigenvalues and associated eigenspaces for the following matrices: 1 -2 @ | 36 Solution: We find the eigenvalues by finding the roots of the characteristic polynomial: det(A— XI) = \3 1A =? onal —x = (1-A)(6- ny (2)(-2) = (6 o +2) 44 = i ; 10 = (A-5)(A 2) So the eigenvalues are 41 = 2 and A, = 5. For A; = 2, we have -1l -27)0 aay = [-} 3] rref 1 2/0 0 010 So we know that a+ 2%. =0 Which is equivalent to By = ~~ 209 Letting z2 = s, we have | 2s | —2 x=] 4 =s 1 ~2 : . So 1 \ is a basis for Ey. For Ap = 5, we have —4 -210 |A-silo) = ; |  So we know that Which is equivalent to Letting 2 = 2s, we have So {| 7 | \ is a basis for Es. | 1 8 1 -1 Solution: We find the eigenvalues by finding the roots of the characteristic polynomial: 1-A & det(A-—AI) = | 1 | = (1-A)(-1—A) = (8)Q) = (-1+3")-8 = V9 = (A-3)(A +3) So the eigenvalues are A; = 3 and Az = —3. For A; = 3, we have —2 8/0 aang = [2 82) ref. 1 -4/0 0 Oj;0 21 — 4%, =0 So we know that Which is equivalent to Letting #2 = s, we have  So { i is a basis for H3. Ag For Az = —3, we have 4 8/0 asarg - [4 8/2] rref 12 0 010 So we know that t +22, =0 Which is equivalent to Ly = —229 Letting rz = s, we have so { 7 | is a basis for B_3. | on w one woe | Solution: We find the eigenvalues by finding the roots of the characteristic polynomial: det(A — AZ) I Oo I x I @ I = — w l Ze a } Zs | S S Se = (-APG-A) So the eigenvalues are A, = 3 and A, = 5. 3. The matrix 7 0 -8 A=|0-1 0 4 0 -5 has eigenvectors 1 1 2 v= -lij,ve=]O01,vs=] 0 1 1 1 What are the eigenvalues of A associated with these eigenvectors? Solution: For v, to be an eigenvector of A we must have Av, = A;v1, so we multiply 7 0 -8 1 { Avy = 0-1 0 -i \ 4 0 -5 1 -1 = 1 ~1 = -V1 So A; = —1 is an eigenvalue of A with eigenvector v1. For ve to be an eigenvector of A we must have Av2 = AzVe2, so we multiply 7 0 +8 1 Avg = 0-1 #0 0 4 0 -5 1 -1 = 0 -1 = -Vv2 So Az = —1 is an eigenvalue of A with eigenvector vo. For v3 to be an eigenvector of A we must have Av3 = A3v3, so we multiply 7 0 -8 2 Avg = |0 -1 0 0 4 0 -65 i 6 = | 0 3 = 3va So \3 = 3 is an eigenvalue of A with eigenvector v3. 4, Use expansion by minors to find the determinant of the following: (a) 3 00-2 4 0.20 0 90 0-10 5 -3 J -4 01 0 6 0-10 38 2 Solution: Expand on the third column to get 30 0-2 4 0 20 0 90 5 0-1 0 5 -3)=(-1) 0 4 01 0 6 0 0-10 3 2 Now expand on the second row 30-2 4 0 2 0 0 Cylg fs 3 [=Cn@ 0-1 3 2 Expand again (first column) 3-2 (-1)(2))}0 5 -3 0 8 Now evaluate the 2 x 2 determinant ye] 3 3 | 0-2 #4 2 0 0 -1 5 +38 -1 3 2 3-2 4 0 5 -3 0 3 2 =-6(10+9) =—114 010 00 000 03 200 40 005 090 100 -1 0 Solution: Expand on the first row to get 010 00 00 03 000 08 20 40 200 4 0/=(-1) 05 090 005 00 10-1 0 jo 0 —1.0  Now expand on the first row 20 40 a0 4 (-1) =(C1-3)|0 5 0 05 00 10-1 10 -1 0 Expand again (second column) 20 4 4 Caealos o/-cay-ae| 7 _f| 10-1 5. Let A be a4 x 4 matrix with det A = —3. Compute the following determinants: (a) det.A? Solution: Since det A? = det A, we have det A? = —3 (b) det A-} Solution: Since det A“! = ztj, we have 1 tA = —5 dei 3 (ec) det AS Solution: Since det A* = (det A)*, we have det A® = (—3)? = —27 (er) © FaR . Ayt (0) A? ~ BT (c) Solution: If A ~ B, then there exists an invertible matrix P so that P-1AP = B, or equivalently AP = PB. Notice that ° A (ep)! =p! (4Py" = (PB) as PRAT = BTpr 6 4; AT(PTY = (PT)-'Bt ws © -y') so AT ~ BT, {c) BoA Solution: This is Theorem 4.21b. (d) AB~ BA Solution: Let a0 Lo a=[To]-#=[0 0] Notice that AB= |} 0 ea |g 0 | 10 00 So if AB ~ BA then there would exist an invertible matrix P so that P(AB)P = BA P-\(AB)P = O PP(AB)P = PO I(AB)P = O (AB)P = O (AB)PP* = OP (AB)I = O AB =O But AB # O, so they cannot be similar.  9. For each matrix, determine if it is diagonalizable. If it is, find a diagonal matrix D and an invertible matrix P so that P-1AP = D. 404 (a) A= 3 1 i —-2 0 2 Solution: We first find the eigenvalues and associated eigenspaces of the ma- trix. We find the eigenvalues by finding the roots of the characteristic polyno- mial: -4-r 0 4 det(A- Ar) = -5 1-A 4 —2 0 2-A = a-a)| “> otal 1=2)((-4= a) =a) - (-2)) = MI-A)A42) So the eigenvalues are \, = 0, Ag = 1 and Ag = —2. For A, = 0, we have [A-orj 0] = [4/0 —-4 0 41/0 = —5 1 4/0 —2 0 2/0 1 9 -1]/0 mf, }o1 -1]0 00 0/0 So we know that xy — £3 0 t—-t3 = 0 Which is equivalent to ty = 4% f = 4% Letting +3 = s, we have  ooo rref a_ -5 [A-zo] = | 5 —2 a i) So we know that Letting 2 = s, we have 0 So | 1 | is a basis for Fy. 0 For Az = —2, we have 20 [A+2ro] = | -5 3 20 So we know that "1-223 = 0 tq—-223 = 0 Which is equivalent to a = 2 ty = 223 (c) A 2-1 2 = 5 -3 3 -1 0 -2 Solution: We first find the eigenvalues and associated eigenspaces of the ma- trix. We find the eigenvalues by finding the roots of the characteristic polyno- mnial: 2-rA -1 2 det(A-— AD) = 5 -8-A 3 -1 0 -2-Xd -1 2 2-dh -1 = | 92, 3 |+-2-9)| 5 sa = (ne) @-3- a) +29 (@-C8-)- = (-1)(-346+4 2A) + (-2-A)(-6 +A 447 +5) = (-1)(8+ 2A) 4 (-2-A)Q?+A-1) = -3—2\-2 242-4 - MHA = —-\M—3\?-3A-1 = —(8 4307 +3A+1) = -(A+1) So the eigenvalue is A, = —1. For Ay = —1, we have 3-1 2]Jo {A+ 2/0] = 5 -2 310 -1 0-1/0 ; 10 1/0 rf, |o11i0 0 0 o|o So we know that a+r; = 0 fg+z3 = 0 Which is equivalent to 1) = —£3 tg = —%3 Letting +3 = s, we have 10. 11. -1 So -1 is a basis for B_y. 1 The geometric multiplicity of 41 = 1, but the algebraic multiplicity of A, = 3, so the matrix is not diagonalizable. Let A be a4 x 4 matrix with eigenvalues 5,2 and —2. If the eigenspace for A = 2 is two-dimensional, can we conclude that A is diagonalizable? Explain. Solution: Yes. Since we have a 4 x 4 matrix, the characteristic polynomial is a fourth degree polynomial. Since we only have three eigenvalues, one of them must have algebraic multiplicity two. We know that the algebraic multiplicity is greater than or equal to the geometric multiplicity, so since the geometric multiplicity of d = 2 is 2, we know its algebraic multiplicity is also 2. The other two eigenvalues have algebraic and geometric multiplicity one, so the matrix is diagonalizable. Let A be a 7x7 matrix with three eigenvalues. If one eigenspace is two-dimensional and the second is three-dimensional, can we conclude that A is diagonalizable? Explain. Solution: No. Since we have a 7 x 7 matrix, the characteristic polynomial is a seventh degree polynomial. Since we only have three eigenvalues, at least one of them must have algebraic multiplicity two or more. We know that the algebraic multiplicity is greater than or equal to the geometric multiplicity, so since the geo- metric multiplicity of A, is 2, we know its algebraic multiplicity is 2 or greater. The geometric multiplicity of Az is 3, 50 we know its algebraic multiplicity is 3 or greater. For A to be diagonalizable, we need the geometric and algebraic multiplicity to be the same, so we need A; to have algebraic multiplicity 2 and A) to have algebraic multiplicity 3. This leaves \3 with algebraic multiplicity 2, (since the sum of all the algebraic multiplicities has to be seven), but Ag can have geometric multiplicity 1 which means A is not guaranteed to be diagonalizable.  1 0 0 0 2 2 Vi= ly 1v2= 1 Va= ] _o 1 -1 2 (a) Show that {vi, v2, v3} form an orthogonal set. Solution: virve = (1)(0) + (0)(2) + )() + A)(-1) = 0 virvs = (1)(0) + (0)(2) + 4)(—2) + (2) = 0 varvs = (0)(0) + (2)(2) + €)(-2) + (DQ) = 0 Since every pair of vectors in the set is orthogonal, the set is orthogonal. (b) If Vis the subspace of R* spanned by v1, v2 and vs, find an orthonormal basis for V. Solution: {v1, v2, v3} forms an orthogonal basis for V. To make it orthonor- mal, we just need to normalize the vectors. So {q,, 42,43} is an orthonormal basis for V if = 1 Yy. a Tal" 1 a It Ss ol SSF o SF  Since this is an orthogonal set of unit vectors, it is an orthonormal set. 14. Find the orthogonal complement W+ of W =span PRR ra Solution: W = col(A) where a ! e so W+ = (col(A))+ = null(A?) Note that rom [1 11 1[0) wer f1 o1 2 ar oj= [7 -14 20] |o 10-4 so any vector in W+ = null(A’) has form 3 a3 — 544 - 1 904 Zs Tq + &4 HH Onesie and a basis for W+ = null(AT) is Oraor BO ih wie 1 15. Find the orthogonal decomposition of v = with respect to -1 2 0 1 3 1 0 1 V= W = span tied a fo} -2 1 -1 1 Solution: We can take vewtwt where w = projy(v) and wt = perpy,(v) Notice that uy >= HERO forms an orthogonal basis for W, so w= Pow = proj,,(v) + projy,(v) + proj,,,(v) = (aan) oe Gea) Gea) ~ (gs + (Fue (Gs | | =. 88 ale GIS SIS ann & wr = perpy({v) = Vv—projy(v) ja oh foo Ea ent a 16. Indicate TRUE if true in all cases or FALSE otherwise. Justify your answer. (a) (b) If an n x n matrix A has two identical rows, then det A = 0. Solution: True. This is Theorem 4.3c. The determinant of an upper triangular matrix equals the product of its diagonal entries. Solution: True. This is Theorem 4.2. 1 ; ; : To For an invertible matrix A, det A” = 44 Solution: False. det AT = det A and det A"? = 74a. If A is ann x n matrix and Ax = 0 for some x # 0, then det A= 0. Solution: True. By the Fundamental Theorem of Invertible matrices, Ax = 0 iff A is not invertible iff det A = 0. If A is a matrix with positive entries, then det A > 0. 12 Solution: False. Let A = E 1 | Then det A=1-4=-3<0. If A and B are n x n invertible matrices, then det(B-!ABT) = det A. Solution: True. det(B1ABT) = det B~'det Adet B™ 1 = TAB det A det B = detA