MATH 304Linear AlgebraLecture 20:Review for Test 1.Topics for Test 1Part I: Elementary linear algebra (Leon 1.1–1.4,2.1–2.2)• Systems of linear equations: elementaryoperatio ns , Gaussian elimination, back substitution.• Matrix of coefficients and augmented matrix.Elementary row operations, row echelon form andreduced row echelon form.• Matrix algebra. Inverse matrix.• Determinants: explicit formulas for 2×2 and3×3 matrices, row and column expansions,elementary row and column op er ations.Topics for Test 1Part II: Abstract linear algebra (Leon 3.1–3.6)• Vector spaces (vectors, matrices, polynomi als,functional spaces).• Subspaces. Nullspace, column space, and rowspace of a matrix.• Span, spanning set. Linear independence.• Bases and dimension. Rank and nullity of amatrix.• Change of coordinates, transition matrix.Sample problems for Test 1Problem 1 (20 pts.) Find the point of intersection of theplanes x + 2y − z = 1, x − 3y = −5, and 2x + y + z = 0 inR3.Problem 2 (30 pts.) Let A =1 −2 4 12 3 2 02 0 −1 12 0 0 1.(i) Evaluate the determinant of the matrix A.(ii) Find the inverse matrix A−1.Problem 3 (20 pts.) Determine which of the following subsetsof R3are subspaces. Briefly explain.(i) The set S1of vectors (x, y, z) ∈ R3such that xyz = 0.(ii) The set S2of vectors (x, y, z) ∈ R3such that x + y + z = 0.(iii) The set S3of vectors (x, y, z) ∈ R3such that y2+ z2= 0.(iv) The set S4of vectors (x, y, z) ∈ R3such that y2− z2= 0.Problem 4 (30 pts.) Let B =0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1.(i) Find the rank and the nullity of the mat rix B.(ii) Find a basis for the row space of B, then extend this basis to abasis for R4.Bonus Problem 5 (20 pts.) Show that the functionsf1(x) = x, f2(x) = xex, and f3(x) = e−xare linearlyindependent in the vector space C∞(R).Bonus Problem 6 (20 pts.) Let V and W be subspaces ofthe vector space Rnsuch that V ∪ W is also a subspace ofRn. Show that V ⊂ W or W ⊂ V .Problem 1. Find the point of i ntersection of the planesx + 2y − z = 1, x − 3y = −5, and 2x + y + z = 0 in R3.The intersection point (x, y , z) is a solution of the systemx + 2y − z = 1,x − 3y = −5,2x + y + z = 0 .To solv e the system, we convert its augmented matrix intoreduced row echelon form using elementary row operations:1 2 −111 −3 0 −52 1 1 0→1 2 −110 −5 1 −62 1 1 0→1 2 −110 −5 1−62 1 10→1 2 −110 −5 1−60 −3 3−2→1 2 −110 −3 3−20 −5 1−6→1 2 −110 1 −1230 −5 1−6→1 2 −110 1 −1230 0 −4−83→1 2 −110 1 −1230 0 123→1 2 −110 1 0430 0 123→1 2 0530 1 0430 0 123→1 0 0−10 1 0430 0 123.Thus the three planes intersect at the point (−1,43,23).Problem 1. Find the point of i ntersection of the planesx + 2y − z = 1, x − 3y = −5, and 2x + y + z = 0 in R3.Alternative solution: The intersection point (x, y , z) is asolution of the systemx + 2y − z = 1,x − 3y = −5,2x + y + z = 0 .Add all three equations: 4x = −4 =⇒ x = −1.Substitute x = −1 into the 2nd equation: =⇒ y =43.Substitute x = −1 and y =43into the 3rd equation:=⇒ z =23.It remains to check that x = −1, y =43, z =23is indeed asolution of the system.Problem 2. Let A =1 −2 4 12 3 2 02 0 −1 12 0 0 1.(i) Ev aluate the determinant of the matrix A.Subtract 2 times the 4th column of A from the 1st column:1 −2 4 12 3 2 02 0 −1 12 0 0 1=−1 −2 4 12 3 2 00 0 −1 10 0 0 1.Expand the determinant by the 4th row:−1 −2 4 12 3 2 00 0 −1 10 0 0 1=−1 −2 42 3 20 0 −1.Expand the determinant by the 3rd row:−1 −2 42 3 20 0 −1= (−1)−1 −22 3= −1.Problem 2. Let A =1 −2 4 12 3 2 02 0 −1 12 0 0 1.(ii) Find the inverse matrix A−1.First we merge the matrix A with the i dentity matrix into one4 × 8 matrix(A | I) =1 −2 4 11 0 0 02 3 2 00 1 0 02 0 −1 10 0 1 02 0 0 1 0 0 0 1.Then we apply elementary row operations to this matrix untilthe l eft part becomes the i dentity matrix.Subtract 2 times the 1st row from the 2nd row:1 −2 4 11 0 0 00 7 −6 −2−2 1 0 02 0 −1 10 0 1 02 0 0 10 0 0 1Subtract 2 times the 1st row from the 3rd row:1 −2 4 11 0 0 00 7 −6 −2−2 1 0 00 4 −9 −1−2 0 1 02 0 0 10 0 0 1Subtract 2 times the 1st row from the 4th row:1 −2 4 11 0 0 00 7 −6 −2 −2 1 0 00 4 −9 −1−2 0 1 00 4 −8 −1 −2 0 0 1Subtract 2 times the 4th row from the 2nd row:1 −2 4 11 0 0 00 −1 10 02 1 0 −20 4 −9 −1−2 0 1 00 4 −8 −1−2 0 0 1Subtract the 4th row from the 3rd row:1 −2 4 11 0 0 00 −1 10 02 1 0 −20 0 −1 00 0 1 −10 4 −8 −1−2 0 0 1Add 4 times the 2nd row to the 4th row:1 −2 4 11 0 0 00 −1 10 0 2 1 0 −20 0 −1 00 0 1 −10 0 32 −1 6 4 0 −7Add 32 times the 3rd row to the 4th row:1 −2 4 11 0 0 00 −1 10 02 1 0 −20 0 −1 00 0 1 −10 0 0 −16 4 32 −39Add 10 times the 3rd row to the 2nd row:1 −2 4 11 0 0 00 −1 0 02 1 10 −120 0 −1 00 0 1 −10 0 0 −16 4 32 −39Add the 4 th row to the 1st row:1 −2 4 07 4 32 −390 −1 0 0 2 1 10 −120 0 −1 00 0 1 −10 0 0 −1 6 4 32 −39Add 4 times the 3rd row to the 1 st row:1 −2 0 07 4 36 −430 −1 0 02 1 10 −120 0 −1 00 0 1 −10 0 0 −16 4 32 −39Subtract 2 times the 2nd row from the 1st row:1 0 0 03 2 16 −190 −1 0 02 1 10 −120 0 −1 00 0 1 −10 0 0 −16 4 32 −39Multiply the 2nd, the 3rd, a nd the 4th rows by −1:1 0 0 03 2 16 −190 1 0 0 −2 −1 −10 120 0 1 0 0 0 −1 10 0 0 1 −6 −4 −32 391 0 0 03 2 16 −190 1 0 0−2 −1 …
View Full Document
Unlocking...