MATH 304Linear AlgebraLecture 19: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 elim ination, back s ubs ti tution.• 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 3.1–3.4, 3.6)• Vector spaces (vectors, matrices, polynomi als,functional spaces).• Subspaces. Nullspace, column space, and rowspace of a matrix.• Span, spanning set. Linear independence.• Basis and dimension.• Rank and nullity of a matrix.Sample problems for Test 1Problem 1 (15 pts.) Find a quadratic polynomial p(x)such that p(1) = 1, p(2) = 3, and p(3) = 7.Problem 2 (25 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 matrix B.(ii) Find a basis for the row space of B, then extend this b asis to abasis for R4.(iii) Find a basis for the nullspace of B.Bonus Problem 5 (15 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 (15 pts.) Let V be a finite-dimensionalvector space and V0be a proper subspace of V (where propermeans that V06= V ). Prove that dim V0< dim V .Problem 1. Find a quadratic polynomial p(x) such thatp(1 ) = 1, p(2) = 3, and p(3) = 7.Let p(x) = ax2+ bx + c. Then p(1) = a + b + c,p(2 ) = 4a + 2b + c, and p(3) = 9a + 3b + c.The coefficients a, b, and c have to be chosen so thata + b + c = 1,4a + 2b + c = 3,9a + 3b + c = 7.We solve this system of linear equations using elementaryoperations:a + b + c = 14a + 2b + c = 39a + 3b + c = 7⇐⇒a + b + c = 13a + b = 29a + 3b + c = 7⇐⇒a + b + c = 13a + b = 29a + 3b + c = 7⇐⇒a + b + c = 13a + b = 28a + 2b = 6⇐⇒a + b + c = 13a + b = 24a + b = 3⇐⇒a + b + c = 13a + b = 2a = 1⇐⇒a + b + c = 1b = −1a = 1⇐⇒c = 1b = −1a = 1Thus the desired polynomial is p(x) = x2− x + 1.Problem 2. Let A =1 −2 4 12 3 2 02 0 −1 12 0 0 1.(i) Evaluate the determinant of the matrix A.Subtract the 4th row of A from the 3rd row:1 −2 4 12 3 2 02 0 −1 12 0 0 1=1 −2 4 12 3 2 00 0 −1 02 0 0 1.Expand the determinant by the 3rd row:1 −2 4 12 3 2 00 0 −1 02 0 0 1= (−1)1 −2 12 3 02 0 1.Expand the determinant by the 3rd column:(−1)1 −2 12 3 02 0 1= (−1)2 32 0+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 ma trix A−1.First we merge the matrix A with the identity 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 left part becomes the identity 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 0 0 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 4th row to the 1st row:1 −2 4 07 4 32 −390 −1 0 0 2 1 10 −120 0 −1 0 0 0 1 −10 0 0 −1 6 4 32 −39Add 4 times the 3rd row to the 1st 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, and the 4th rows by −1:1 0 0 03 2 16 −190 1 0 0 −2 −1 −10 120 0 1 00 0 −1 10 0 0 1 −6 −4 −32 391 0 0 03 2 16 −190 1 0 0−2 −1 −10 120 0 1 00 0 −1 10 0 0 1−6 −4 −32 39= (I | A−1)Finally the left part o f our 4 × 8 matrix is transformed into theidentity matrix. Therefore the current right part is the inversematrix of A. ThusA−1=1 −2 4 12 3 2 02 0 −1 12 0 0 1−1=3 2 16 −19−2 −1 −10 120 0 −1 1−6 −4 −32 39.Problem 2. Let A =1 −2 …
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