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- Rather than use the standard definitions of addition and scalar multiplication in R2, let these two operations be defined as shown below. (x1,y1)+(x2,y2)=(x1+x2,y1+y2)c(x,y)=(cx,y) (x1,y1)+(x2,y2)=(x1,0)c(x,y)=(cx,cy) (x1,y1)+(x2,y2)=(x1+x2,y1+y2)c(x,y)=(cx,cy) With each of these new definitions, is R2 a vector space? Justify your answers.arrow_forward(3) Let V be R2, the set of all ordered pairs (x, y) of real numbers. Define an operation of "addition" by (u, v) (x, y) = (u+x+1,v+y+1) for all (u, v) and (x, y) in V. Define an operation of "scalar multiplication" by a(r, y) = (ar, ay) for all a ER and (x, y) = V. Under the operations and the set V is not a vector space. The vector space axioms (see 5.1.1 (1)-(10)) which fail to hold are andarrow_forwardInstead of using the standard definitions of addition and scalar multiplication in R³, suppose these two operations are defined as follows. (x1, Y1, 21) + (x2, Y2, 2) = (x1 + x2 + 1, y1 + y2 + 1, 21 + 22 + 1) c(x, y, z) = (cx +c - 1, cy + e – 1, cz + e – 1) With these new definitions, Show that it satisfies the Axioms 5,6, and 7arrow_forward
- (a) Let V be R², and the set of all ordered pairs (x, y) of real numbers. Define an addition by (a, b) + (c,d) = (a + c, 1) for all (a, b) and (c,d) in V. Define a scalar multiplication by k · (a, b) = (ka, b) for all k E R and (a, b) in V. . Verify the following axioms: (i) k(u + v) = ku + kv (ii) u + (-u) = 0arrow_forwardLet A ∈ Rm×n, B ∈ Rn×r, and x, y ∈ Rn. Supposethat the product AxyTB is computed in the followingways:(i) (A(xyT ))B (ii) (Ax)(yTB)(iii) ((Ax)yT )B Compare the number of scalar additions and multiplications for each of the three methods when m = 5, n = 4, and r = 3. Which method is most efficient in this case?arrow_forwardplz provide answer q1arrow_forward
- (u1 + v1, U2 + v2) and au = (au1, 2), respectively. Let V = {(T,y)r, Y ER}, with addition and scalar multiplication defined as u + v = (u1, U2) and v = (v1, V2) and scalars a. Which of the following is true? where u = O A. k(lu) = (kl)u and 1u = u for all E V O B. (0,0) is the zero of V and (kl)u = k(lu) for all scalars k, l and u E V || O C. Addition is associative and k(u + v) ku + kv for all u, v E V and for all scalars k O D. 3(u + v) : 3u + 3v and the negative of (-2, 3) is (2, –3) O E. k(u + v) = ku + kv for u, v E V and scalar k, and addition is commutative Reset Selectionarrow_forward7. Let V={(x,y) x,y e R}. Suppose addition and scalar multiplication are defined using the following non-standard rules where c is any real number. (x,y) + (x₂,₂)=(x₁-x₂₂0) c(x,y)=(-x,₁,Scy,) a. Find the result of (3,-2) + (-4,-3) under the above operations. b. Find the result of -3(2,-4) under the above operations. c. Show that V, with respect to these operations of addition and scalar multiplication, is not a vector space by showing that one of the vector space axioms does not hold. Clearly identify the axiom you have chosen.arrow_forwardThis is a problem from Linear Algebra:arrow_forward
- Show that the function f: R² → R defined by f(x₁, x₂) = x² + 2x² + 3x₁x₂ + 4x₁ + 5x₂ + 6 is a vector quadratic function. HINT: Start by saying how 'vector quadratic function' is defined.arrow_forwardT: P2 P2 defined by T(a + bx + cx²) = a + b(x + 8) + b(x + 8)² Find the following images of the function for vectors p₁(x) = a₁ + b₁x + c₁x² and p₂(x) = a₂ + b₂x + c₂x2 in P₂ and the scalar k. (Give all answers in terms of a₁, b₁, C₁, 22, b₂, C₂ and c.) T(P₁) + T(P₂): = T(P1 + P₂) = cT (P1) = T(CP1) = Determine whether T is a linear transformation. 4 Olinear transformation O not a linear transformationarrow_forwardProve that: (5a + 6b | a, b = Z} = Zarrow_forward
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