in problem1-14 , a subset of some n-space r^n is defined by means of a given con
ID: 2981391 • Letter: I
Question
in problem1-14 , a subset of some n-space r^n is defined by means of a given condition imposed on the typical vector(x1,x2,...,xn). apply theorem 1 to determine wherther or not W is a subspace of R^n. 1)W is the set of all vectors in R^3 such that x1=5x^2 2)w is the set of all vectors in r^3 such that x2=1. 3)w is the set of all vectors in R^4 such that x1=3x3 and x2=4x4. in problem 15-18 apply method of example 5 to find 2 solutions vector u and v such that the solution space is the set of all linear comnations of the form su+tv 4)x1-4x2-3x3-7x4=0 2x1-x2+x3+7x4=0 x1+2x2+3x3+11x4=0 5) x1+3x2+8x3-x4=0 x1-3x2-10x3+5x5=0 x1+4x2+11x3-2x4=0 please post the answers in ready to copy form, got like 30 minutes to copy stuff.. as soon as possible.. and the theorem 1 is the if u and v are in vector pace W then U+V is also in vector spce W. and, if u is in W and c is scalar, then vector Cu is also in WExplanation / Answer
A subset W of some n-space Rn is defined by means of a given condition imposed on on the typical vector (x1,x2,...,xn). Apply the theorem to determine whether or not W is a subspace of Rn Theorem: The nonempty subset W of the vector space V is a subspace of V iff it satisfies the following two conditions: (i)If u and v are vectors in W, the u + v is also in W. (ii)If u is in W and c is a scalar, the the vector cu is also in W. W is the set of all vectors in R4 such that x1 + x2 = x3 + x4. LET ANY ELEMENT X IN U BE X=[X1,X2,X3,X4] AND ANYELEMENT Y IN V BE Y=[Y1,Y2,Y3,Y4] SO WE HAVE FROM THE GIVEN PROPERTY THAT X1+X2=X3+X4...............................................................1 Y1+Y2=Y3+Y4.......................................................2 NOW LET US APPLY THE 2 TESTS NEEDED FOR THEM TO FORM A SUB SPACE WE GET BEFORE APPLYING THE TESTS WE NOTE THAT W IS NON EMPTY SET SINCE WE CAN FIND SEVERAL VECTORS IN R4 such that x1 + x2 = x3 + x4....VIZ.... [3,2,4,1].....3+2=4+1=5 ETC...SO OK ... (i)If u and v are vectors in W, the u + v is also in W. X+Y=[X1,X2,X3,X4]+[Y1,Y2,Y3,Y4] = [X1+Y1,X2+Y2,X3+Y3,X4+Y4] IN THE RESULTANT VECTOR WE FIND FROM 1 AND 2 THAT (X1+Y1)+(X2+Y2) = (X1+X2)+(Y1+Y2)= (X3+X4)+(Y3+Y4)=(X3+Y3)+(X4+Y4) SO IT SATISFIES THE GIVEN REQUIREMENT AND HENCE IS AN ELEMENT OF W ...OK.. (ii)If u is in W and c is a scalar, the the vector cu is also in W. CX=C[X1,X2,X3,X4]=[CX1,CX2,CX3,CX4] IN THE RESULTANT VECTOR WE FIND FROM 1 THAT CX1+CX2=C[X1+X2]=C[X3+X4]=CX3+CX4... SO IT SATISFIES THE GIVEN REQUIREMENT AND HENCE IS AN ELEMENT OF W ...OK.. HENCE W IS SUB SPACE............PROVED
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