#1 : Solve the 3 linear equations in 5 unknowns, r1,T2,X3Ju,25, given by KLK? Le
ID: 3116310 • Letter: #
Question
#1 : Solve the 3 linear equations in 5 unknowns, r1,T2,X3Ju,25, given by KLK? Le! + 2x2-1x3 + 2x5 = 4 in the following manner The associated augment ed matrix is just 1 2 2 909 B= 1216118 1 2 -1 0 24 Go to www.wolfram.com and enter in the box the expression: row reduce 1,2, 2,9,0,93, {1,2, 1,6,1,8), {1,2,-1,0, 2, 4 ), one gets the output, the row reduce form 1 2 0 3 03 0 013 0 3- Now answer the following questions a) Where are the pivots in Xjust circle them b) Place the variables above the corresponding columns in X. What are the basis variables? What are the free variables? c) The solution expresses the basic variables in terms of the free variables. What is the general solution to these equations? d) What is the dimension of the row space of B, that is the span of the rows? e) What is a basis for the row space of B? f) What is the dimension of the column space of B, that is the span of the columns? g) What is a basis for the column space of B?Explanation / Answer
a). We know that a pivot position in a matrix A is a position in the matrix that corresponds to a row–leading 1 in the reduced row echelon form of the matrix A. Here, the pivots in X are the a11, a23,and a35 entries in X(aij denotes the entry in the ith row, jth column).
b). The variables are placed above the corresponding columns in X.
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The given system of linear equations is equivalent to x1+2x2+3x4=3 or,x1=3-2x2-3x4, x3+3x4=3 or,x3=3-3x4 and x5 = 2. Thus, x2 and x4 are free variables as x1 and x3 can be determined in terms of x2and x4.
c). The general solution is X = (x1,x2,x3,x4,x5)T =(3-2x2-3x4, x2,3-3x4,x4,2)T = (3-2r-3t, r,3-3t,t,2)T= (3,0,3,0,2)T +r(-2, 1,0,0,0)T+t(-3,0,-3,1,0)T where r(=x2) and t(=x4) are arbitrary real numbers.
d). The dimension of row(B), being equal to the number of non-zero rows in X, is 3.
e). A basis for the row space of B is {(1,2,2,9,0),(1,2,1,6,1),(1,2,-1,2)}.
f). The dimension of col(B) is 3. (the dimensions of row space and column space are equal)
g). A basis for col(B) is { (1,1,1)T,(2,1,-1)T,(0,1,2)T}. As is evident fro X, the RREF of B,the 2nd and the 4th columns of B are linear combinations of its 1st, 3rd and 5th columns.
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