Let A = [1 1 2 -1 1 7 6 13 1 15 3 3 6 5 11 1 0 1 -1 1]. Find the rank of A. Find
ID: 3037514 • Letter: L
Question
Let A = [1 1 2 -1 1 7 6 13 1 15 3 3 6 5 11 1 0 1 -1 1]. Find the rank of A. Find a basis for the row space of A. What is the dimension of the row space of A? Find a basis for the column space of A. What is the dimension of the column space of A? Find a basis for t he solution space of A. What is the dimension of the solution space of A? Feel free to use a computer algebra system for the row reduction in this question - only this question. The syntax for Mathematica (and Wolfram Alpha) is RowReduce[{r_1, r_2, ..., r_n}] where the row r_1 = (1, 1, 2, -1, 1) is entered as {1, 1, 2, -1, 1}. To display the results in convenient matrix notation, use the command MatrixForm[].Explanation / Answer
(a) The rank of A is the number of non-zero rows in its RREF. We will therefore, reduce A to its RREF as under:
Add -7 times the 1st row to the 2nd row; Add -3 times the 1st row to the 3rd row
Add -1 times the 1st row to the 4th row; Multiply the 2nd row by -1
Add 1 times the 2nd row to the 4th row; Multiply the 3rd row by 1/8
Add 8 times the 3rd row to the 4th row; Add 8 times the 3rd row to the 2nd row
Add 1 times the 3rd row to the 1st row; Add -1 times the 2nd row to the 1st row
Then the RREF of A is
1
0
1
0
2
0
1
1
0
0
0
0
0
1
1
0
0
0
0
0
Thus, Rank(A) = 3.
(b). In view of the RREF of A, a basis for Row (A) is {(1,1,2,-1,1),(7,6,13,1,15),(3,3,6,5,11)}. dim (Row(A)) = 3.
(c). In view of the RREF of A, a basis for Col(A) is {(1,7,3,1)T,(1,6,3,0)T,(-1,1,5,-1)T } as the 3rd and 5th columns are linear combinations of 1st, 2nd and 4th columns.dim(Col(A)) = 3.
(d). If, by the solutoion space of A, we mean that the set of solutions of the equation AX = 0, and if X = (x1,x2,x3,x4,x5)T, then the equation AX=0 is equivalent to x1+x3+2x5= 0, x2+x3= 0 and x4+x5= 0 so that x1 = -x3 -2x5, x2 = -x3 and x4 = -x5 so that X = (-x3 -2x5,-x3,x3,-x5,x5)T= x3( -1,-1,1,0,0)T + x5(-2,0,-1,1)T.Thus, a basis for the solution space of Ax = 0 is {(-1,-1,1,0,0)T, (-2,0,-1,1)T}.
1
0
1
0
2
0
1
1
0
0
0
0
0
1
1
0
0
0
0
0
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