Use the fact that matrices are A and B are row-equivalent. (a) Find the rank and

Question

Use the fact that matrices are A and B are row-equivalent.

(a) Find the rank and nullity of A.

(b) Find a basis for the nullspace of A.

(c) Find a basis for the row space of A.

(d) Find a basis for the column space of A.

(e) Determine whether the rows of A are linearly independent.

(f) Let the columns of A be denoted by a1, a2, a3, a4 and a5. Detrmine whether each set is linearly independent.

(i) {a1, a2, a4} (ii) {a1, a2, a3} (iii) {a1, a3, a5}

A=[[-2,-5,8,0,-17],[1,3,-5,1,5],[3,11,-19,7,1],[1,7,-13,5,-3]]

B=[[1,0,1,0,1],[0,1,-2,0,3],[0,0,0,1,-5],[0,0,0,0,0]]

Explanation / Answer

The matrix B is the RREF of the matrix A and the matrices A and B are, therefore, row equivalent.

a. The rank of A is the number of non-zero rows in its RREF, i.e. B. Since B has 3 non-zero rows, the rank of A is 3. Now, as per the rank-nullity theorem, the rank(A) + nullity(A) = number of columns of A = 5. Hence, the nullity of A = 5-3 = 2.

b. The null space of A is the set of solutions of the equation AX = 0. If X = (x1,x2,x3,x4,x5 )T , then, since B is the RREF of A, the equation AX = 0 is equivalent to the linear system x1 +x3 +x5 = 0,        x2 -2x3 +3x5 = 0 , and x4 -5x5 = 0. Let x3 = r and x5 = t. Then x1 = -x3 –x5 = -r-t , x2 = 2x3 -3x5 = 2r –t and x4 = 5x5 = 5t. Then X = ( -r-t, 2r-t, r, 5t,t)T = r( -1, 2, 1 ,0,0)T +t ( -1, -1,0,5,1)T . Then a basis for the null space of A is {( -1, 2, 1 ,0,0)T , ( -1, -1,0,5,1)T }.

c. Since B is the RREF of A, a basis for the row space of A is {(-2,-5,8,0,-17),(1,3,-5,1,5),( 3,11,-19,7,1)}.

d. B is the RREF of A, and it is apparent that a3 = a1 -2a2 and a5 = a1 +3a2 -5a4 . Hence A has only 3 linearly independent columns i.e. a1 , a2 and a4. Hence a basis for Col(A) is {(-2,1,3,1)T,(-5,3,11,7)T, (0,1,7,5)T}.

e. Since B is the RREF of A and since B has 1 zero row, hence the rows of A are not linearly independent.

f. From (d) above, it is apparent that the set(i) i.e.{ a1,a2,a4) is linearly independent, set(ii) i.e.{ a1,a2,a3} is linearly dependent( as a3 = a1 -2a2 ) and the set (iii) i.e. {a1 , a3 , a5} is linearly independent.

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