Matrices
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SAMPLE ANSWER
=> add -1 times the 1st row to the 3rd row =>
=> add -2 times the 2nd row to the 1st row=>
=> add -4 times the 1st row to the 2nd row =>
=> add -7 times the 1st row to the 3rd row =>
=> multiply the 2nd row by -1/3 =>
=> add 6 times the 2nd row to the 3rd row =>
=> add -2 times the 2nd row to the 1st row =>
3rd row is free for both A and B.
Consider the augmented matrix
We can convert this to reduced echelon form by subtracting twice row 1 from row 3 and
subtracting 3 times row 2 from row 3:
In order for this system to be consistent, it must be the case that
There are no constraints on b1 and b2, the possible right-hand sides of the equation are vectors of the form
for any real numbers b1 and b2. In other words, the column space of A is the plane containing the vectors and . Looking at the reduced echelon form of the matrix, we see that it is of rank 2 and that a particular solution of the given equation is .
=> Switch rows 1 and 3 =>
Divide row 1 by 2, row 2 by 3 =>
Multiply row 2 by 3 and subtract from row 1 =>
=> Switch rows 1 and 3 =>
Divide row 1 by 2, row 2 by 3 =>
Multiply row 2 by 3 and subtract from row 1 =>
=> Switch rows 1 and 3 =>
Divide row 1 by 2, row 2 by 3 =>
Multiply row 2 by 3 and subtract from row 1 =>
Let . Then . Thus .
So .
By the Rank-Nullity Theorem,
and
So it suffices to show that, therefore,
Form the augmented matrix [A b]:
Then subtracting row 1 from rows 2 and 3 and multiplying row 1 by 1 2 yields
Next, subtracting twice row 2 from row 1 and adding row 2 to row 3 gives
This is now in reduced echelon form, so we can answer the question. Notice that the pivot columns are the first and second columns; hence, the column space of A is the span of the first two columns of A, namely and .
Geometrically, this is just the plane containing these two vectors. Returning the the reduced echelon form of the augmented matrix, notice that we must have
x1 = 4 − x3 + 2x4
x2 = −1 − x3 − 2x4
so the special solutions are of the form
for some real numbers x3 and x4. Hence, the nullspace of A consists precisely of such linear combinations.
Finally, all solutions to the equation Ax = b are of the form
where the first term is a particular solution and the latter two terms comprise the special (or homogeneous) solutions.
- Solution:
Solution: Any four vectors in R3 are dependent (since the dimension of R3 is 3), and the vectors v1, v2, v3 are independent because they are columns of an invertible 3 by 3 matrix. Solving the system Ax = 0 is finding the nullspace of the matrix
Special solution here is x4 = 1, x3 = −4, x2 = 1, x1 = 1, so the nullspace consists of all (c1, c2, c3, c4) = x4(1, 1, −4, 1).
- Solution:
Since v4 = v2 − v1, v5 = v3 − v1, and v6 = v3 − v2, there are at most three independent vectors among these: furthermore, applying row reduction to the matrix [v1 v2 v3] gives three pivots, showing that v1, v2, and v3 are independent.
The echelon matrix has only a single pivot, in the second column . The second column is therefore a basis for the column space . (The third column of is equal to four times the second column.) The dimension of is 1 (same as the rank of ).
, all vectors with the last coordinate equal to 0.
, all vectors with the first coordinate equal to 0.
, all vectors with the last coordinate equal to 0.
, all vectors with the first coordinate equal to 0.
transformed into echelon form
=> add -1 times the 1st row to the 2nd row =>
=> add -1 times the 2nd row to the 3rd row =>
=> add -3 times the 2nd row to the 1st row =>
The second and the fourth variables are the pivots. The first, third, and the fifth variables are free variables. The row operation matrix E (R = EA) is
Nullspace: The null space is as follow:
for .
Column space: Since the pivot columns (the second and the fourth) of A span the column space, we have
for
Row space: it is the same as the row space of R as R is obtained by invertible row operations. So
for
Left nullspace: It has a basis given by the rows of E for which the corresponding rows of R are all zero. That is to say, we need to take the last row of E. Thus,
for
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