To solve this problem, we need to follow these steps:

(a) To find the condition on k ∈ R such that Ax = b, x ∈ R4 is solvable, we need to form the augmented matrix [A|b] and then reduce it to row echelon form.

The augmented matrix [A|b] is:

[2 1 6 1 | 1]
[3 1 7 2 | 1]
[4 1 8 3 | k]
[5 1 9 3 | 4]

We can start by subtracting the first row from the others to eliminate the first column in the second, third and fourth rows:

[2 1 6 1 | 1]
[1 0 1 1 | 0]
[2 0 2 2 | k-1]
[3 0 3 2 | 3]

Next, we can subtract 2 times the second row from the third row and 3 times the second row from the fourth row:

[2 1 6 1 | 1]
[1 0 1 1 | 0]
[0 0 0 0 | k-1]
[1 0 1 0 | 3]

Finally, we can subtract the second row from the fourth row:

[2 1 6 1 | 1]
[1 0 1 1 | 0]
[0 0 0 0 | k-1]
[0 0 0 -1 | 3]

From the third row, we can see that the system is solvable if and only if k = 1.

(b) When k = 1, the system is solvable and we can find the solutions by back substitution. From the fourth row, we have -x4 = 3, so x4 = -3. Substituting x4 into the second row, we get x2 + x3 - 3 = 0, so x2 = 3 - x3. Substituting x2 and x4 into the first row, we get 2x1 + (3 - x3) + 6x3 - 3 = 1, so x1 = 1 - 3 + 3/2 = 1/2. Therefore, the solutions are x = [1/2, 3 - x3, x3, -3] for any real number x3.

Question

To solve this problem, we need to follow these steps:

(a) To find the condition on k ∈ R such that Ax = b, x ∈ R4 is solvable, we need to form the augmented matrix [A|b] and then reduce it to row echelon form.

The augmented matrix [A|b] is:

[2 1 6 1 | 1]
    [3 1 7 2 | 1]
    [4 1 8 3 | k]
    [5 1 9 3 | 4]

We can start by subtracting the first row from the others to eliminate the first column in the second, third and fourth rows:

[2 1 6 1 | 1]
    [1 0 1 1 | 0]
    [2 0 2 2 | k-1]
    [3 0 3 2 | 3]

Next, we can subtract 2 times the second row from the third row and 3 times the second row from the fourth row:

[2 1 6 1 | 1]
    [1 0 1 1 | 0]
    [0 0 0 0 | k-1]
    [1 0 1 0 | 3]

Finally, we can subtract the second row from the fourth row:

[2 1 6 1 | 1]
    [1 0 1 1 | 0]
    [0 0 0 0 | k-1]
    [0 0 0 -1 | 3]

From the third row, we can see that the system is solvable if and only if k = 1.

(b) When k = 1, the system is solvable and we can find the solutions by back substitution. From the fourth row, we have -x4 = 3, so x4 = -3. Substituting x4 into the second row, we get x2 + x3 - 3 = 0, so x2 = 3 - x3. Substituting x2 and x4 into the first row, we get 2x1 + (3 - x3) + 6x3 - 3 = 1, so x1 = 1 - 3 + 3/2 = 1/2. Therefore, the solutions are x = [1/2, 3 - x3, x3, -3] for any real number x3.

Knowee AI · Accepted Answer