Question 1

The system of linear equations:
2x + 3y + z = 5
4x + 6y + 2z = 10
3x + 5y + z = 8

has:

Accepted Answer

Setup: We use the rank method to determine consistency and uniqueness of solutions for a system of linear equations.

Solution:
Write the augmented matrix [A|B]:
⎡2  3  1 | 5⎤
⎢4  6  2 | 10⎥
⎣3  5  1 | 8⎦

Row operations:
R₂ → R₂ - 2R₁:
⎡2  3  1 | 5⎤
⎢0  0  0 | 0⎥
⎣3  5  1 | 8⎦

R₃ → R₃ - (3/2)R₁:
⎡2  3  1 | 5⎤
⎢0  0  0 | 0⎥
⎣0  1/2  -1/2 | 1/2⎦

Swap R₂ and R₃:
⎡2  3  1 | 5⎤
⎢0  1/2  -1/2 | 1/2⎥
⎣0  0  0 | 0⎦

Rank of coefficient matrix A = 2
Rank of augmented matrix [A|B] = 2
n (number of variables) = 3

Since rank(A) = rank([A|B]) < n, the system is consistent but has infinitely many solutions (one free variable).

Trap explanation: Students often confuse the condition for unique solutions (rank = n) with infinitely many solutions (rank < n). Some may incorrectly conclude no solution exists if they miscount the rank.

Question 2

If the system of equations:
px + qy = 1
rx + sy = 1

has no solution, then which of the following must be true?

Accepted Answer

A system Ax = b has no solution when rank(A) < rank([A|b]), meaning the coefficient matrix is singular but the augmented matrix has higher rank.

Step 1: Write the coefficient matrix and augmented matrix.
A = [[p, q], [r, s]], [A|b] = [[p, q, 1], [r, s, 1]]

Step 2: Condition for no solution.
No solution exists when:
- det(A) = 0 (the coefficient matrix is singular), AND
- The rows of A are proportional, but the augmented rows are NOT proportional.

Step 3: Analyze the proportionality condition.
If det(A) = ps - qr = 0, then the rows [p, q] and [r, s] are proportional.
This means [r, s] = k[p, q] for some scalar k ≠ 0.

Step 4: Check consistency.
For no solution, we need the augmented rows to NOT be proportional.
If [r, s] = k[p, q], then for the augmented rows to be proportional, we would need [r, s, 1] = k[p, q, 1], which implies 1 = k(1), so k = 1.
But if k = 1, then [r, s] = [p, q], meaning the two equations are identical, giving infinitely many solutions.

For no solution, we need [r, s] = k[p, q] with k ≠ 1.
This means ps - qr = 0 (so det(A) = 0) AND the constant terms are NOT equal (1 ≠ 1 is false, but the proportionality breaks at the constant).

Actually, let's reconsider: If [r, s] = k[p, q] and both constant terms are 1, then:
- If k = 1: equations are identical → infinitely many solutions.
- If k ≠ 1: the augmented rows are NOT proportional (because 1 ≠ k·1 when k ≠ 1) → no solution.

Step 5: Conclusion.
No solution exists when ps - qr = 0 AND the rows of the coefficient matrix are proportional but NOT identical.
Equivalently: ps - qr = 0 AND NOT (p = r AND q = s).
Or more directly: ps - qr = 0 AND (p ≠ r OR q ≠ s).

Question 3

The system of linear equations:
2x + 3y = 5
4x + 6y = 10

has:

Accepted Answer

We analyze this system using the coefficient matrix and augmented matrix.

Coefficient matrix A = [[2, 3], [4, 6]]
Augmented matrix [A|B] = [[2, 3, 5], [4, 6, 10]]

Step 1: Calculate det(A) = 2(6) - 3(4) = 12 - 12 = 0

Step 2: Since det(A) = 0, the matrix is singular. We check consistency by row reduction.
R₂ → R₂ - 2R₁:
[[2, 3, 5], [0, 0, 0]]

Step 3: The second row becomes [0, 0, 0], which is consistent (0 = 0 is true).

Step 4: We have rank(A) = rank([A|B]) = 1 < number of unknowns (2).

By the rank theorem: When rank(A) = rank([A|B]) < n (number of variables), the system has infinitely many solutions.

Verification: The second equation 4x + 6y = 10 is exactly 2 times the first equation 2x + 3y = 5, so they represent the same line. Every point on the line 2x + 3y = 5 is a solution.

Question 4

Using Cramer's rule, find the value of x in the system:
x + 2y = 5
3x - y = 4

Accepted Answer

Cramer's rule states: For Ax = b, if det(A) ≠ 0, then x = det(A_x) / det(A), where A_x is the matrix formed by replacing the first column of A with the constant vector b.

Step 1: Identify the coefficient matrix and constant vector.
A = [[1, 2], [3, -1]], b = [5, 4]

Step 2: Calculate det(A).
det(A) = 1(-1) - 2(3) = -1 - 6 = -7

Step 3: Form A_x by replacing the first column of A with b.
A_x = [[5, 2], [4, -1]]

Step 4: Calculate det(A_x).
det(A_x) = 5(-1) - 2(4) = -5 - 8 = -13

Step 5: Apply Cramer's rule.
x = det(A_x) / det(A) = -13 / -7 = 13/7

Verification: Substitute x = 13/7 into first equation: 13/7 + 2y = 5 ⟹ 2y = 35/7 - 13/7 = 22/7 ⟹ y = 11/7.
Check second equation: 3(13/7) - 11/7 = 39/7 - 11/7 = 28/7 = 4 ✓

Question 5

For the system of equations:
ax + by = 1
cx + dy = 1

to have a unique solution, which condition must be satisfied?

Accepted Answer

A system Ax = b has a unique solution if and only if the coefficient matrix A is invertible, which occurs when det(A) ≠ 0.

Step 1: Write the coefficient matrix.
A = [[a, b], [c, d]]

Step 2: Calculate the determinant.
det(A) = ad - bc

Step 3: Apply the invertibility criterion.
A is invertible ⟺ det(A) ≠ 0 ⟺ ad - bc ≠ 0

Step 4: Conclusion.
For a unique solution to exist, we require ad - bc ≠ 0.

Note: If ad - bc = 0, the system either has no solution or infinitely many solutions (depending on whether the equations are consistent). The constant vector [1, 1] does not affect the uniqueness condition—only the coefficient matrix determinant matters.

Question 6

The system of equations:
2x + y + z = 3
x + 2y + z = 4
x + y + 2z = 5

has the solution (x, y, z) equal to:

Accepted Answer

We solve this 3×3 system using the matrix method or Gaussian elimination.

Step 1: Write the augmented matrix.
[A|b] = [[2, 1, 1, 3], [1, 2, 1, 4], [1, 1, 2, 5]]

Step 2: Perform row operations to achieve row echelon form.
R₁ ↔ R₂ (swap rows 1 and 2 for easier computation):
[[1, 2, 1, 4], [2, 1, 1, 3], [1, 1, 2, 5]]

R₂ → R₂ - 2R₁:
[[1, 2, 1, 4], [0, -3, -1, -5], [1, 1, 2, 5]]

R₃ → R₃ - R₁:
[[1, 2, 1, 4], [0, -3, -1, -5], [0, -1, 1, 1]]

R₂ ↔ R₃ (swap for convenience):
[[1, 2, 1, 4], [0, -1, 1, 1], [0, -3, -1, -5]]

R₃ → R₃ - 3R₂:
[[1, 2, 1, 4], [0, -1, 1, 1], [0, 0, -4, -8]]

Step 3: Back substitution.
From row 3: -4z = -8 ⟹ z = 2

From row 2: -y + z = 1 ⟹ -y + 2 = 1 ⟹ y = 1

From row 1: x + 2y + z = 4 ⟹ x + 2(1) + 2 = 4 ⟹ x = 0

Step 4: Verification.
Equation 1: 2(0) + 1 + 2 = 3 ✓
Equation 2: 0 + 2(1) + 2 = 4 ✓
Equation 3: 0 + 1 + 2(2) = 5 ✓

Question 7

The system of linear equations
2x + 3y = 5
4x + 6y = 10
has how many solutions?

Accepted Answer

To determine the number of solutions, we write the system in matrix form AX = B, where A = [[2, 3], [4, 6]], X = [[x], [y]], B = [[5], [10]].

Step 1: Calculate det(A) = 2(6) − 3(4) = 12 − 12 = 0.

Step 2: Since det(A) = 0, the system is either inconsistent (no solution) or dependent (infinitely many solutions).

Step 3: Check consistency. Notice that the second equation is exactly 2 times the first equation: 2(2x + 3y) = 2(5) gives 4x + 6y = 10. Both equations represent the same line.

Step 4: Since the equations are identical (one is a scalar multiple of the other), the system is consistent and dependent.

Conclusion: The system has infinitely many solutions (all points on the line 2x + 3y = 5).

Question 8

Using Cramer's rule, find the value of x in the system:
x + 2y = 5
3x − y = 4

Accepted Answer

Cramer's rule states: x = det(A_x) / det(A), where A_x is the matrix formed by replacing the first column of A with the constant vector B.

Step 1: Write coefficient matrix A = [[1, 2], [3, −1]] and constant vector B = [[5], [4]].

Step 2: Calculate det(A) = 1(−1) − 2(3) = −1 − 6 = −7.

Step 3: Form A_x by replacing the first column of A with B:
A_x = [[5, 2], [4, −1]].

Step 4: Calculate det(A_x) = 5(−1) − 2(4) = −5 − 8 = −13.

Step 5: Apply Cramer's rule: x = det(A_x) / det(A) = −13 / −7 = 13/7.

Verification: Substitute x = 13/7 into first equation: 13/7 + 2y = 5 → 2y = 35/7 − 13/7 = 22/7 → y = 11/7. Check in second equation: 3(13/7) − 11/7 = 39/7 − 11/7 = 28/7 = 4 ✓

Question 9

For the system of equations
2x + y + z = 3
x − y + 2z = 5
3x + 2y − z = 2
the coefficient matrix A has determinant det(A) = −15. Using the matrix method, which statement is correct?

Accepted Answer

Step 1: The coefficient matrix is A = [[2, 1, 1], [1, −1, 2], [3, 2, −1]].

Step 2: We are given that det(A) = −15 ≠ 0.

Step 3: By the fundamental theorem for systems of linear equations: if det(A) ≠ 0, then the system AX = B has a unique solution given by X = A⁻¹B.

Step 4: The solution can be found using either:
- Matrix inversion method: X = A⁻¹B
- Cramer's rule: x = det(A_x)/det(A), y = det(A_y)/det(A), z = det(A_z)/det(A)

Step 5: Since det(A) = −15 ≠ 0, the system is consistent and has exactly one unique solution.

Conclusion: The system has a unique solution that can be computed using A⁻¹ or Cramer's rule.

Question 10

The system of linear equations
x + 2y = 3
2x + 4y = 5
has which of the following properties?

Accepted Answer

Step 1: Write the coefficient matrix A = [[1, 2], [2, 4]] and augmented matrix [A|B] = [[1, 2, 3], [2, 4, 5]].

Step 2: Calculate det(A) = 1(4) − 2(2) = 4 − 4 = 0. The matrix is singular.

Step 3: Since det(A) = 0, the system either has no solution or infinitely many solutions. We must check consistency.

Step 4: Examine the equations. The second equation's left side is 2x + 4y = 2(x + 2y), which is 2 times the left side of the first equation.

Step 5: However, the right side: 2 times the first equation's right side would be 2(3) = 6, but the second equation has 5 on the right side.

Step 6: This means the coefficients are proportional (second row = 2 × first row) but the constants are NOT proportional (5 ≠ 2 × 3). This is an inconsistent system.

Conclusion: The system has no solution (the lines are parallel but distinct).

Question 11

If the system of equations
ax + by = 1
cx + dy = 2
has the unique solution x = 1, y = 1, then which of the following must be true?

Accepted Answer

Step 1: Given that x = 1, y = 1 is the unique solution, substitute into both equations:
a(1) + b(1) = 1 → a + b = 1
c(1) + d(1) = 2 → c + d = 2

Step 2: For the solution to be unique, the coefficient matrix A = [[a, b], [c, d]] must be invertible, which requires det(A) ≠ 0.

Step 3: Calculate det(A) = ad − bc.

Step 4: From the constraints a + b = 1 and c + d = 2, we know:
- a = 1 − b
- c = 2 − d

Step 5: Substitute into det(A):
det(A) = (1 − b)d − b(2 − d) = d − bd − 2b + bd = d − 2b

Step 6: For uniqueness, we need det(A) ≠ 0, which means d − 2b ≠ 0, or equivalently d ≠ 2b.

Step 7: However, the question asks what MUST be true. From the given information alone (that x = 1, y = 1 is the unique solution), we can definitively conclude:
- a + b = 1 (from substituting into first equation)
- c + d = 2 (from substituting into second equation)
- ad − bc ≠ 0 (for uniqueness)

The most direct statement that MUST be true is: a + b = 1 and c + d = 2, and ad ≠ bc.

Question 12

The system of linear equations:
2x + 3y = 5
4x + 6y = 10
has how many solutions?

Accepted Answer

To determine the number of solutions, we examine the coefficient matrix A and augmented matrix [A|B].

Coefficient matrix: A = [[2, 3], [4, 6]]
Augmented matrix: [A|B] = [[2, 3, 5], [4, 6, 10]]

Step 1: Calculate rank(A).
Row reduce A: R₂ → R₂ - 2R₁ gives [[2, 3], [0, 0]]
rank(A) = 1

Step 2: Calculate rank([A|B]).
Row reduce [A|B]: R₂ → R₂ - 2R₁ gives [[2, 3, 5], [0, 0, 0]]
rank([A|B]) = 1

Step 3: Apply consistency theorem.
Since rank(A) = rank([A|B]) = 1 < n = 2 (number of variables), the system is consistent with infinitely many solutions.

Verification: The second equation 4x + 6y = 10 is exactly 2 times the first equation 2x + 3y = 5, so they represent the same line. All points on the line 2x + 3y = 5 satisfy both equations.

Question 13

The system of equations:
ax + by = 1
cx + dy = 1
has a unique solution if and only if:

Accepted Answer

A system of linear equations Ax = b has a unique solution if and only if the coefficient matrix A is invertible, which occurs when det(A) ≠ 0.

Step 1: Identify the coefficient matrix.
A = [[a, b], [c, d]]

Step 2: Calculate the determinant.
det(A) = ad - bc

Step 3: Apply the uniqueness condition.
For a unique solution to exist, the system must be consistent and the coefficient matrix must be non-singular.
This requires: det(A) ≠ 0, which means ad - bc ≠ 0.

Step 4: Verify the logic.
- If ad - bc = 0, then det(A) = 0, so A is singular (not invertible). The system either has no solution or infinitely many solutions.
- If ad - bc ≠ 0, then det(A) ≠ 0, so A is invertible. The system has exactly one solution x = A⁻¹b.

Therefore, the condition for a unique solution is ad - bc ≠ 0.

Question 14

If the system:
2x + 3y + z = 6
4x + 6y + 2z = 12
x + y + z = 5
has a solution, what is the rank of the augmented matrix [A|B]?

Accepted Answer

Step 1: Write the augmented matrix [A|B].
[A|B] = [[2, 3, 1, 6], [4, 6, 2, 12], [1, 1, 1, 5]]

Step 2: Perform row reduction.
R₂ → R₂ - 2R₁:
[[2, 3, 1, 6], [0, 0, 0, 0], [1, 1, 1, 5]]

Swap R₂ and R₃:
[[2, 3, 1, 6], [1, 1, 1, 5], [0, 0, 0, 0]]

R₁ → R₁ - 2R₂:
[[0, 1, -1, -4], [1, 1, 1, 5], [0, 0, 0, 0]]

Swap R₁ and R₂:
[[1, 1, 1, 5], [0, 1, -1, -4], [0, 0, 0, 0]]

Step 3: Identify pivot rows.
The matrix is now in row echelon form. There are 2 non-zero rows.
rank([A|B]) = 2

Step 4: Verify consistency.
The last row [0, 0, 0, 0] indicates no contradiction. The system is consistent (has a solution), confirming that rank(A) = rank([A|B]) = 2.

Question 15

For what value of k does the system:
x + 2y = 3
2x + 4y = k
have infinitely many solutions?

Accepted Answer

For a system to have infinitely many solutions, the two equations must represent the same line. This occurs when one equation is a scalar multiple of the other.

Step 1: Examine the coefficients.
First equation: x + 2y = 3
Second equation: 2x + 4y = k

Step 2: Check if the coefficient rows are proportional.
The second row [2, 4] is exactly 2 times the first row [1, 2].
Coefficient ratio: 2/1 = 4/2 = 2

Step 3: Apply the proportionality condition to constants.
For the equations to represent the same line, the constant terms must also be in the same ratio.
If the coefficient ratio is 2, then: k/3 = 2
Therefore: k = 6

Step 4: Verify.
When k = 6, the second equation becomes 2x + 4y = 6, which is exactly 2·(x + 2y = 3).
Both equations represent the same line, so infinitely many solutions exist.

Alternative verification using determinant:
det(A) = 1·4 - 2·2 = 4 - 4 = 0 for all k.
For infinitely many solutions, rank(A) = rank([A|B]).
This requires the augmented matrix rows to be proportional: [2, 4, k] = 2·[1, 2, 3]
So k = 2·3 = 6.

Question 16

Consider the matrix equation AX = B, where:
A = [1  2]
    [3  4]

B = [5]
    [11]

If X = [x], find x + y using the matrix method (inverse method).
      [y]

Which of the following is correct?

Accepted Answer

Setup: To solve AX = B, we compute X = A⁻¹B, where A⁻¹ is the inverse of A.

Step 1: Calculate det(A).
det(A) = 1(4) − 2(3) = 4 − 6 = −2

Step 2: Form the adjugate matrix adj(A).
For a 2×2 matrix [a b; c d], adj(A) = [d −b; −c a].
adj(A) = [4  −2]
         [−3  1]

Step 3: Calculate A⁻¹.
A⁻¹ = (1/det(A)) × adj(A) = (1/−2) × [4  −2]
                                      [−3  1]

A⁻¹ = [−2   1]
      [3/2  −1/2]

Step 4: Compute X = A⁻¹B.
X = [−2   1  ] × [5 ]
    [3/2  −1/2]   [11]

X = [−2(5) + 1(11)  ]
    [3/2(5) − 1/2(11)]

X = [−10 + 11]
    [15/2 − 11/2]

X = [1]
    [2]

Step 5: Find x + y.
x = 1, y = 2
x + y = 1 + 2 = 3

Verification: AX = [1 2][1] = [1 + 4] = [5] ✓
                   [3 4][2]   [3 + 8]   [11]

Trap Explanation: Students frequently make errors in computing the adjugate matrix (sign errors or transposition confusion), or they incorrectly compute A⁻¹ by inverting the determinant sign or forgetting to multiply by 1/det(A).

Question 17

If the system of equations
ax + by = 1
cx + dy = 2

has a unique solution, which condition must hold?

Accepted Answer

Setup: Testing the determinant condition for uniqueness of solutions in a 2×2 linear system.

Solution:
The system can be written in matrix form:
⎡ a  b ⎤ ⎡ x ⎤   ⎡ 1 ⎤
⎢ c  d ⎥ ⎢ y ⎥ = ⎢ 2 ⎥
⎣     ⎦ ⎣   ⎦   ⎣   ⎦

Let A = ⎡ a  b ⎤
        ⎢ c  d ⎥
        ⎣     ⎦

For a unique solution to exist, the coefficient matrix A must be invertible.
A is invertible if and only if det(A) ≠ 0.

det(A) = ad - bc

Therefore, the condition is: ad - bc ≠ 0, or equivalently, ad ≠ bc.

Question 18

Using Cramer's rule, solve for y in the system:
3x + 2y = 7
5x - y = 4

What is the value of y?

Accepted Answer

Setup: Cramer's rule application for solving a 2×2 system. By Cramer's rule, y = det(Aᵧ) / det(A), where Aᵧ is the matrix with the y-column replaced by the constant column.

Solution:
Step 1: Find det(A) where A = ⎡ 3  2 ⎤
                                ⎢ 5 -1 ⎥
                                ⎣     ⎦
det(A) = 3(-1) - 2(5) = -3 - 10 = -13

Step 2: Find det(Aᵧ) where Aᵧ = ⎡ 3  7 ⎤ (replace y-column with constants)
                                  ⎢ 5  4 ⎥
                                  ⎣     ⎦
det(Aᵧ) = 3(4) - 7(5) = 12 - 35 = -23

Step 3: Apply Cramer's rule:
y = det(Aᵧ) / det(A) = -23 / -13 = 23/13

Verification: Substitute x and y back into original equations to confirm (optional but recommended).

Question 19

The system of equations:
2x − y + z = 3
x + 2y − z = 1
3x + y + αz = β

has infinitely many solutions. Which of the following pairs (α, β) is correct?

Accepted Answer

Setup: A system of three equations in three unknowns has infinitely many solutions when the coefficient matrix and the augmented matrix have the same rank, and this rank is less than 3 (i.e., the system is underdetermined and consistent).

Step 1: Write the augmented matrix [A|B]:
[2  −1   1 | 3]
[1   2  −1 | 1]
[3   1   α | β]

Step 2: Perform row reduction. Use R₁ ↔ R₂ to simplify:
[1   2  −1 | 1]
[2  −1   1 | 3]
[3   1   α | β]

Step 3: Eliminate below the first pivot:
R₂ → R₂ − 2R₁:
[1   2  −1 | 1]
[0  −5   3 | 1]
[3   1   α | β]

R₃ → R₃ − 3R₁:
[1   2  −1 | 1]
[0  −5   3 | 1]
[0  −5  α+3 | β−3]

Step 4: For infinitely many solutions, the third row must be a linear combination of the first two rows. Specifically, R₃ should be proportional to R₂:
R₃ = R₂ ⟹ −5 = −5 (✓), 3 = α + 3, and 1 = β − 3

From 3 = α + 3: α = 0
From 1 = β − 3: β = 4

Step 5: Verify consistency:
With α = 0 and β = 4, the third equation becomes:
3x + y = 4

Check if this is a linear combination of the first two equations:
Equation 1: 2x − y + z = 3
Equation 2: x + 2y − z = 1

Equation 1 + Equation 2: 3x + y = 4 ✓

This confirms that Equation 3 is the sum of Equations 1 and 2, so the system is consistent with infinitely many solutions.

Question 20

For what value of k does the system
x + 2y + 3z = 4
2x + 4y + 6z = 8
3x + 6y + 9z = k

have infinitely many solutions?

Accepted Answer

Setup: Testing consistency and linear dependence in a 3×3 system. A system has infinitely many solutions when rank(A) = rank([A|b]) < n (number of variables).

Solution:
Observe the coefficient matrix:
Row 1: [1, 2, 3]
Row 2: [2, 4, 6] = 2 × Row 1
Row 3: [3, 6, 9] = 3 × Row 1

All three rows are scalar multiples of the first row, so rank(A) = 1.

For the augmented matrix [A|b]:
Row 1: [1, 2, 3 | 4]
Row 2: [2, 4, 6 | 8] = 2 × Row 1 ✓ (consistent)
Row 3: [3, 6, 9 | k]

For Row 3 to be consistent with Row 1, we need:
[3, 6, 9 | k] = 3 × [1, 2, 3 | 4]
This gives: [3, 6, 9 | 12]

Therefore, k = 12.

When k = 12:
- rank(A) = 1
- rank([A|b]) = 1
- rank < n (1 < 3), so infinitely many solutions exist (a 2-parameter family).

When k ≠ 12:
- rank(A) = 1
- rank([A|b]) = 2
- rank(A) ≠ rank([A|b]), so the system is inconsistent.

Question 21

The system of equations
2x + 3y - z = 1
4x + 6y - 2z = 2
x + y + z = 3

can be written in matrix form as AX = B. What is the rank of the augmented matrix [A|B]?

Accepted Answer

Setup: Computing the rank of an augmented matrix by row reduction to identify consistency and the dimension of the solution space.

Solution:
Augmented matrix [A|B]:
⎡ 2  3  -1 | 1 ⎤
⎢ 4  6  -2 | 2 ⎥
⎣ 1  1   1 | 3 ⎦

Row operations:
R₂ → R₂ - 2R₁:
⎡ 2  3  -1 | 1 ⎤
⎢ 0  0   0 | 0 ⎥
⎣ 1  1   1 | 3 ⎦

R₃ → R₃ - (1/2)R₁:
⎡ 2  3  -1 | 1 ⎤
⎢ 0  0   0 | 0 ⎥
⎣ 0 -1/2 3/2 | 5/2 ⎦

Swap R₂ and R₃:
⎡ 2  3  -1 | 1 ⎤
⎢ 0 -1/2 3/2 | 5/2 ⎥
⎣ 0  0   0 | 0 ⎦

The matrix is now in row echelon form. Count non-zero rows: 2.

Therefore, rank([A|B]) = 2.

Note: rank(A) = 2 as well (same non-zero rows in coefficient part), so the system is consistent. Since rank < 3 (number of variables), infinitely many solutions exist.

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