Matrices & Determinants Question - None

Question

If the system of linear equations
$x + ay + z = 3$
$x + 2y + 2z = 6$
$x + 5y + 3z = b$
has no solution, then

Accepted Answer

Step 1: Write the augmented matrix and row reduce:

$$\begin{bmatrix} 1 & a & 1 & | & 3 \ 1 & 2 & 2 & | & 6 \ 1 & 5 & 3 & | & b \end{bmatrix}$$

Step 2: Perform R₂ - R₁ and R₃ - R₁:

$$\begin{bmatrix} 1 & a & 1 & | & 3 \ 0 & 2-a & 1 & | & 3 \ 0 & 5-a & 2 & | & b-3 \end{bmatrix}$$

Step 3: For the system to have no solution, we need rank(A) < rank(A|B). Multiply R₂ by (5-a) and R₃ by (2-a):

From R₃ - (rows must be dependent): If 2-a ≠ 0, perform R₃ - 2R₂/(2-a)·(5-a)

The third row becomes: 0 = k(b-3) where k ≠ 0

Step 4: For no solution: The coefficients must vanish but RHS must not.

When a = 3: Row 2 becomes [0, -1, 1 | 3] and Row 3 becomes [0, 2, 2 | b-3]

Then 2R₂ + R₃ gives: [0, 0, 4 | 6+b-3] = [0, 0, 4 | b+3]

For no solution: b+3 ≠ 0, so a = 3 and b ≠ -3

∴ Answer: B

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