Suppose the vectors $$x_1, x_2$$ and $$x_3$$ are the solutions of the system of linear equations, $$Ax = b$$ when the vector $$b$$ on the right side is equal to $$b_1, b_2$$ and $$b_3$$ respectively. If $$x_1 = \begin{bmatrix}1\\1\\1\end{bmatrix}$$, $$x_2 = \begin{bmatrix}0\\2\\1\end{bmatrix}$$, $$x_3 = \begin{bmatrix}0\\0\\1\end{bmatrix}$$; $$b_1 = \begin{bmatrix}1\\0\\0\end{bmatrix}$$, $$b_2 = \begin{bmatrix}0\\2\\0\end{bmatrix}$$, $$b_3 = \begin{bmatrix}0\\0\\2\end{bmatrix}$$, then the determinant of $$A$$ is equal to

We are told that the same unknown matrix $$A$$ satisfies three different systems

$$A\,x_1 = b_1,\qquad A\,x_2 = b_2,\qquad A\,x_3 = b_3.$$

We collect the three solution vectors as the columns of one matrix

$$X = \begin{bmatrix} 1 & 0 & 0\\[2pt] 1 & 2 & 0\\[2pt] 1 & 1 & 1 \end{bmatrix},$$

and the corresponding right-hand sides as the columns of another matrix

$$B = \begin{bmatrix} 1 & 0 & 0\\[2pt] 0 & 2 & 0\\[2pt] 0 & 0 & 2 \end{bmatrix}.$$

The three given relations combine into the single matrix equation

$$A\,X = B.$$

If a square matrix multiplies another square matrix on the right and gives a square result, then, provided the middle matrix is invertible, we may isolate the left matrix. The formula we use is

$$\text{If } A\,X = B \text{ and } X \text{ is invertible, then } A = B\,X^{-1}.$$

The determinant of a product equals the product of the determinants, so

$$\det(A) = \dfrac{\det(B)}{\det(X)}.$$

Now we evaluate the two determinants separately.

Determinant of $$B$$

The matrix $$B$$ is diagonal with entries $$1, 2, 2$$ along the diagonal, hence

$$\det(B) = 1 \times 2 \times 2 = 4.$$

Determinant of $$X$$

For the matrix

$$X = \begin{bmatrix} 1 & 0 & 0\\ 1 & 2 & 0\\ 1 & 1 & 1 \end{bmatrix},$$

we expand about the first row:

$$\det(X) = 1 \times \det\begin{bmatrix} 2 & 0\\ 1 & 1 \end{bmatrix} \;-\; 0 \times (\text{something}) \;+\; 0 \times (\text{something}).$$

The remaining $$2 \times 2$$ determinant is

$$\det\begin{bmatrix} 2 & 0\\ 1 & 1 \end{bmatrix} = 2\cdot1 - 0\cdot1 = 2.$$

So

$$\det(X) = 1 \times 2 = 2.$$

Putting the values together

Substituting into the earlier ratio, we get

$$\det(A) = \dfrac{\det(B)}{\det(X)} = \dfrac{4}{2} = 2.$$

Hence, the correct answer is Option B.