Step-by-step Solution:

We want to check the system \[ \begin{cases} x + y + z = 4,\\ 2x - y + 2z = 5,\\ x - 2y - z = -3 \end{cases} \] (Assertion A) and the statement > If \(A\) is a \(3 \times 3\) matrix and \(\mathbf{B}\) is a \(3 \times 1\) non-zero column matrix, then the equation \(A\mathbf{X} = \mathbf{B}\) has a unique solution if \(A\) is non-singular (Reason R). --- Rewriting the system in matrix form \[ A \mathbf{X} = \mathbf{B}, \quad\text{where}\quad A = \begin{pmatrix} 1 & 1 & 1\\ 2 & -1 & 2\\ 1 & -2 & -1 \end{pmatrix}, \quad \mathbf{X} = \begin{pmatrix} x\\[6pt] y\\[6pt] z \end{pmatrix}, \quad \mathbf{B} = \begin{pmatrix} 4\\[6pt] 5\\[6pt] -3 \end{pmatrix}. \] Checking whether \(A\) is non-singular Compute \(\det(A)\). Using the first row for expansion, \[ \det(A) = 1 \cdot \begin{vmatrix} -1 & 2\\ -2 & -1 \end{vmatrix} \;-\; 1 \cdot \begin{vmatrix} 2 & 2\\ 1 & -1 \end{vmatrix} \;+\; 1 \cdot \begin{vmatrix} 2 & -1\\ 1 & -2 \end{vmatrix}. \] 1. \(\displaystyle \begin{vmatrix} -1 & 2\\ -2 & -1 \end{vmatrix} = (-1)(-1)\;-\;(2)(-2) = 1 + 4 = 5. \) 2. \(\displaystyle \begin{vmatrix} 2 & 2\\ 1 & -1 \end{vmatrix} = (2)(-1)\;-\;(2)(1) = -2 - 2 = -4. \) 3. \(\displaystyle \begin{vmatrix} 2 & -1\\ 1 & -2 \end{vmatrix} = (2)(-2)\;-\;(-1)(1) = -4 + 1 = -3. \) Putting these together: \[ \det(A) = 1 \times 5 \;-\; 1 \times (-4) \;+\; 1 \times (-3) = 5 + 4 - 3 = 6. \] Since \(\det(A) = 6 \neq 0\), \(A\) is non-singular. By the standard result (Reason R), a \(3 \times 3\) system \(A\mathbf{X} = \mathbf{B}\) with \(A\) non-singular always has a unique solution. Hence: - Assertion A (the given system has a unique solution) is true. - Reason R (the standard criterion for uniqueness when \(A\) is non-singular) is also true, and it correctly explains why the system in A has a unique solution. Therefore, the correct choice is that both A and R are correct, and R is the correct explanation of A.