For the system of equations 2x + z = 3 x + y = 2 y + z = 2 let A be t…

The correct answer is Option (3) → (A), (B) and (D) only

The coefficient matrix and its determinant (expanding along the first row):

    \( \displaystyle A = \begin{bmatrix}2 & 0 & 1\\ 1 & 1 & 0\\ 0 & 1 & 1\end{bmatrix} \)

    \( |A| = 2(1 - 0) - 0 + 1(1 - 0) = 3 \)

Since \(|A| \ne 0\), the system has exactly one solution. The cofactors give

    \( \displaystyle \mathrm{adj}\,A = \begin{bmatrix}1 & 1 & -1\\ -1 & 2 & 1\\ 1 & -2 & 2\end{bmatrix} \)

    \( \displaystyle X = \frac{1}{3}\begin{bmatrix}1 & 1 & -1\\ -1 & 2 & 1\\ 1 & -2 & 2\end{bmatrix}\begin{bmatrix}3\\ 2\\ 2\end{bmatrix} = \frac{1}{3}\begin{bmatrix}3\\ 3\\ 3\end{bmatrix} = \begin{bmatrix}1\\ 1\\ 1\end{bmatrix} \)

For order 3, \(|\mathrm{adj}\,A| = |A|^{2} = 9\).

So \(|A| = 3\), the only solution is x = 1, y = 1, z = 1 (with x + y + z = 3), and \(|\mathrm{adj}\,A| = 9\).

Checking the statements

(A) A is non-singular. — Correct. \(|A| = 3 \ne 0\).

(B) The system has the unique solution \(x = 1,\ y = 1,\ z = 1\). — Correct. \(|A| \ne 0\), and \(2 + 1 = 3\), \(1 + 1 = 2\), \(1 + 1 = 2\).

(C) For the solution of the system, \(x + y + z = 4\). — Incorrect. \(1 + 1 + 1 = 3\).

(D) \(|\mathrm{adj}\,A| = 9\). — Correct. \(|A|^{2} = 3^{2} = 9\).