Solution

Let first, second and third numbers be x, y and z, respectively. Then, we have \(x + y + z = 6\) \(y + 3z = 11\) \(x + z = 2y\) or \(x - 2y + z = 0\) This system can be written as \(AX = B\), where \(A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1 \end{bmatrix}\), \(X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\) and \(B = \begin{bmatrix} 6 \\ 11 \\ 0 \end{bmatrix}\). Here \(|A| = 1(1+6) - 1(0-3) + 1(0-1) = 7+3-1 = 9 \neq 0\). Now we find adj A. \(A_{11} = 7, A_{12} = 3, A_{13} = -1\) \(A_{21} = -3, A_{22} = 0, A_{23} = 3\) \(A_{31} = 2, A_{32} = -3, A_{33} = 1\) Hence, \(adj A = \begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix}\). Thus, \(A^{-1} = \frac{1}{|A|}adj(A) = \frac{1}{9}\begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix}\). Since \(X = A^{-1}B\) \(X = \frac{1}{9}\begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix}\begin{bmatrix} 6 \\ 11 \\ 0 \end{bmatrix}\) \(\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{9}\begin{bmatrix} 42-33+0 \\ 18+0+0 \\ -6+33+0 \end{bmatrix} = \frac{1}{9}\begin{bmatrix} 9 \\ 18 \\ 27 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\). Thus, \(x=1, y=2, z=3\).