Question

Sol. [Key Concept: Recover A's columns from the given matrix–vector products, form $A^{2} + A$, then $det(adj\ M) = (det\ M)^{n - 1}$.]

Let $A = \begin{pmatrix} \alpha_{1} & \alpha_{2} & \alpha_{3} \\ \beta_{1} & \beta_{2} & \beta_{3} \\ \gamma_{1} & \gamma_{2} & \gamma_{3} \end{pmatrix}$

Now $A = \begin{pmatrix} \alpha_{1} & \beta_{1} & \gamma_{1} \\ \alpha_{2} & \beta_{2} & \gamma_{2} \\ \alpha_{3} & \beta_{3} & \gamma_{3} \end{pmatrix}\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}$

$$\begin{array}{r} \Rightarrow \begin{pmatrix} \gamma_{1} \\ \gamma_{2} \\ \gamma_{3} \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}\#(1) \end{array}$$

Now $\begin{pmatrix} \alpha_{1} & \alpha_{2} & \alpha_{3} \\ \beta_{1} & \beta_{2} & \beta_{3} \\ \gamma_{1} & \gamma_{2} & \gamma_{3} \end{pmatrix}\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}$

$$\begin{array}{r} \Rightarrow \begin{pmatrix} \alpha_{3} \\ \beta_{3} \\ \gamma_{3} \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}\#(2) \end{array}$$

$${A^{T}\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} - A^{T}\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \\ 2 \end{pmatrix} - \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix} }{\Rightarrow A^{T}\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}}$$

$$\begin{array}{r} \Rightarrow \begin{pmatrix} \alpha_{1} \\ \alpha_{2} \\ \alpha_{3} \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\#(3) \end{array}$$

Now $A\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} - A\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \\ 4 \end{pmatrix} - \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}$

$$\Rightarrow A\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}$$

$$\begin{array}{r} \Rightarrow \begin{pmatrix} \alpha_{1} \\ \beta_{1} \\ \gamma_{1} \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}\#(4) \end{array}$$

$${A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & \beta_{2} & 3 \\ 3 & 1 & 1 \end{pmatrix} }{|A| = 2\left( \beta_{2} - 3 \right) - 1(1 - 9) + 1\left( 1 - 3\beta_{2} \right) = 1 }{\Rightarrow 2\beta_{2} - 6 + 8 + 1 - 3\beta_{2} = 1 }{\Rightarrow \beta_{2} = 2 }$$So $A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{pmatrix}$

$${A^{2} = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{pmatrix}\begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{pmatrix} = \begin{pmatrix} 8 & 5 & 6 \\ 13 & 8 & 10 \\ 10 & 6 & 7 \end{pmatrix} }{A^{2} + A = \begin{pmatrix} 10 & 6 & 7 \\ 14 & 10 & 13 \\ 13 & 7 & 8 \end{pmatrix} \Rightarrow det\left( A^{2} + A \right) = 8 }{\Rightarrow Det\left( adj\left( A^{2} + A \right) \right) = 8^{2} = 64 }$$