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Step-by-Step Solution
Step 1: Set up the matrix equation
We have a system of linear equations represented by Ax = b. The matrix A is a 3×3 matrix:
$A = \begin{bmatrix}
a_{1} & a_{2} & a_{3} \\
a_{4} & a_{5} & a_{6} \\
a_{7} & a_{8} & a_{9}
\end{bmatrix}.$
We are given three specific solutions x1, x2, x3 corresponding to three different right-hand side vectors b1, b2, b3 respectively:
$x_{1} = \begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}, \quad b_{1} = \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}$
$x_{2} = \begin{bmatrix}0 \\ 2 \\ 1\end{bmatrix}, \quad b_{2} = \begin{bmatrix}0 \\ 2 \\ 0\end{bmatrix}$
$x_{3} = \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}, \quad b_{3} = \begin{bmatrix}0 \\ 0 \\ 2\end{bmatrix}$
Step 2: Write down the system from Ax1 = b1
Using $x_{1} = \begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}$ and $b_{1} = \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}$, we get:
$A \, x_{1} = \begin{bmatrix}
a_{1} & a_{2} & a_{3} \\
a_{4} & a_{5} & a_{6} \\
a_{7} & a_{8} & a_{9}
\end{bmatrix}
\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}
=
\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}.$
Equating each row’s product to the corresponding entry of $b_{1}$ gives the system:
$a_{1} + a_{2} + a_{3} = 1 \quad (1)$
$a_{4} + a_{5} + a_{6} = 0 \quad (2)$
$a_{7} + a_{8} + a_{9} = 0 \quad (3)$
Step 3: Write down the system from Ax2 = b2
Using $x_{2} = \begin{bmatrix}0 \\ 2 \\ 1\end{bmatrix}$ and $b_{2} = \begin{bmatrix}0 \\ 2 \\ 0\end{bmatrix}$, we get:
$A \, x_{2} = \begin{bmatrix}
a_{1} & a_{2} & a_{3} \\
a_{4} & a_{5} & a_{6} \\
a_{7} & a_{8} & a_{9}
\end{bmatrix}
\begin{bmatrix}0 \\ 2 \\ 1\end{bmatrix}
=
\begin{bmatrix}0 \\ 2 \\ 0\end{bmatrix}.$
This yields:
$2a_{2} + a_{3} = 0 \quad (4)$
$2a_{5} + a_{6} = 2 \quad (5)$
$2a_{8} + a_{9} = 0 \quad (6)$
Step 4: Write down the system from Ax3 = b3
Using $x_{3} = \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}$ and $b_{3} = \begin{bmatrix}0 \\ 0 \\ 2\end{bmatrix}$, we get:
$A \, x_{3} = \begin{bmatrix}
a_{1} & a_{2} & a_{3} \\
a_{4} & a_{5} & a_{6} \\
a_{7} & a_{8} & a_{9}
\end{bmatrix}
\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}
=
\begin{bmatrix}0 \\ 0 \\ 2\end{bmatrix}.$
This gives us directly:
$a_{3} = 0$
$a_{6} = 0$
$a_{9} = 2$
Step 5: Substitute back into the earlier equations
From $a_{3} = 0$ in (4): $2a_{2} + 0 = 0 \implies a_{2} = 0.$
From $a_{6} = 0$ in (5): $2a_{5} + 0 = 2 \implies a_{5} = 1.$
From $a_{9} = 2$ in (6): $2a_{8} + 2 = 0 \implies a_{8} = -1.$
From $a_{1} + a_{2} + a_{3} = 1$ (i.e., (1)): $a_{1} + 0 + 0 = 1 \implies a_{1} = 1.$
From $a_{4} + a_{5} + a_{6} = 0$ (i.e., (2)): $a_{4} + 1 + 0 = 0 \implies a_{4} = -1.$
From $a_{7} + a_{8} + a_{9} = 0$ (i.e., (3)): $a_{7} - 1 + 2 = 0 \implies a_{7} = -1.$
Step 6: Form the matrix A
Putting all these values together, we get:
$A = \begin{bmatrix}
1 & 0 & 0 \\
-1 & 1 & 0 \\
-1 & -1 & 2
\end{bmatrix}.$
Step 7: Compute the determinant of A
The determinant $|A|$ can be computed by expansion or by noticing the block-like structure. Using expansion along the first row:
$|A| = 1 \times \begin{vmatrix} 1 & 0 \\ -1 & 2 \end{vmatrix}
- 0 \times \begin{vmatrix} -1 & 0 \\ -1 & 2 \end{vmatrix}
+ 0 \times \begin{vmatrix} -1 & 1 \\ -1 & -1 \end{vmatrix}.
$
So,
$
|A| = 1 \times (1 \cdot 2 - 0 \cdot (-1)) = 2.
$
Therefore, the determinant of A is 2.
