6. (a) 已知 $A = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 2 & 3 & 1 \end{pmatrix}$ 的逆矩阵为 $A^{-1} = \begin{pmatrix} -1 & -1 & 1 \\ 1 & m & n \\ -1 & \frac{1}{2} & \frac{1}{2} \end{pmatrix}$
(i) 求 $m + n$ 的值。 (3%)
(ii) 利用 (i) 的结果，解方程组 $\begin{cases} x + 2y = 5 \\ y + z = -1 \\ 2x + 3y + z = 5 \end{cases}$ (3%)
1
To find the values of $m$ and $n$, we use the identity $A A^{-1} = I$, where $I$ is the identity matrix.
2
Multiplying the second row of $A$ by the second column of $A^{-1}$ gives the $(2,2)$ element of $I$, which is 1: $(0)(-1) + (1)(m) + (1)(\frac{1}{2}) = 1 \Rightarrow m + \frac{1}{2} = 1 \Rightarrow m = \frac{1}{2}$.
3
Multiplying the second row of $A$ by the third column of $A^{-1}$ gives the $(2,3)$ element of $I$, which is 0: $(0)(1) + (1)(n) + (1)(\frac{1}{2}) = 0 \Rightarrow n + \frac{1}{2} = 0 \Rightarrow n = -\frac{1}{2}$.
4
Thus, $m + n = \frac{1}{2} + (-\frac{1}{2}) = 0$.
5
The system of equations in part (ii) can be written in matrix form as $A \mathbf{x} = \mathbf{b}$, where $\mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 5 \\ -1 \\ 5 \end{pmatrix}$.
6
Solving for $\mathbf{x}$ gives $\mathbf{x} = A^{-1} \mathbf{b}$. Substituting $m = \frac{1}{2}$ and $n = -\frac{1}{2}$ into $A^{-1}$, we get:
7
$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -1 & -1 & 1 \\ 1 & \frac{1}{2} & -\frac{1}{2} \\ -1 & \frac{1}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} 5 \\ -1 \\ 5 \end{pmatrix}$
8
Performing the matrix multiplication:
9
$x = (-1)(5) + (-1)(-1) + (1)(5) = -5 + 1 + 5 = 1$
10
$y = (1)(5) + (\frac{1}{2})(-1) + (-\frac{1}{2})(5) = 5 - 0.5 - 2.5 = 2$
11
$z = (-1)(5) + (\frac{1}{2})(-1) + (\frac{1}{2})(5) = -5 - 0.5 + 2.5 = -3$
12
The final solution is $x = 1, y = 2, z = -3$.
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