題組內容

(a) The matrix of coefficients A and column matrix of constants B are

$A = \begin{bmatrix} 1 & 3 & 1 \\ 2 & 5 & 1 \\ 1 & 2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} -2 \\ -5 \\ 6 \end{bmatrix}$

It is found that |A| = –3 ≠ 0. Thus Cramer’s rule can be applied. We get

\(A_1 = \begin{bmatrix} -2 & 3 & 1 \\ -5 & 5 & 1 \\ 6 & 2 & 3 \end{bmatrix}\) \(A_2 = \begin{bmatrix} 1 & -2 & 1 \\ 2 & -5 & 1 \\ 1 & 6 & 3 \end{bmatrix}\) \(A_3 = \begin{bmatrix} 1 & 3 & -2 \\ 2 & 5 & -5 \\ 1 & 2 & 6 \end{bmatrix}\)

giving \(|A_1| = -3, |A_2| = 6, |A_3| = -9\)  。Cramer’s rule now gives

\(x_1 = \frac{|A_1|}{|A|} = \frac{-3}{-3} = 1, x_2 = \frac{|A_2|}{|A|} = \frac{6}{-3} = -2, x_3 = \frac{|A_3|}{|A|} = \frac{-9}{-3} = 3\)