线性代数三部曲(一)·行列式
Part1:从解方程组谈起 栗子 :试讨论以下方程的解. \[\begin{cases} a_{11}x_1+a_{12}x_2=b_1\qquad(1)\\ a_{21}x_1+a_{22}x_2=b_2\qquad(2) \end{cases} \] 解 :将 \((1)\) 乘以 \(a_{21}\) , \((2)\) 乘以 \(a_(11)\) 有 \[\begin{cases} a_{11}a_{21}x_1+a_{12}a_{21}x_2=a_{21}b_1\qquad(3)\\ a_{11}a_{21}x_1+a_{22}a_{11}x_2=a_{11}b_2\qquad(4) \end{cases} \] 消去 \(x_1\) 有 \[(a_{11}a_{22}-a_{12}a_{21})x_2=a_{11}b_2-a_{21}b_1 \] 即 \[x_2=\frac{a_{11}b_2-a_{21}b_1}{a_{11}a_{22}-a_{12}a_{21}} \] 重复对 \(x_2\) 消元,有 \[x_1=\frac{a_{22}b_1-a_{12}b_2}{a_{11}a_{22}-a_{12}a_{21}} \] 当 \(a_{11}a_{22}-a_{12}a_{21}\ne0\) 时,方程组有唯一解; 当 \(a_{11}a_{22}-a_{12