作业帮a[11]*x[1] + a[12]*x[2] = b[1]a[21]*x[1] + a[22]*x[2] = b[2]这两个方程,怎麼利用加减消元法得到(a[11]*a[22] - a[12]*a[21])*x[1] = b[1]*a[22] - b[2]*a[12](a[11]*a[22] - a[12]*a[21])*z[2] = b[2]*a[11] - b[1]*a[21]为什麼要用加减消
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/12 07:05:48
![a[11]*x[1] + a[12]*x[2] = b[1]a[21]*x[1] + a[22]*x[2] = b[2]这两个方程,怎麼利用加减消元法得到(a[11]*a[22] - a[12]*a[21])*x[1] = b[1]*a[22] - b[2]*a[12](a[11]*a[22] - a[12]*a[21])*z[2] = b[2]*a[11] - b[1]*a[21]为什麼要用加减消](/uploads/image/z/3697183-55-3.jpg?t=a%5B11%5D%2Ax%5B1%5D+%2B+a%5B12%5D%2Ax%5B2%5D+%3D+b%5B1%5Da%5B21%5D%2Ax%5B1%5D+%2B+a%5B22%5D%2Ax%5B2%5D+%3D+b%5B2%5D%E8%BF%99%E4%B8%A4%E4%B8%AA%E6%96%B9%E7%A8%8B%2C%E6%80%8E%E9%BA%BC%E5%88%A9%E7%94%A8%E5%8A%A0%E5%87%8F%E6%B6%88%E5%85%83%E6%B3%95%E5%BE%97%E5%88%B0%28a%5B11%5D%2Aa%5B22%5D+-+a%5B12%5D%2Aa%5B21%5D%29%2Ax%5B1%5D+%3D+b%5B1%5D%2Aa%5B22%5D+-+b%5B2%5D%2Aa%5B12%5D%28a%5B11%5D%2Aa%5B22%5D+-+a%5B12%5D%2Aa%5B21%5D%29%2Az%5B2%5D+%3D+b%5B2%5D%2Aa%5B11%5D+-+b%5B1%5D%2Aa%5B21%5D%E4%B8%BA%E4%BB%80%E9%BA%BC%E8%A6%81%E7%94%A8%E5%8A%A0%E5%87%8F%E6%B6%88)
a[11]*x[1] + a[12]*x[2] = b[1]a[21]*x[1] + a[22]*x[2] = b[2]这两个方程,怎麼利用加减消元法得到(a[11]*a[22] - a[12]*a[21])*x[1] = b[1]*a[22] - b[2]*a[12](a[11]*a[22] - a[12]*a[21])*z[2] = b[2]*a[11] - b[1]*a[21]为什麼要用加减消
a[11]*x[1] + a[12]*x[2] = b[1]
a[21]*x[1] + a[22]*x[2] = b[2]
这两个方程,怎麼利用加减消元法得到
(a[11]*a[22] - a[12]*a[21])*x[1] = b[1]*a[22] - b[2]*a[12]
(a[11]*a[22] - a[12]*a[21])*z[2] = b[2]*a[11] - b[1]*a[21]
为什麼要用加减消元法?怎麼得到的?
[] 为的下标.
a[11]*x[1] + a[12]*x[2] = b[1]a[21]*x[1] + a[22]*x[2] = b[2]这两个方程,怎麼利用加减消元法得到(a[11]*a[22] - a[12]*a[21])*x[1] = b[1]*a[22] - b[2]*a[12](a[11]*a[22] - a[12]*a[21])*z[2] = b[2]*a[11] - b[1]*a[21]为什麼要用加减消
对于这种二元方程组,都是用消元法解的,目的是将相同的未知数系数化为一样,方便两式相减后消除：第一个方程*a[22]-第二个方程*a[12]可以将x2消除,化为只含x1的方程(a[11]*a[22] - a[12]*a[21])*x[1] = b[1]*a[22] - b[2]*a[12]
第二个方程*a[11]- 第一个方程*a[21],化为只含x2的方程(a[11]*a[22] - a[12]*a[21])*z[2] = b[2]*a[11] - b[1]*a[21]