Question

考虑未知数为 $x$ 、 $y$ 、 $z$ 的三元一次方程组：其中每个系数都是 $0$ 或 $1$ ，且方程组有不同于 $x=y=z=0$ 的解。例如，下面就是一个这样的方程组：它有非零解 $(x,y,z) = (1, -1, 1)$ 。这样的方程组共有多少个？（一个方程组中的方程不必互不相同；同样的方程以不同顺序出现时，视为不同方程组。） $\begin{cases} a_1 x + b_1 y + c_1 z & = 0 \\ a_2 x + b_2 y + c_2 z & = 0 \\ a_3 x + b_3 y + c_3 z & = 0 \end{cases}$ $\begin{cases} 1 x + 1 y + 0 z & = 0 \\ 0 x + 1 y + 1 z & = 0 \\ 0 x + 0 y + 0 z & = 0 \end{cases}$

Consider systems of three linear equations with unknowns $x,$ $y,$ and $z,$ $\begin{cases} a_1 x + b_1 y + c_1 z & = 0 \\ a_2 x + b_2 y + c_2 z & = 0 \\ a_3 x + b_3 y + c_3 z & = 0 \end{cases}$ where each of the coefficients is either $0$ or $1$ and the system has a solution other than $x=y=z=0.$ For example, one such system is $\begin{cases} 1 x + 1 y + 0 z & = 0 \\ 0 x + 1 y + 1 z & = 0 \\ 0 x + 0 y + 0 z & = 0 \end{cases}$ with a nonzero solution of $(x,y,z) = (1, -1, 1).$ How many such systems of equations are there? (The equations in a system need not be distinct, and two systems containing the same equations in a different order are considered different.)

$\ 302$

$\ 338$

$\ 340$

$\ 343$

$\ 344$

答案：B

知识点：方程组补集计数分类讨论

难度评级：1970

解答：

共有 $2^9=512$ 种配置。现在用补集计数来判断其中有多少种有多于一个解。

如果某个配置有个不包含冗余信息的方程，那么它只有一个解。 $7\cdot6\cdot5=210$

零向量不能出现，也就是不能有 $3+3=6$ 的行，且三行必须互不相同，所以有 $3!=6$ 、、种有序选择。这给出种配置。不过有些配置仍可能有冗余信息。如果两个方程相加得到另一个方程，就有冗余。其中还要去掉三行线性相关的情况。情况个方程满足，另一个方程恰有个变量系数为，第三个方程恰有 $2$ 个变量系数为。有 $3$ 种方法选择哪个方程所有变量系数都是。然后有种方法选择哪个方程只有一个变量系数为，而这个方程有 $3$ 种方法选择哪个变量系数为。这种情况要排除种配置。情况个方程有个变量系数为，另外两个方程各自只有一个变量系数为，这两个变量彼此不同，并且其中一个变量也出现在第一个方程中。有 $3$ 种方法选择哪个方程有个变量系数为，有种方法选择哪些变量系数为，还有种方法安排另外两个方程的顺序。这种情况也要排除种配置。

因而只有零解的方程组有 $210-6\cdot6=174$ 个，所以有非零解的方程组数为。 $512-174=338.$

所以正确答案是 B。

There are $2^9=512$ ordered binary coefficient matrices. A homogeneous system has only the zero solution exactly when its three row vectors are linearly independent, so we count those matrices and subtract.

An independent matrix must have three distinct nonzero rows. There are $7\cdot6\cdot5=210$ ordered choices of such rows. Among three distinct nonzero binary vectors, dependence occurs exactly when one is the ordinary sum of the other two; the two summands must have disjoint nonempty supports.

If the sum has support of size $2,$ choose its two coordinates in $3$ ways; its summands are the two corresponding unit vectors. If the sum has support of size $3,$ choose which one coordinate forms one summand in $3$ ways, with the other two coordinates forming the other summand. Thus there are $3+3=6$ unordered dependent triples, each with $3!=6$ row orders.

Hence the number of independent matrices is $210-6\cdot6=174.$ The desired number of singular matrices, and therefore of systems with a nonzero solution, is $512-174=338.$

Thus, the answer is B .

2022 AMC 10B 第 18 题

答案：B

难度评级：1970

解答：

其他年份的第 18 题