Question

考虑未知数为 $x$ 、 $y$ 和 $z$ 的三元一次方程组 $\begin{cases} a_1 x + b_1 y + c_1 z & = 0 \\ a_2 x + b_2 y + c_2 z & = 0 \\ a_3 x + b_3 y + c_3 z & = 0 \end{cases}$ 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，一个这样的方程组是 $\begin{cases} 1 x + 1 y + 0 z & = 0 \\ 0 x + 1 y + 1 z & = 0 \\ 0 x + 0 y + 0 z & = 0 \end{cases}$ 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。）

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

小提示：

三个不同的非零相关行向量中，一个向量是另外两个不相交支撑向量的普通和

For three distinct nonzero dependent rows, one row is the ordinary sum of two rows with disjoint supports

大提示：

数出所有二进制系数矩阵，再减去非奇异矩阵

Count all binary coefficient matrices and subtract nonsingular ones

解答：

共有 $2^9=512$ 个有序二进制系数矩阵。齐次方程组只有零解，当且仅当它的三个行向量线性无关，因此我们数出这些矩阵后再从总数中减去。

线性无关矩阵的三行必须是互不相同的非零向量。这样的行有 $7\cdot6\cdot5=210$ 种有序选法。在三个互不相同的非零二进制向量中，线性相关恰好发生在一个向量是另外两个向量的普通和时；这两个加数的支撑必须非空且互不相交。

若这个和的支撑大小为 $2$ ，有 $3$ 种方法选择它的两个坐标，而两个加数就是对应的两个单位向量。若这个和的支撑大小为 $3$ ，有 $3$ 种方法选择一个加数所占的单个坐标，其余两个坐标组成另一个加数。因此，共有 $3+3=6$ 组无序相关三元组，每一组又有 $3!=6$ 种行的排列顺序。

所以线性无关矩阵有 $210-6\cdot6=174$ 个。所求的奇异矩阵数，也就是有非零解的方程组数，为 $512-174=338\text{。}$

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

小提示：

大提示：

解答：

其他年份的第 18 题