Olympiad Maths Prep

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Problem 230

AMC 10/12, early questions
Algebra Difficulty 3.9 Find the answer

Consider systems of three linear equations with unknowns xx, yy, and zz,
\begin{align*} 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{align*}
where each of the coefficients is either 00 or 11 and the system has a solution other than x=y=z=0x=y=z=0.
For example, one such system is {1x+1y+0z=0,0x+1y+1z=0,0x+0y+0z=0}\{ 1x + 1y + 0z = 0, 0x + 1y + 1z = 0, 0x + 0y + 0z = 0 \}
with a nonzero solution of {x,y,z}={1,−1,1}\{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.)
(A) 302(B) 338(C) 340(D) 343(E) 344\textbf{(A)}\ 302 \qquad\textbf{(B)}\ 338 \qquad\textbf{(C)}\ 340 \qquad\textbf{(D)}\ 343 \qquad\textbf{(E)}\ 344

Official solution

Let M1=[a1b1c1],M2=[a2b2c2],M_1=\begin{bmatrix}a_1 & b_1 & c_1\end{bmatrix}, M_2=\begin{bmatrix}a_2 & b_2 & c_2\end{bmatrix}, and M3=[a3b3c3].M_3=\begin{bmatrix}a_3 & b_3 & c_3\end{bmatrix}.
We wish to count the ordered triples (M1,M2,M3)(M_1,M_2,M_3) of row matrices. We perform casework:

M1=M2=M3.M_1=M_2=M_3.
There are 23=82^3=8 options for M1.M_1. Once M1M_1 is chosen, M2M_2 and M3M_3 are uniquely determined.
In this case, we have 8\boldsymbol{8} ordered triples (M1,M2,M3).\boldsymbol{(M_1,M_2,M_3).}
Exactly two of M1,M2,M_1,M_2, and M3M_3 are equal.
For M1=M2≠M3,M_1=M_2\neq M_3, there are 23=82^3=8 options for M1.M_1. Once M1M_1 is chosen, M2M_2 is uniquely determined, and M3M_3 has 23−1=72^3-1=7 options. So, there are 8⋅7=568\cdot7=56 ordered triples (M1,M2,M3).(M_1,M_2,M_3).
Similarly, for each of M1=M3≠M2M_1=M_3\neq M_2 and M2=M3≠M1,M_2=M_3\neq M_1, there are 5656 ordered triples (M1,M2,M3).(M_1,M_2,M_3).
In this case, we have 56⋅3=168\boldsymbol{56\cdot3=168} ordered triples (M1,M2,M3).\boldsymbol{(M_1,M_2,M_3).}
All of M1,M2,M_1,M_2, and M3M_3 are different.
There are two subcases:

Exactly one of M1,M2,M_1,M_2, and M3M_3 is [000].\begin{bmatrix}0 & 0 & 0\end{bmatrix}.
For M1=[000],M_1=\begin{bmatrix}0 & 0 & 0\end{bmatrix}, there are 23−1=72^3-1=7 options for M2M_2 and 23−2=62^3-2=6 options for M3.M_3. So, there are 7⋅6=427\cdot6=42 ordered triples (M1,M2,M3).(M_1,M_2,M_3).
Similarly, for each of M2=[000]M_2=\begin{bmatrix}0 & 0 & 0\end{bmatrix} and M3=[000],M_3=\begin{bmatrix}0 & 0 & 0\end{bmatrix}, there are 4242 ordered triples (M1,M2,M3).(M_1,M_2,M_3).
In this subcase, we have 42⋅3=126\boldsymbol{42\cdot3=126} ordered triples (M1,M2,M3).\boldsymbol{(M_1,M_2,M_3).}
The sum of two of M1,M2,M_1,M_2, and M3M_3 is equal to the third matrix.
For M1+M2=M3:M_1+M_2=M_3:

If M1=[100],M_1=\begin{bmatrix}1 & 0 & 0\end{bmatrix}, then M2∈{[011],[010],[001]}.M_2\in\{\begin{bmatrix}0 & 1 & 1\end{bmatrix},\begin{bmatrix}0 & 1 & 0\end{bmatrix},\begin{bmatrix}0 & 0 & 1\end{bmatrix}\}.
More generally, if M1M_1 consists of one 11 and two 00's, then M2M_2 has 33 options, and M3M_3 is uniquely determined. So, there are (31)⋅3=9\binom31\cdot3=9 ordered triples (M1,M2,M3).(M_1,M_2,M_3).
If M1=[110],M_1=\begin{bmatrix}1 & 1 & 0\end{bmatrix}, then M2=[001].M_2=\begin{bmatrix}0 & 0 & 1\end{bmatrix}.
More generally, if M1M_1 consists of two 11's and one 0,0, then M2M_2 and M3M_3 are uniquely determined. So, there are (32)=3\binom32=3 ordered triples (M1,M2,M3).(M_1,M_2,M_3).

There are 9+3=129+3=12 ordered triples (M1,M2,M3).(M_1,M_2,M_3).
Similarly, for each of M1+M3=M2M_1+M_3=M_2 and M2+M3=M1,M_2+M_3=M_1, there are 1212 ordered triples (M1,M2,M3).(M_1,M_2,M_3).
In this subcase, we have 12⋅3=36\boldsymbol{12\cdot3=36} ordered triples (M1,M2,M3).\boldsymbol{(M_1,M_2,M_3).}

Together, the answer is 8+168+126+36=(B) 338.8+168+126+36=\boxed{\textbf{(B)}\ 338}.
~MRENTHUSIASM

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.