If n is a real number, then the simultaneous system nx+y=1, ny+z=1, x+nz=1 has no solution if and only if n is equal to

1973 AMC 12 Problem 10

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10.

n

is a real number, then the simultaneous system

\begin{cases} nx+y=1,\\ ny+z=1,\\ x+nz=1 \end{cases}

has no solution if and only if

n

is equal to

-1

0

1

0

1

\frac12

Add all three equations and inspect the coefficient of

x+y+z

For every remaining value of

n,

try the symmetric choice

x=y=z

Adding the equations gives

(n+1)(x+y+z)=3.

When

n=-1,

this becomes

0=3,

so no solution exists. If

n\ne-1,

the symmetric assignment

x=y=z=\frac1{n+1}

satisfies all three equations. Hence the system has no solution exactly when

n=-1.

Therefore, the correct answer is A.