Q1
If $\det \mathbf{A} = -1,$ then find $\det (\mathbf{7A}).$
GT
In general, $\det (k \mathbf{A}) = k^2 \det \mathbf{A}.$ Thus,
[\det (7 \mathbf{A}) = 7^2 (-1) = \boxed{-49}.]
Problem & Correction
The GT answer is only true if A is 2x2, but that's not stated or implied in the question.
The general solution is -7^n, where n is the dimension of the matrix.
Q2
In the diagram below, $AB = AC = 115,$ $AD = 38,$ and $CF = 77.$ Compute $\frac{[CEF]}{[DBE]}.$
[asy]
unitsize(0.025 cm);
pair A, B, C, D, E, F;
B = (0,0);
C = (80,0);
A = intersectionpoint(arc(B,115,0,180),arc(C,115,0,180));
D = interp(A,B,38/115);
F = interp(A,C,(115 + 77)/115);
E = extension(B,C,D,F);
draw(C--B--A--F--D);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, NE);
label("$D$", D, W);
label("$E$", E, SW);
label("$F$", F, SE);
[/asy]
GT
\begin{align*} \frac{[CEF]}{[DBE]} &= \frac{\frac{1}{2} \cdot EF \cdot CE \cdot \sin \angle CEF}{\frac{1}{2} \cdot DE \cdot BE \cdot \sin \angle BED} \ &= \frac{EF}{DE} \cdot \frac{CE}{BE} \cdot \frac{\sin \angle CEF}{\sin \angle BED} \ &= \boxed{\frac{19}{96}}. \end{align*}
(In case it doesn't render, the final answer says 19/96)
Problem & Correction
$\frac{[CEF]}{[DBE]} = \frac{EF}{DE} \cdot \frac{CE}{BE} = 1 \cdot \frac{96}{19} = \frac{96}{19}$
In other words, the GT answer is inverted
Q3
The real number $x$ satisfies
[3x + \frac{1}{2x} = 3.]
Find
[64x^6 + \frac{1}{729x^6}.]
GT
Multiplying both sides of $3x + \frac{1}{2x} = 3$ by $\frac{2}{3},$ we get
[2x + \frac{1}{3x} = 2.]
Squaring both sides, we get
[4x^2 + \frac{4}{3} + \frac{1}{9x^2} = 4,]
so
[4x^2 + \frac{1}{9x^2} = \frac{8}{3}.]
Cubing both sides, we get
[64x^3 + 3 \cdot \frac{(4x^2)^2}{9x^2} + 3 \cdot \frac{4x^2}{(9x^2)^2} + \frac{1}{729x^6} = \frac{512}{27}.]
Then
\begin{align*}
64x^3 + \frac{1}{729x^6} &= \frac{512}{27} - \frac{3 \cdot 4x^2}{9x^2} \left( 4x^2 + \frac{1}{9x^2} \right) \
&= \frac{512}{27} - \frac{3 \cdot 4}{9} \cdot \frac{8}{3} \
&= \boxed{\frac{416}{27}}.
\end{align*}
Problem & Correction
The first term in the cubed expression should be 64x^6, not 64x^3.
Q4
Let $a,$ $b,$ $c$ be distinct complex numbers such that
[\frac{a}{1 - b} = \frac{b}{1 - c} = \frac{c}{1 - a} = k.]
Find the sum of all possible values of $k.$
GT
From the given equation,
\begin{align*}
a &= k(1 - b), \
b &= k(1 - c), \
c &= k(1 - a).
\end{align*}Then
\begin{align*}
a &= k(1 - b) \
&= k(1 - k(1 - c)) \
&= k(1 - k(1 - k(1 - a))).
\end{align*}Expanding, we get $ak^3 + a - k^3 + k^2 - k = 0,$ which factors as
[(k^2 - k + 1)(ak + a - k) = 0.]If $ak + a - k = 0,$ then $a = \frac{k}{k + 1},$ in which case $b = c = \frac{k}{k + 1}.$ This is not allowed, as $a,$ $b,$ and $c$ are distinct, so $k^2 - k + 1 = 0.$ The sum of the roots is $\boxed{1}.$
Note: The roots of $k^2 - k + 1 = 0$ are
[\frac{1 \pm i \sqrt{3}}{2}.]For either value of $k,$ we can take $a = 0,$ $b = 1,$ and $c = k.$
Problem & Solution
$a = 0,$ $b = 1,$ and $c = k$ is not permitted since it would make $\frac{a}{1 - b}$ undefined
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