Find 6 real numbers that satisfy these systems of equations and then compute a value. 
$a \neq 17$ should be $r \neq 17$. 
I came across this problem in a past math competition test and wanted to tackle it just for fun. 
How would you solve this? 
 A: The equations given are:
\begin{gather}
17x+ sy + tz=0 \\
rx+34y+tz=0 \\
rx+sy+87z=0
\end{gather}
When we attempt to rewrite this is matrix form, we should remind ourselves of how matrices are multiplied with vectors:
$$
\begin{pmatrix}a & b & c \\ d & e & f \\ g & h & i\end{pmatrix} \begin{pmatrix} x \\y\ \\z\end{pmatrix} = \begin{pmatrix} ax+by+cz \\dx+ey+fz\ \\gx+hy+iz\end{pmatrix}
$$
You can imagine it like this: the rows on the left matrix, sink one-by-one (like divers) into the vector on the right (like an ocean), so that they come side by side, like $\begin{pmatrix} ax \\by\ \\cz\end{pmatrix}$. Then, we add all the products to get the entry, in this case $ax+by+cz$. Do the same with the other rows.
Now,with this background, I request you to verify that the above equations can be rewritten in the following form:
$$
\begin{pmatrix}17 & s & t \\ r & 34 & t \\ r & s & 87\end{pmatrix} \begin{pmatrix} x \\y\ \\z\end{pmatrix} = \begin{pmatrix} 0 \\0\ \\0\end{pmatrix}
$$
The above exercise will be beneficial to you, you can familiarize yourself with how matrices are multiplied.
There is something very special about the relation between equation solving and matrix representations of equations. This is the determinant. Every square matrix has associated to it, something called a determinant, which is a real number.
While the determinant of a matrix can be defined in terms of it's entries directly (I request you to read this up from Wikipedia, but if you don't understand please get back), I'll get to the salient features of the determinant:

1) Suppose you have a matrix, and you subtract a row from another row, leaving the subtracted row unchanged. Then, the determinant of the new matrix is the same. In fact, if you add/subtract multiples of rows from one another, the determinant remains unchanged.
2) If you interchange the rows of a matrix, the determinant changes sign.
3) Suppose there is a constant $c$ that divides a whole row or column of  a matrix. Then,suppose we have a new matrix where we divide that row by the constant and leave everything else unchanged. Then, the ratio of the two determinants is $c$.

And the most important property: invertibility.
We have the number $1$, which if you multiply with any number,it remains unchanged. Similarly, we have the identity matrix, which if you multiply with any matrix, leaves it unchanged. This matrix has $1$ s on the diagonal, and zero elsewhere, like:
$$
I=\begin{pmatrix} 1&0&0 \\0&1&0\\ 0&0&1\end{pmatrix}
$$
Now, for every real number $a \neq 0$, we have this reciprocal $\frac{1}{a}$, which when we multiply by $a$, we get $1$, right?So we ask a similar question: Given a matrix $A$, can we find a matrix $B$, such that $B \cdot A = A \cdot B = I$? (i.e. does $\frac{I}{A}$ exist, in an odd sense?). $A$ is said to be invertible if this is possible. The answer to that question is non-trivial in the least:

4)  A matrix is invertible if and only if it's determinant is non-zero.

That means, that the determinant actually gives a description of invertible matrices!
Finally, we return to our question after the preliminaries. Our question is:
$$
\begin{pmatrix}17 & s & t \\ r & 34 & t \\ r & s & 87\end{pmatrix} \begin{pmatrix} x \\y\ \\z\end{pmatrix} = \begin{pmatrix} 0 \\0\ \\0\end{pmatrix}
$$
Suppose that the large square matrix (call it $A$)is invertible. Then, there is some matrix, let's call it $\frac{I}{A}$, such that $frac{I}{A} \cdot A = A \cdot \frac{I}{A} = I$. Let us multiply by this matrix on both sides of the equation:
$$
\frac{I}{A}\begin{pmatrix}17 & s & t \\ r & 34 & t \\ r & s & 87\end{pmatrix} \begin{pmatrix} x \\y\ \\z\end{pmatrix} = \frac{I}{A}\begin{pmatrix} 0 \\0\ \\0\end{pmatrix} = \begin{pmatrix} 0 \\0\ \\0\end{pmatrix}
$$
(Zero matrix times any matrix is zero matrix, you can check this very easily from the "diving into the ocean" argument I gave above).
Now, by definition of $\frac{I}{A}$, the right side simplifies to:
$$
\begin{pmatrix} x \\y \\z\end{pmatrix}
=\begin{pmatrix} 0 \\0\ \\0\end{pmatrix}$$
But this means that $x=0$! We have assumed that is not the case. Hence, $A$ is not invertible, hence it's determinant is zero. We write $\det$ for determinant.
Hence:
$$
0=\det \begin{pmatrix}17 & s & t \\ r & 34 & t \\ r & s & 87\end{pmatrix} 
$$
Subtract the third row from the second:
$$
0=\det \begin{pmatrix}17 & s & t \\ 0 & 34-s & t-87 \\ r & s & 87\end{pmatrix} 
$$
Now, subtract the first row from the third:
$$
0=\det \begin{pmatrix}17 & s & t \\ 0 & 34-s & t-87 \\ r-17 & 0 & 87-t\end{pmatrix} 
$$
Now, we can remove $17-r$,$34-s$ and $87-t$ outside, while ensuring that their columns have been respectively divided by the right constants. Since the left side is zero, and we have assumed $r \neq 17, s \neq 34, t \neq 87$, we can remove these constants from the equation, because they are non-zero. Hence, we would get:
$$
0=\det \begin{pmatrix}\frac{17}{17-r} & \frac{s}{34-s} & \frac t{87-t} \\ 0 & 1 & -1 \\ -1 & 0 & 1\end{pmatrix} 
$$
Now, we can evaluate the determinant directly, using any rule we know (read up the ordinary rule from Wikipedia, if you like).
Ishall expand by the first row:
$$
\det \begin{pmatrix}\frac{17}{17-r} & \frac{s}{34-s} & \frac t{87-t} \\ 0 & 1 & -1 \\ -1 & 0 & 1\end{pmatrix}  \\ = \frac{17}{17-r} \det \begin{pmatrix}1&-1 \\0&1\end{pmatrix} - \frac{s}{34-s} \det \begin{pmatrix}0&-1 \\-1&1\end{pmatrix} + \frac{t}{87-t} \det \begin{pmatrix}0&1 \\-1&0\end{pmatrix}
$$
Simplifying the $2 \times 2$ determinants (exercise, if you like), we get:
$$
2 = \frac{17}{17-r} + \frac{34}{34-s} + \frac{87}{87-t}
$$
Hence the answer follows. If you do not understand any of the above steps, please get back, I am very happy to help.
A: Here is a more symmetric calculation than the one in Aston Villa's answer. It is not really easier, but it bothered me that the previous calculation treated the columns differently:
With $A=17$, $B=34$ and $C=87$ we have
$$ Ax+sy+tz = 0 \\
rx+By+tz = 0 \\
rx+sy+Cz = 0 $$
Since it is specified that $r\ne A$ and $x\ne 0$ we know that $Ax\ne rx$.  Now if $s$ were $B$, both of $Ax$ and $rx$ would need to equal $(-By+tz)$, which is impossible. So $s\ne B$. Similarly, we get $t\ne C$, so the divisons in the sought quantity make sense.
As Aston Villa explains, because $x\ne 0$, the determinant
$$\left|\begin{matrix}A & s & t \\ r & B & t \\ r & s & C \end{matrix}\right| $$
needs to be $0$. We can compute this determinant directly using Sarrus's rule:
$$ \tag{1} 0 = ABC + 2rst - Ast - rBt - rsC $$
The quantity we want to find is
$$ \frac{A}{A-r}+\frac{B}{B-s}+\frac{C}{C-t} $$
which, putting on a common denominator is
$$ \frac{A(B-s)(C-t) + (A-r)B(C-t) + (A-r)(B-s)C}{(A-r)(B-s)(C-t)} $$
Multiplying out and collecting like terms in numerator and denominator, this equals as
$$ \frac{3ABC - 2(ABt+AsC+rBC) + (Ast+rBt+rsC)}{ABC-(ABt+AsC+rBC)+(Ast+rBt+rsC)-rst} $$
Change one of the $ABC$s in the numerator to $(Ast+rBt+rsC)-2rst$ using $\text{(1)}$, and we get
$$ \frac{2ABC - 2(ABt+AsC+rBC) + 2(Ast+rBt+rsC) - 2rst}{ABC-(ABt+AsC+rBC)+(Ast+rBt+rsC)-rst} = 2 $$
