Determinant With Real Numbers
Let $ a, $ $ b, $ $ c, $ $ p, $ $ q, $ and $ r $ be real numbers such that
\[\begin{vmatrix} p & b & c \\ a & q & c \\ a & b & r \end{vmatrix} = 0.\]Assuming that $ a \neq p, $ $ b \neq q, $ and $ c \neq r, $ find the value of $ \frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} $.
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- $\frac{a}{b}$
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- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
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- $\sin{}$
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- $\cap$
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- $\infty$
Solution
Expanding the determinant, we get
\begin{align*}
\begin{vmatrix} p & b & c \\ a & q & c \\ a & b & r \end{vmatrix} &= p \begin{vmatrix} q & c \\ b & r \end{vmatrix} - b \begin{vmatrix} a & c \\ a & r \end{vmatrix} + c \begin{vmatrix} a & q \\ a & b \end{vmatrix} \\
&= p(qr - bc) - b(ar - ac) + c(ab - aq) \\
&= pqr - bpc - abr + abc + abc - acq \\
&= 2abc - abr - acq - bcp + pqr.\end{align*}Let $ x = p - a, $ $ y = q - b, $ and $ z = r - c $. Then $ p = a + x, $ $ q = b + y, $ and $ r = c + z $. Substituting, we get
\[2abc - ab(c + z) - ac(b + y) - bc(a + x) + (a + x)(b + y)(c + z) = 0.\]This simplifies to $ ayz + bxz + cxy + xyz = 0 $.
Then
\begin{align*}
\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} &= \frac{a + x}{x} + \frac{b + y}{y} + \frac{c + z}{z} \\
&= \frac{a}{x} + 1 + \frac{b}{y} + 1 + \frac{c}{z} + 1 \\
&= \frac{a}{x} + \frac{b}{y} + \frac{c}{z} + 3 \\
&= \frac{ayz + bxz + cxy}{xyz} + 3 \\
&= \frac{-xyz}{xyz} + 3 = \boxed{2}.\end{align*}
Freeflows