Question

If \[\left| \begin{matrix}
   a & b-y & c-z \\
   a-x & b & c-z \\
   a-x & b-y & c \\
\end{matrix} \right|=0\], then using properties of determinants find the value of \[\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}\], where \[x,y,z\ne 0\].

Answer 1

Hint: We simplify the given determinant by performing row operations \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\] and \[{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}\] inside it. We then apply the definition of determinant to the terms present inside it. We then make necessary arrangements and divide the obtained expression with $xyz$ to get the desired value of \[\left( \dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z} \right)\].

Complete step by step answer:
According to the problem, we are given a determinant \[\left| \begin{matrix}
   a & b-y & c-z \\
   a-x & b & c-z \\
   a-x & b-y & c \\
\end{matrix} \right|=0\], we need to find the value of $\left( \dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z} \right)$, where $x,y,z\ne 0$ using the properties of determinants.
We are asked to solve it by using the properties of determinants. Let us simplify the determinant, which is given to us.
\[\begin{matrix}
   {{R}_{1}}\to \\
   {{R}_{2}}\to \\
   {{R}_{3}}\to \\
\end{matrix}\left| \begin{matrix}
   a & b-y & c-z \\
   a-x & b & c-z \\
   a-x & b-y & c \\
\end{matrix} \right|=0\].
Let us write the expressions as,
 Thus, by applying the conditions ${{R}_{2}}\to {{R}_{2}}-{{R}_{1}}$, ${{R}_{3}}\to {{R}_{3}}-{{R}_{1}}$ we get the determinant as,
\[\left| \begin{matrix}
   a & b-y & c-z \\
   \left( a-x \right)-a & b-\left( b-y \right) & \left( c-z \right)-\left( c-z \right) \\
   \left( a-x \right)-a & \left( b-y \right)-\left( b-y \right) & c-\left( c-z \right) \\
\end{matrix} \right|=0\].
Let us simplify the above,
\[\left| \begin{matrix}
   a & b-y & c-z \\
   -x & +y & 0 \\
   -x & 0 & z \\
\end{matrix} \right|=0\] - (1).
Let us recall the definition of determinant which is defined below,
\[\left| \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right|=a\left( ei-fh \right)-b\left( di-fg \right)+c\left( dh-eg \right)\].
We use this in equation (1).
\[\Rightarrow a\left( yz-0 \right)-\left( b-y \right)\left( -xz-0 \right)+\left( c-z \right)\left( +xy-0 \right)=0\].
\[\Rightarrow ayz+xz\left( b-y \right)+xy\left( c-z \right)=0\].
Now let us simplify the equation,
\[\Rightarrow ayz+bxz-xyz+xyc-xyz=0\].
\[\Rightarrow ayz+bxz+xyc=2xyz\].
Now let us divide the entire expression by xyz:
\[\dfrac{ayz}{xyz}+\dfrac{bxz}{xyz}+\dfrac{xyc}{xyz}=\dfrac{2xyz}{xyz}\].
\[\Rightarrow \dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=2\].

So, we got the required value of \[\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=2\].

Note: We need not always calculate the determinant in the given form as the determinant obtained will be long, complex and confusing. We can make use of row or column operations in order to simplify the terms in the determinant and make the process of solving the determinant easy. We should not make calculation mistakes while solving this problem.