Question

Prove that the given determinant:
$\left| \begin{matrix}
   ax & by & cz \\
   {{x}^{2}} & {{y}^{2}} & {{z}^{2}} \\
   1 & 1 & 1 \\
\end{matrix} \right|=\left| \begin{matrix}
   a & b & c \\
   x & y & z \\
   yz & xz & xy \\
\end{matrix} \right|$ .

Answer 1

Hint: We will apply the formula of expanding determinant of a matrix to both, left hand side and right hand side, of the given equation and then form the same expression on both sides to prove it.

Complete Step-by-Step solution:
To every square matrix, there is a number associated with it which is called its determinant.
For any square matrix of the form:
$\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right)$
Its determinant is represented as:
$\left| \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right|$
This determinant is the sum of product of elements of any one of the row or columns with its corresponding cofactors. Where, cofactor of ${{a}_{ij}}^{th}$ element, represented as ${{C}_{ij}}$, is the product of ${{\left( -1 \right)}^{i+j}}$ with the determinant that we will get by removing ${{i}^{th}}$ row and ${{j}^{th}}$ column from given determinant. For e.g., in this determinant, cofactor of ${{a}_{11}}$ will be the product of ${{\left( -1 \right)}^{1+1}}$ and $\left| \begin{matrix}
   {{a}_{22}} & {{a}_{23}} \\
   {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right|$, that is,
${{C}_{11}}={{\left( -1 \right)}^{1+1}}\left| \begin{matrix}
   {{a}_{22}} & {{a}_{23}} \\
   {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right|$.
 Now, let us evaluate this determinant along its first row, as given below:
$\left| \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right|$
$={{a}_{11}}{{C}_{11}}+{{a}_{12}}{{C}_{12}}+{{a}_{13}}{{C}_{13}}$
$={{a}_{11}}{{(-1)}^{1+1}}\left| \begin{matrix}
   {{a}_{22}} & {{a}_{23}} \\
   {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right|+{{a}_{12}}{{(-1)}^{1+2}}\left| \begin{matrix}
   {{a}_{21}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{33}} \\
\end{matrix} \right|+{{a}_{13}}{{(-1)}^{1+3}}\left| \begin{matrix}
   {{a}_{21}} & {{a}_{22}} \\
   {{a}_{31}} & {{a}_{32}} \\
\end{matrix} \right|$
$={{(-1)}^{2}}{{a}_{11}}({{a}_{22}}\times {{a}_{33}}-{{a}_{32}}\times {{a}_{23}})+{{(-1)}^{3}}{{a}_{12}}({{a}_{21}}\times {{a}_{33}}-{{a}_{31}}\times {{a}_{23}})+{{(-1)}^{4}}{{a}_{13}}({{a}_{21}}\times {{a}_{32}}-{{a}_{31}}\times {{a}_{22}})$
$={{a}_{11}}({{a}_{22}}\times {{a}_{33}}-{{a}_{32}}\times {{a}_{23}})-{{a}_{12}}({{a}_{21}}\times {{a}_{33}}-{{a}_{31}}\times {{a}_{23}})+{{a}_{13}}({{a}_{21}}\times {{a}_{32}}-{{a}_{31}}\times {{a}_{22}})$.
We can also expand it along any row or column.
Now, let us apply this formula to left hand side of the given equation as below:
L.H.S. = $\left| \begin{matrix}
   ax & by & cz \\
   {{x}^{2}} & {{y}^{2}} & {{z}^{2}} \\
   1 & 1 & 1 \\
\end{matrix} \right|$
$=ax({{y}^{2}}\times 1-1\times {{z}^{2}})-by({{x}^{2}}\times 1-1\times {{z}^{2}})+cz({{x}^{2}}\times 1-1\times {{y}^{2}})$
$=ax({{y}^{2}}-{{z}^{2}})-by({{x}^{2}}-{{z}^{2}})+cz({{x}^{2}}-{{y}^{2}})$
$=ax{{y}^{2}}-ax{{z}^{2}}-by{{x}^{2}}+by{{z}^{2}}+cz{{x}^{2}}-cz{{y}^{2}}$
$=ax{{y}^{2}}-by{{x}^{2}}-ax{{z}^{2}}+cz{{x}^{2}}+by{{z}^{2}}-cz{{y}^{2}}$
$=xy(ay-bx)-xz(az+cx)+yz(bz-cy)\cdots \cdots \left( i \right)$
 Now, let us apply determinant formula to right hand side of given equation as below:
R.H.S. = $\left| \begin{matrix}
   a & b & c \\
   x & y & z \\
   yz & xz & xy \\
\end{matrix} \right|$
$=a(y\times xy-xz\times z)-b(x\times xy-yz\times z)+c(x\times xz-yz\times z)$
$=a(x{{y}^{2}}-x{{z}^{2}})-b(y{{x}^{2}}-y{{z}^{2}})+c(z{{x}^{2}}-y{{z}^{2}})$
$=ax{{y}^{2}}-ax{{z}^{2}}-by{{x}^{2}}+by{{z}^{2}}+cz{{x}^{2}}-cy{{z}^{2}}$
$=ax{{y}^{2}}-by{{x}^{2}}-ax{{z}^{2}}+cz{{x}^{2}}+by{{z}^{2}}-cy{{z}^{2}}$
$=xy(ay-bx)-xz(az+cx)+yz(bz-cy)\cdots \cdots \left( ii \right)$
From equation $\left( i \right)$ and equation $\left( ii \right)$, we get,
L.H.S.=R.H.S.
That is, $\left| \begin{matrix}
   ax & by & cz \\
   {{x}^{2}} & {{y}^{2}} & {{z}^{2}} \\
   1 & 1 & 1 \\
\end{matrix} \right|=\left| \begin{matrix}
   a & b & c \\
   x & y & z \\
   yz & xz & xy \\
\end{matrix} \right|$.
Hence the given equation is proved.

Note: Don’t confuse determinant with matrix and try to solve this question by equating terms of both the determinants. That will be wrong. Also avoid making mistakes while placing minus sign in determinant formula.