Question

Show that \left| {\begin{array}{*{20}{c}} 1&{yz}&{y + z} \\\ 1&{zx}&{z + x} \\\ 1&{xy}&{x + y} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 1&x;&{{x^2}} \\\ 1&y;&{{y^2}} \\\ 1&z;&{{z^2}} \end{array}} \right|.

Answer 1

First take the L.H.S. and subtract row 1 (${R_1}$) from row 2 (${R_2}$) and row 3 (${R_3}$). After that, take common from row 2 (${R_2}$) and row 3 (${R_3}$). Now, solve the determinant along column 1 (${C_1}$). Then, take the R.H.S. and subtract r and subtract row 1 (${R_1}$) from row 2 (${R_2}$) and row 3 (${R_3}$). After that, take common from row 2 (${R_2}$) and row 3 (${R_3}$). Now, solve the determinant along column 1 (${C_1}$). Now check if the value of L.H.S. is equal to R.H.S. or not.

Complete step-by-step answer:
Take L.H.S. and simplify it.
\left| {\begin{array}{*{20}{c}} 1&{yz}&{y + z} \\\ 1&{zx}&{z + x} \\\ 1&{xy}&{x + y} \end{array}} \right|
Subtract row 1 from row 2 and row 3,
$\begin{gathered} {R_2} \to {R_2} - {R_1} \\\ {R_3} \to {R_3} - {R_1} \\\ \end{gathered}$
\left| {\begin{array}{*{20}{c}} 1&{yz}&{y + z} \\\ 0&{zx - yz}&{\left( {z + x} \right) - \left( {y + z} \right)} \\\ 0&{xy - yz}&{\left( {x + y} \right) - \left( {y + z} \right)} \end{array}} \right|
Take $z$ common from 2nd column of 2nd row, $y$ from 2nd column of 3rd row and simplify the 3rd column,
\left| {\begin{array}{*{20}{c}} 1&{yz}&{y + z} \\\ 0&{z\left( {x - y} \right)}&{x - y} \\\ 0&{y\left( {x - z} \right)}&{x - z} \end{array}} \right|
Take out $\left( {x - y} \right)$ common from 2nd row and $\left( {x - z} \right)$ from 3rd row.
\left( {x - y} \right)\left( {x - z} \right)\left| {\begin{array}{*{20}{c}} 1&{yz}&{y + z} \\\ 0&z;&1 \\\ 0&y;&1 \end{array}} \right|
Solve the determinant along the first column (${C_1}$),
\left( {x - y} \right)\left( {x - z} \right)\left[ {1\left( {\left| {\begin{array}{*{20}{c}} z&1 \\\ y&1 \end{array}} \right|} \right) - 0 + 0} \right]
Solve the determinant inside the bracket,
$\left( {x - y} \right)\left( {x - z} \right)\left( {z - y} \right)$ ….. (1)
Now, take L.H.S. and simplify it.
\left| {\begin{array}{*{20}{c}} 1&x;&{{x^2}} \\\ 1&y;&{{y^2}} \\\ 1&z;&{{z^2}} \end{array}} \right|
Subtract row 1 from row 2 and row 3,
$\begin{gathered} {R_2} \to {R_2} - {R_1} \\\ {R_3} \to {R_3} - {R_1} \\\ \end{gathered}$
\left| {\begin{array}{*{20}{c}} 1&x;&{{x^2}} \\\ 0&{y - x}&{{y^2} - {x^2}} \\\ 0&{z - x}&{{z^2} - {x^2}} \end{array}} \right|
Take out $\left( {y - x} \right)$ common from 2nd row and $\left( {z - x} \right)$ from 3rd row.
\left( {y - x} \right)\left( {z - x} \right)\left| {\begin{array}{*{20}{c}} 1&x;&{{x^2}} \\\ 0&1&{y + x} \\\ 0&1&{z + x} \end{array}} \right|
Solve the determinant along the first column (${C_1}$),
\left( {y - x} \right)\left( {z - x} \right)\left[ {1\left( {\left| {\begin{array}{*{20}{c}} 1&{y + x} \\\ 1&{z + x} \end{array}} \right|} \right) - 0 + 0} \right]
Solve the determinant inside the bracket,
$\left( {x - y} \right)\left( {x - z} \right)\left[ {\left( {z + x} \right) - \left( {y + x} \right)} \right]$
Solve the variables inside the bracket,
$\left( {x - y} \right)\left( {x - z} \right)\left( {z - y} \right)$ ….. (2)
Since the value of equation (1) and (2) are equal.
Hence, \left| {\begin{array}{*{20}{c}} 1&{yz}&{y + z} \\\ 1&{zx}&{z + x} \\\ 1&{xy}&{x + y} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 1&x;&{{x^2}} \\\ 1&y;&{{y^2}} \\\ 1&z;&{{z^2}} \end{array}} \right|.

Note: Determinant, in linear and multilinear algebra, a value, denoted det A, associated with a square matrix A of n rows and n columns.
The determinant of a matrix A is commonly denoted det(A) or |A|
A $k \times k$ determinant \left| {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}& \cdots &{{a_{1k}}} \\\ {{a_{21}}}&{{a_{22}}}& \cdots &{{a_{2k}}} \\\ \vdots & \vdots & \ddots & \vdots \\\ {{a_{k1}}}&{{a_{k2}}}& \cdots &{{a_{kk}}} \end{array}} \right| can be expanded "by minors" to obtain

{{a_{22}}}&{{a_{23}}}& \cdots &{{a_{2k}}} \\\ \vdots & \vdots & \ddots & \vdots \\\ {{a_{k2}}}&{{a_{k3}}}& \cdots &{{a_{kk}}} \end{array}} \right| - {a_{12}}\left| {\begin{array}{*{20}{c}} {{a_{21}}}&{{a_{23}}}& \cdots &{{a_{2k}}} \\\ \vdots & \vdots & \ddots & \vdots \\\ {{a_{k1}}}&{{a_{k3}}}& \cdots &{{a_{kk}}} \end{array}} \right| + \ldots \pm {a_{1k}}\left| {\begin{array}{*{20}{c}} {{a_{21}}}&{{a_{22}}}& \cdots &{{a_{2\left( {k - 1} \right)}}} \\\ \vdots & \vdots & \ddots & \vdots \\\ {{a_{k1}}}&{{a_{k2}}}& \cdots &{{a_{k\left( {k - 1} \right)}}} \end{array}} \right|$$