Question

Prove that: \[\left| {\begin{array}{*{20}{c}}
  x&{{x^2}}&{1 + p{x^3}} \\
  y&{{y^2}}&{1 + p{y^3}} \\
  z&{{z^2}}&{1 + p{z^3}}
\end{array}} \right| = \left( {1 + pxyz} \right)\left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)\]

Answer 1

Hint: Here we use the property of splitting the determinant on the basis of the column that contains a sum of elements that can be put up as separate columns in two determinants. Take common values that exist in every element of a row. Create the second determinant identical and equal to the first determinant by shifting the rows and columns. Use row transformations to create factors similar to that in RHS of the equation. In the end calculate the determinant using a general method.
* Property of determinant: A determinant can be split into a sum of two determinants along any row or column.
* Determinant of a matrix \[\left[ {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right] = a(ei - hf) - b(di - fg) + c(dh - eg)\]
* If a row in the matrix contains elements which all have a common factor say p, then we can bring out the factor from the matrix.
\[\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  {pd}&{pe}&{pf} \\
  g&h&i
\end{array}} \right| = p\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right|\]

Complete step by step answer:
Here we name each row as a variable:
First row is \[{R_1}\]
Second row is \[{R_2}\]
Third row is \[{R_3}\]
Similarly, we name each column with a variable:
First column is \[{C_1}\]
Second column is \[{C_2}\]
Third column is \[{C_3}\]
We have the determinant
\[\left| {\begin{array}{*{20}{c}}
  x&{{x^2}}&{1 + p{x^3}} \\
  y&{{y^2}}&{1 + p{y^3}} \\
  z&{{z^2}}&{1 + p{z^3}}
\end{array}} \right|\]
Since the third column has elements in addition form, we can split the given determinant into two determinants having different third columns, which in addition give the third column of the given determinant.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  x&{{x^2}}&{1 + p{x^3}} \\
  y&{{y^2}}&{1 + p{y^3}} \\
  z&{{z^2}}&{1 + p{z^3}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  x&{{x^2}}&1 \\
  y&{{y^2}}&1 \\
  z&{{z^2}}&1
\end{array}} \right| + \left| {\begin{array}{*{20}{c}}
  x&{{x^2}}&{p{x^3}} \\
  y&{{y^2}}&{p{y^3}} \\
  z&{{z^2}}&{p{z^3}}
\end{array}} \right|\]
Now we know if an element is common in any row or any column, we can bring out that element from the determinant and write the remaining factors inside the determinant.
So, take ‘x’, ‘y’ and ‘z’ common from \[{R_1},{R_2},{R_3}\]respectively and ‘p’ from \[{C_3}\]of the second determinant in RHS
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  x&{{x^2}}&{1 + p{x^3}} \\
  y&{{y^2}}&{1 + p{y^3}} \\
  z&{{z^2}}&{1 + p{z^3}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  x&{{x^2}}&1 \\
  y&{{y^2}}&1 \\
  z&{{z^2}}&1
\end{array}} \right| + (pxyz)\left| {\begin{array}{*{20}{c}}
  1&x&{{x^2}} \\
  1&y&{{y^2}} \\
  1&z&{{z^2}}
\end{array}} \right|\]
We can see that both determinants have the same elements just not arranged the same way. So we shuffle the second determinant in RHS by shifting its columns to create an equal determinant to the first determinant. Always write (-1) as a factor of determinant when shifting any row or column.
Use column transformation \[{C_1} \leftrightarrow {C_2}\] in second determinant in RHS
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  x&{{x^2}}&{1 + p{x^3}} \\
  y&{{y^2}}&{1 + p{y^3}} \\
  z&{{z^2}}&{1 + p{z^3}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  x&{{x^2}}&1 \\
  y&{{y^2}}&1 \\
  z&{{z^2}}&1
\end{array}} \right| + ( - 1)(pxyz)\left| {\begin{array}{*{20}{c}}
  x&1&{{x^2}} \\
  y&1&{{y^2}} \\
  z&1&{{z^2}}
\end{array}} \right|\]
Again use column transformation \[{C_2} \leftrightarrow {C_3}\]in second determinant in RHS
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  x&{{x^2}}&{1 + p{x^3}} \\
  y&{{y^2}}&{1 + p{y^3}} \\
  z&{{z^2}}&{1 + p{z^3}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  x&{{x^2}}&1 \\
  y&{{y^2}}&1 \\
  z&{{z^2}}&1
\end{array}} \right| + ( - 1)( - 1)(pxyz)\left| {\begin{array}{*{20}{c}}
  x&{{x^2}}&1 \\
  y&{{y^2}}&1 \\
  z&{{z^2}}&1
\end{array}} \right|\]
Write multiplication of two negative numbers as a positive number
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  x&{{x^2}}&{1 + p{x^3}} \\
  y&{{y^2}}&{1 + p{y^3}} \\
  z&{{z^2}}&{1 + p{z^3}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  x&{{x^2}}&1 \\
  y&{{y^2}}&1 \\
  z&{{z^2}}&1
\end{array}} \right| + (pxyz)\left| {\begin{array}{*{20}{c}}
  x&{{x^2}}&1 \\
  y&{{y^2}}&1 \\
  z&{{z^2}}&1
\end{array}} \right|\]
Take the determinant common
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  x&{{x^2}}&{1 + p{x^3}} \\
  y&{{y^2}}&{1 + p{y^3}} \\
  z&{{z^2}}&{1 + p{z^3}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  x&{{x^2}}&1 \\
  y&{{y^2}}&1 \\
  z&{{z^2}}&1
\end{array}} \right|(1 + pxyz)\] … (1)
Now to form the remaining factors we will apply row transformations to the determinant in equation (1)
Now we use the row transformation \[{R_1} \to {R_1} - {R_2};{R_2} \to {R_2} - {R_3}\]
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  x&{{x^2}}&{1 + p{x^3}} \\
  y&{{y^2}}&{1 + p{y^3}} \\
  z&{{z^2}}&{1 + p{z^3}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  {x - y}&{{x^2} - {y^2}}&{1 - 1} \\
  {y - z}&{{y^2} - {z^2}}&{1 - 1} \\
  z&{{z^2}}&1
\end{array}} \right|(1 + pxyz)\]
Use the identity \[{a^2} - {b^2} = (a - b)(a + b)\]to write terms in second column
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  x&{{x^2}}&{1 + p{x^3}} \\
  y&{{y^2}}&{1 + p{y^3}} \\
  z&{{z^2}}&{1 + p{z^3}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  {x - y}&{(x - y)(x + y)}&0 \\
  {y - z}&{(y - z)(y + z)}&0 \\
  z&{{z^2}}&1
\end{array}} \right|(1 + pxyz)\]
Take \[(x - y)\]common from\[{R_1}\] and \[(y - z)\]common from \[{R_2}\]
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  x&{{x^2}}&{1 + p{x^3}} \\
  y&{{y^2}}&{1 + p{y^3}} \\
  z&{{z^2}}&{1 + p{z^3}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  1&{(x + y)}&0 \\
  1&{(y + z)}&0 \\
  z&{{z^2}}&1
\end{array}} \right|(x - y)(y - z)(1 + pxyz)\] … (2)
Calculate the determinant using general method.
\[ \Rightarrow \]Determinant of \[\left| {\begin{array}{*{20}{c}}
  1&{(x + y)}&0 \\
  1&{(y + z)}&0 \\
  z&{{z^2}}&1
\end{array}} \right| = 1\left( {y + z - 0} \right) - \left( {x + y} \right)\left( {1 - 0} \right) + 0\]
\[ \Rightarrow \]Determinant of \[\left| {\begin{array}{*{20}{c}}
  1&{(x + y)}&0 \\
  1&{(y + z)}&0 \\
  z&{{z^2}}&1
\end{array}} \right| = \left( {y + z} \right) - \left( {x + y} \right)\]
\[ \Rightarrow \]Determinant of \[\left| {\begin{array}{*{20}{c}}
  1&{(x + y)}&0 \\
  1&{(y + z)}&0 \\
  z&{{z^2}}&1
\end{array}} \right| = y + z - x - y\]
\[ \Rightarrow \]Determinant of \[\left| {\begin{array}{*{20}{c}}
  1&{(x + y)}&0 \\
  1&{(y + z)}&0 \\
  z&{{z^2}}&1
\end{array}} \right| = (z - x)\] … (3)
Substitute the value of determinant from equation (3) in equation (2)
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  x&{{x^2}}&{1 + p{x^3}} \\
  y&{{y^2}}&{1 + p{y^3}} \\
  z&{{z^2}}&{1 + p{z^3}}
\end{array}} \right| = (z - x)(x - y)(y - z)(1 + pxyz)\]
\[\therefore \left| {\begin{array}{*{20}{c}}
  x&{{x^2}}&{1 + p{x^3}} \\
  y&{{y^2}}&{1 + p{y^3}} \\
  z&{{z^2}}&{1 + p{z^3}}
\end{array}} \right| = (1 + pxyz)(x - y)(y - z)(z - x)\]
Hence Proved

Note: Students many times make mistake while splitting the determinant and they write one determinant empty as they think adding the elements from second determinant will give us the sum of the determinant, this is wrong as they confuse splitting of determinant with matrix addition. Also, keep in mind we always take one negative sign for one shift of column or a row.