In the given $3 \times 3$ matrix, $1$, $a$ and $a^2$ are the entries in the first row, $1$, $b$ and $b^2$ are the elements in the second row and $1$, $c$ and $c^2$ are the entries in the third row.

${\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \\ \end{vmatrix}}$

The determinant of the matrix of the order $3$ has to calculate in this matrix problem. So, let’s learn how to evaluate the determinant of this square matrix.

$1$ is an entry in the first row and the first column. $1$ is also an element in the second row and the first column. Subtract the elements in the second row from the corresponding elements in the first row and substitute the difference of the entries in the respective positions of first row in the matrix.

$R_1-R_2 \,\to\, R_1$

It makes the element in the first row and the first column to become $0$.

$=\,\,\,$ ${\begin{vmatrix} 1-1 & a-b & a^2-b^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \\ \end{vmatrix}}$

$=\,\,\,$ ${\begin{vmatrix} 0 & a-b & a^2-b^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \\ \end{vmatrix}}$

In the first row of $3 \times 3$ matrix, $0$, $a-b$ and $a^2-b^2$ are the entries in the matrix. The element $a^2-b^2$ represents the difference of the squares and they can be expressed in factor form as per the difference rule of squares.

$=\,\,\,$ ${\begin{vmatrix} 0 & a-b & (a-b)(a+b) \\ 1 & b & b^2 \\ 1 & c & c^2 \\ \end{vmatrix}}$

In the first row, $a-b$ is a factor in both second and third columns but there is no factor in the first column. However, it can be written as follows for our convenience.

$=\,\,\,$ ${\begin{vmatrix} 0 \times (a-b) & 1 \times (a-b) & (a-b) \times (a+b) \\ 1 & b & b^2 \\ 1 & c & c^2 \\ \end{vmatrix}}$

Now, $a-b$ is a common factor in each entry of the first row in this $3 \times 3$ square matrix. So, it can be taken out common from the entries of the first row.

$=\,\,\,$ $(a-b){\begin{vmatrix} 0 & 1 & a+b \\ 1 & b & b^2 \\ 1 & c & c^2 \\ \end{vmatrix}}$

$1$ is an element in the second row and the first column. $1$ is also an entry in the third row and the first column. Find the subtraction of the entries in the third row from the corresponding elements in the second row of the matrix.

$R_2-R_3 \,\to\, R_2$

Substitute them in their respective positions of the second row. The subtraction process makes the entry to become $0$ in the second row and the first column.

$=\,\,\,$ $(a-b){\begin{vmatrix} 0 & 1 & a+b \\ 1-1 & b-c & b^2-c^2 \\ 1 & c & c^2 \\ \end{vmatrix}}$

$=\,\,\,$ $(a-b){\begin{vmatrix} 0 & 1 & a+b \\ 0 & b-c & b^2-c^2 \\ 1 & c & c^2 \\ \end{vmatrix}}$

In the second row of the square matrix of order $3$, the elements are $0$, $b-c$ and $b^2-c^2$. The element $b^2-c^2$ expresses the difference of two terms. So, the difference of the terms can be written in factor form by using the difference rule of the squares in factor form.

$=\,\,\,$ $(a-b){\begin{vmatrix} 0 & 1 & a+b \\ 0 & b-c & (b-c)(b+c) \\ 1 & c & c^2 \\ \end{vmatrix}}$

In the second row, $b-c$ is a factor in both second and third elements but there is no factor like that in the first column. So, the factor $b-c$ can be included as follows for a cause.

$=\,\,\,$ $(a-b){\begin{vmatrix} 0 & 1 & a+b \\ 0 \times (b-c) & 1 \times (b-c) & (b-c) \times (b+c) \\ 1 & c & c^2 \\ \end{vmatrix}}$

In all three elements of the second row, $b-c$ is a common factor and it can be taken out common from them.

$=\,\,\,$ $(a-b)(b-c){\begin{vmatrix} 0 & 1 & a+b \\ 0 & 1 & b+c \\ 1 & c & c^2 \\ \end{vmatrix}}$

$1$ is an element in the first row and the second column. $1$ is also an entry in the second row and the second column. Let’s find the difference of them by subtracting the entries in the second row from the respective elements in the first row. Substitute the difference of the elements in the respective positions in the first row.

$R_1-R_2 \,\to\, R_1$

The idea behind finding the subtraction of the elements is to make the entry to become $0$ in the first row and the second column.

$=\,\,\,$ $(a-b)(b-c){\begin{vmatrix} 0-0 & 1-1 & a+b-(b+c) \\ 0 & 1 & b+c \\ 1 & c & c^2 \\ \end{vmatrix}}$

$=\,\,\,$ $(a-b)(b-c){\begin{vmatrix} 0 & 0 & a+b-b-c \\ 0 & 1 & b+c \\ 1 & c & c^2 \\ \end{vmatrix}}$

$=\,\,\,$ $(a-b)(b-c){\begin{vmatrix} 0 & 0 & a+\cancel{b}-\cancel{b}-c \\ 0 & 1 & b+c \\ 1 & c & c^2 \\ \end{vmatrix}}$

$=\,\,\,$ $(a-b)(b-c){\begin{vmatrix} 0 & 0 & a-c \\ 0 & 1 & b+c \\ 1 & c & c^2 \\ \end{vmatrix}}$

In the square matrix of the order $3$, the first and second elements are $0$. So, the determinant of the simplified matrix can be evaluated by the third element.

$=\,\,\,$ $(a-b)(b-c)\Big((-1)^{1+3} \times (a-c) \times (0 \times c\,-\,1 \times 1)\Big)$

$=\,\,\,$ $(a-b)(b-c)\Big((-1)^{4} \times (a-c) \times (0\,-\,1)\Big)$

$=\,\,\,$ $(a-b)(b-c)\Big(1 \times (a-c) \times (0\,-\,1)\Big)$

$=\,\,\,$ $(a-b)(b-c)\Big((a-c) \times (-1)\Big)$

$=\,\,\,$ $(a-b)(b-c)\Big((-1) \times (a-c)\Big)$

$=\,\,\,$ $(a-b)(b-c)(-a+c)$

$=\,\,\,$ $(a-b)(b-c)(c-a)$

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