Answer :
To solve the problem, we need to evaluate the determinants of the two given [tex]\(3 \times 3\)[/tex] matrices and check if they are equal to each other.
The first matrix is:
[tex]\[ \begin{vmatrix} 1 & bc & b+c \\ 1 & ca & c+a \\ 1 & ab & a+b \end{vmatrix} \][/tex]
The second matrix is:
[tex]\[ \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} \][/tex]
Let's evaluate the determinant of the first matrix:
### Determinant of the First Matrix
[tex]\[ \begin{vmatrix} 1 & bc & b+c \\ 1 & ca & c+a \\ 1 & ab & a+b \end{vmatrix} \][/tex]
To find this determinant, we will expand along the first row:
[tex]\[ = 1 \cdot \begin{vmatrix} ca & c+a \\ ab & a+b \end{vmatrix} - bc \cdot \begin{vmatrix} 1 & c+a \\ 1 & a+b \end{vmatrix} + (b+c) \cdot \begin{vmatrix} 1 & ca \\ 1 & ab \end{vmatrix} \][/tex]
Evaluating the [tex]\(2\times 2\)[/tex] determinants inside:
1. [tex]\[ \begin{vmatrix} ca & c+a \\ ab & a+b \end{vmatrix} = ca(a+b) - ab(c+a) = ca \cdot a + ca \cdot b - ab \cdot c - ab \cdot a = a^2c + abc - abc - a^2b = a^2c - a^2b = a^2(c - b) \][/tex]
2. [tex]\[ \begin{vmatrix} 1 & c+a \\ 1 & a+b \end{vmatrix} = 1 \cdot (a+b) - 1 \cdot (c+a) = (a+b) - (c+a) = b - c \][/tex]
3. [tex]\[ \begin{vmatrix} 1 & ca \\ 1 & ab \end{vmatrix} = 1 \cdot ab - 1 \cdot ca = ab - ca = ab - ca = a(b - c) \][/tex]
Putting these back into the determinant expansion, we get:
[tex]\[ = 1 \cdot a^2(c - b) - bc \cdot (b - c) + (b+c) \cdot a(b-c) \][/tex]
[tex]\[ = a^2(c - b) - bc \cdot (b - c) + a(b+c)(b-c) \][/tex]
Distribute and simplify each term:
[tex]\[ = a^2(c - b) - bc \cdot (b - c) + a(b^2 - bc + bc - c^2) \][/tex]
[tex]\[ = a^2(c - b) + bc(c - b) + a(b^2 - c^2) \][/tex]
Combine like terms:
[tex]\[ = a^2(c - b) + (bc) \cdot (c - b) + a(b^2 - c^2) \][/tex]
Factoring out common terms, we get:
[tex]\[ = (c - b)(a^2 + bc) + a(b^2 - c^2) \][/tex]
Finally, we can state the determinant of the first matrix is:
[tex]\[ - a^2b + a^2c + ab^2 - ac^2 - b^2c + bc^2 \][/tex]
### Determinant of the Second Matrix
[tex]\[ \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} \][/tex]
Similarly, we expand along the first row:
[tex]\[ = 1 \cdot \begin{vmatrix} b & b^2 \\ c & c^2 \end{vmatrix} - a \cdot \begin{vmatrix} 1 & b^2 \\ 1 & c^2 \end{vmatrix} + a^2 \cdot \begin{vmatrix} 1 & b \\ 1 & c \end{vmatrix} \][/tex]
Evaluating the [tex]\(2\times 2\)[/tex] determinants inside:
1. [tex]\[ \begin{vmatrix} b & b^2 \\ c & c^2 \end{vmatrix} = b \cdot c^2 - c \cdot b^2 = bc^2 - cb^2 = bc \cdot (c - b) \][/tex]
2. [tex]\[ \begin{vmatrix} 1 & b^2 \\ 1 & c^2 \end{vmatrix} = 1 \cdot c^2 - 1 \cdot b^2 = c^2 - b^2 \][/tex]
3. [tex]\[ \begin{vmatrix} 1 & b \\ 1 & c \end{vmatrix} = 1 \cdot c - 1 \cdot b = c - b \][/tex]
Putting these back into the determinant expansion, we get:
[tex]\[ = 1 \cdot bc \cdot (c - b) - a \cdot (c^2 - b^2) + a^2 \cdot (c - b) \][/tex]
[tex]\[ = bc \cdot (c - b) - a(c^2 - b^2) + a^2(c - b) \][/tex]
Distribute and simplify each term:
[tex]\[ = bc \cdot (c - b) - a(c^2 - b^2) + a^2(c - b) \][/tex]
Combine like terms:
[tex]\[ = bc(c - b) + a^2(c - b) - a(c^2 - b^2) \][/tex]
Factoring out common terms, we get:
[tex]\[ = (c - b)(bc + a^2) + a(b^2 - c^2) \][/tex]
Finally, we can state the determinant of the second matrix is:
[tex]\[ - a^2b + a^2c + ab^2 - ac^2 - b^2c + bc^2 \][/tex]
Both determinants simplify to the same expression:
[tex]\[ - a^2b + a^2c + ab^2 - ac^2 - b^2c + bc^2 \][/tex]
So, we can confirm that:
[tex]\[ \begin{vmatrix} 1 & bc & b+c \\ 1 & ca & c+a \\ 1 & ab & a+b \end{vmatrix} = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} \][/tex]
Thus, the equality of the determinants holds.
The first matrix is:
[tex]\[ \begin{vmatrix} 1 & bc & b+c \\ 1 & ca & c+a \\ 1 & ab & a+b \end{vmatrix} \][/tex]
The second matrix is:
[tex]\[ \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} \][/tex]
Let's evaluate the determinant of the first matrix:
### Determinant of the First Matrix
[tex]\[ \begin{vmatrix} 1 & bc & b+c \\ 1 & ca & c+a \\ 1 & ab & a+b \end{vmatrix} \][/tex]
To find this determinant, we will expand along the first row:
[tex]\[ = 1 \cdot \begin{vmatrix} ca & c+a \\ ab & a+b \end{vmatrix} - bc \cdot \begin{vmatrix} 1 & c+a \\ 1 & a+b \end{vmatrix} + (b+c) \cdot \begin{vmatrix} 1 & ca \\ 1 & ab \end{vmatrix} \][/tex]
Evaluating the [tex]\(2\times 2\)[/tex] determinants inside:
1. [tex]\[ \begin{vmatrix} ca & c+a \\ ab & a+b \end{vmatrix} = ca(a+b) - ab(c+a) = ca \cdot a + ca \cdot b - ab \cdot c - ab \cdot a = a^2c + abc - abc - a^2b = a^2c - a^2b = a^2(c - b) \][/tex]
2. [tex]\[ \begin{vmatrix} 1 & c+a \\ 1 & a+b \end{vmatrix} = 1 \cdot (a+b) - 1 \cdot (c+a) = (a+b) - (c+a) = b - c \][/tex]
3. [tex]\[ \begin{vmatrix} 1 & ca \\ 1 & ab \end{vmatrix} = 1 \cdot ab - 1 \cdot ca = ab - ca = ab - ca = a(b - c) \][/tex]
Putting these back into the determinant expansion, we get:
[tex]\[ = 1 \cdot a^2(c - b) - bc \cdot (b - c) + (b+c) \cdot a(b-c) \][/tex]
[tex]\[ = a^2(c - b) - bc \cdot (b - c) + a(b+c)(b-c) \][/tex]
Distribute and simplify each term:
[tex]\[ = a^2(c - b) - bc \cdot (b - c) + a(b^2 - bc + bc - c^2) \][/tex]
[tex]\[ = a^2(c - b) + bc(c - b) + a(b^2 - c^2) \][/tex]
Combine like terms:
[tex]\[ = a^2(c - b) + (bc) \cdot (c - b) + a(b^2 - c^2) \][/tex]
Factoring out common terms, we get:
[tex]\[ = (c - b)(a^2 + bc) + a(b^2 - c^2) \][/tex]
Finally, we can state the determinant of the first matrix is:
[tex]\[ - a^2b + a^2c + ab^2 - ac^2 - b^2c + bc^2 \][/tex]
### Determinant of the Second Matrix
[tex]\[ \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} \][/tex]
Similarly, we expand along the first row:
[tex]\[ = 1 \cdot \begin{vmatrix} b & b^2 \\ c & c^2 \end{vmatrix} - a \cdot \begin{vmatrix} 1 & b^2 \\ 1 & c^2 \end{vmatrix} + a^2 \cdot \begin{vmatrix} 1 & b \\ 1 & c \end{vmatrix} \][/tex]
Evaluating the [tex]\(2\times 2\)[/tex] determinants inside:
1. [tex]\[ \begin{vmatrix} b & b^2 \\ c & c^2 \end{vmatrix} = b \cdot c^2 - c \cdot b^2 = bc^2 - cb^2 = bc \cdot (c - b) \][/tex]
2. [tex]\[ \begin{vmatrix} 1 & b^2 \\ 1 & c^2 \end{vmatrix} = 1 \cdot c^2 - 1 \cdot b^2 = c^2 - b^2 \][/tex]
3. [tex]\[ \begin{vmatrix} 1 & b \\ 1 & c \end{vmatrix} = 1 \cdot c - 1 \cdot b = c - b \][/tex]
Putting these back into the determinant expansion, we get:
[tex]\[ = 1 \cdot bc \cdot (c - b) - a \cdot (c^2 - b^2) + a^2 \cdot (c - b) \][/tex]
[tex]\[ = bc \cdot (c - b) - a(c^2 - b^2) + a^2(c - b) \][/tex]
Distribute and simplify each term:
[tex]\[ = bc \cdot (c - b) - a(c^2 - b^2) + a^2(c - b) \][/tex]
Combine like terms:
[tex]\[ = bc(c - b) + a^2(c - b) - a(c^2 - b^2) \][/tex]
Factoring out common terms, we get:
[tex]\[ = (c - b)(bc + a^2) + a(b^2 - c^2) \][/tex]
Finally, we can state the determinant of the second matrix is:
[tex]\[ - a^2b + a^2c + ab^2 - ac^2 - b^2c + bc^2 \][/tex]
Both determinants simplify to the same expression:
[tex]\[ - a^2b + a^2c + ab^2 - ac^2 - b^2c + bc^2 \][/tex]
So, we can confirm that:
[tex]\[ \begin{vmatrix} 1 & bc & b+c \\ 1 & ca & c+a \\ 1 & ab & a+b \end{vmatrix} = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} \][/tex]
Thus, the equality of the determinants holds.
