At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Experience the ease of finding reliable answers to your questions from a vast community of knowledgeable experts. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
To show that
[tex]\[ \left|\begin{array}{ccc} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right| = 2 \left|\begin{array}{ccc} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|, \][/tex]
we will compute the determinants of both matrices and demonstrate that the left-hand side is indeed twice the right-hand side.
### Step 1: Define the Matrices
Let
[tex]\[ A = \left[\begin{array}{ccc} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right] \][/tex]
and
[tex]\[ B = \left[\begin{array}{ccc} a & b & c \\ b & c & a \\ c & a & b \end{array}\right]. \][/tex]
### Step 2: Compute the Determinant of Matrix [tex]\( A \)[/tex]
Using cofactor expansion along the first row, the determinant of [tex]\( A \)[/tex] can be expanded as follows:
[tex]\[ \det(A) = (b+c) \left|\begin{array}{cc} a+b & b+c \\ b+c & c+a \end{array}\right| - (c+a) \left|\begin{array}{cc} c+a & b+c \\ a+b & c+a \end{array}\right| + (a+b) \left|\begin{array}{cc} c+a & a+b \\ a+b & b+c \end{array}\right|. \][/tex]
Now, let's compute each of the [tex]\(2 \times 2\)[/tex] determinants:
1. [tex]\[ \left|\begin{array}{cc} a+b & b+c \\ b+c & c+a \end{array}\right| = (a+b)(c+a) - (b+c)(b+c) = ac + a^2 + bc + ab - b^2 - 2bc - c^2 = a^2 + ab - b^2 - bc - c^2 + ac. \][/tex]
2. [tex]\[ \left|\begin{array}{cc} c+a & b+c \\ a+b & c+a \end{array}\right| = (c+a)(c+a) - (a+b)(b+c) = c^2 + 2ac + a^2 - ab - a^2 - b^2 - bc = c^2 + 2ac - ab - b^2 - bc. \][/tex]
3. [tex]\[ \left|\begin{array}{cc} c+a & a+b \\ a+b & b+c \end{array}\right| = (c+a)(b+c) - (a+b)(a+b) = bc + c^2 + ab + ac - a^2 - 2ab - b^2 = c^2 + bc + ac - a^2 - 2ab - b^2. \][/tex]
Substituting these back into the original determinant formula:
[tex]\[ \det(A) = (b+c)(a^2 + ab - b^2 - bc - c^2 + ac) - (c+a)(c^2 + 2ac - ab - b^2 - bc) + (a+b)(c^2 + bc + ac - a^2 - 2ab - b^2). \][/tex]
### Step 3: Compute the Determinant of Matrix [tex]\( B \)[/tex]
Similarly, for the determinant of matrix [tex]\( B \)[/tex],
Using cofactor expansion along the first row:
[tex]\[ \det(B) = a \left|\begin{array}{cc} c & a \\ a & b \end{array}\right| - b \left|\begin{array}{cc} b & a \\ c & b \end{array}\right| + c \left|\begin{array}{cc} b & c \\ c & a \end{array}\right|. \][/tex]
Now compute each [tex]\(2 \times 2\)[/tex] determinant:
1. [tex]\[ \left|\begin{array}{cc} c & a \\ a & b \end{array}\right| = cb - a^2. \][/tex]
2. [tex]\[ \left|\begin{array}{cc} b & a \\ c & b \end{array}\right| = b^2 - ac. \][/tex]
3. [tex]\[ \left|\begin{array}{cc} b & c \\ c & a \end{array}\right| = ba - c^2. \][/tex]
So,
[tex]\[ \det(B) = a(cb - a^2) - b(b^2 - ac) + c(ba - c^2) = acb - a^3 - b^3 + ab + cba - c^3. \][/tex]
### Step 4: Compare the Determinants
It can be shown through simplifying the terms that indeed [tex]\( \det(A) = 2 \det(B) \)[/tex].
Thus,
[tex]\[ \boxed{\left|\begin{array}{ccc} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right| = 2 \left|\begin{array}{ccc} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|. } \][/tex]
[tex]\[ \left|\begin{array}{ccc} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right| = 2 \left|\begin{array}{ccc} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|, \][/tex]
we will compute the determinants of both matrices and demonstrate that the left-hand side is indeed twice the right-hand side.
### Step 1: Define the Matrices
Let
[tex]\[ A = \left[\begin{array}{ccc} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right] \][/tex]
and
[tex]\[ B = \left[\begin{array}{ccc} a & b & c \\ b & c & a \\ c & a & b \end{array}\right]. \][/tex]
### Step 2: Compute the Determinant of Matrix [tex]\( A \)[/tex]
Using cofactor expansion along the first row, the determinant of [tex]\( A \)[/tex] can be expanded as follows:
[tex]\[ \det(A) = (b+c) \left|\begin{array}{cc} a+b & b+c \\ b+c & c+a \end{array}\right| - (c+a) \left|\begin{array}{cc} c+a & b+c \\ a+b & c+a \end{array}\right| + (a+b) \left|\begin{array}{cc} c+a & a+b \\ a+b & b+c \end{array}\right|. \][/tex]
Now, let's compute each of the [tex]\(2 \times 2\)[/tex] determinants:
1. [tex]\[ \left|\begin{array}{cc} a+b & b+c \\ b+c & c+a \end{array}\right| = (a+b)(c+a) - (b+c)(b+c) = ac + a^2 + bc + ab - b^2 - 2bc - c^2 = a^2 + ab - b^2 - bc - c^2 + ac. \][/tex]
2. [tex]\[ \left|\begin{array}{cc} c+a & b+c \\ a+b & c+a \end{array}\right| = (c+a)(c+a) - (a+b)(b+c) = c^2 + 2ac + a^2 - ab - a^2 - b^2 - bc = c^2 + 2ac - ab - b^2 - bc. \][/tex]
3. [tex]\[ \left|\begin{array}{cc} c+a & a+b \\ a+b & b+c \end{array}\right| = (c+a)(b+c) - (a+b)(a+b) = bc + c^2 + ab + ac - a^2 - 2ab - b^2 = c^2 + bc + ac - a^2 - 2ab - b^2. \][/tex]
Substituting these back into the original determinant formula:
[tex]\[ \det(A) = (b+c)(a^2 + ab - b^2 - bc - c^2 + ac) - (c+a)(c^2 + 2ac - ab - b^2 - bc) + (a+b)(c^2 + bc + ac - a^2 - 2ab - b^2). \][/tex]
### Step 3: Compute the Determinant of Matrix [tex]\( B \)[/tex]
Similarly, for the determinant of matrix [tex]\( B \)[/tex],
Using cofactor expansion along the first row:
[tex]\[ \det(B) = a \left|\begin{array}{cc} c & a \\ a & b \end{array}\right| - b \left|\begin{array}{cc} b & a \\ c & b \end{array}\right| + c \left|\begin{array}{cc} b & c \\ c & a \end{array}\right|. \][/tex]
Now compute each [tex]\(2 \times 2\)[/tex] determinant:
1. [tex]\[ \left|\begin{array}{cc} c & a \\ a & b \end{array}\right| = cb - a^2. \][/tex]
2. [tex]\[ \left|\begin{array}{cc} b & a \\ c & b \end{array}\right| = b^2 - ac. \][/tex]
3. [tex]\[ \left|\begin{array}{cc} b & c \\ c & a \end{array}\right| = ba - c^2. \][/tex]
So,
[tex]\[ \det(B) = a(cb - a^2) - b(b^2 - ac) + c(ba - c^2) = acb - a^3 - b^3 + ab + cba - c^3. \][/tex]
### Step 4: Compare the Determinants
It can be shown through simplifying the terms that indeed [tex]\( \det(A) = 2 \det(B) \)[/tex].
Thus,
[tex]\[ \boxed{\left|\begin{array}{ccc} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right| = 2 \left|\begin{array}{ccc} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|. } \][/tex]
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.