Answer on Question #44408 – Math – Linear Algebra
Question. Use the properties of determinants to evaluate the following determinant:
∣ ( b + c ) 2 a 2 a 2 b 2 ( c + a ) 2 b 2 c 2 c 2 ( a + b ) 2 ∣ . \left| \begin{array}{ccc} (b + c)^2 & a^2 & a^2 \\ b^2 & (c + a)^2 & b^2 \\ c^2 & c^2 & (a + b)^2 \end{array} \right|. ∣ ∣ ( b + c ) 2 b 2 c 2 a 2 ( c + a ) 2 c 2 a 2 b 2 ( a + b ) 2 ∣ ∣ .
Solution. Multiply the second column by − 1 -1 − 1
∣ ( b + c ) 2 a 2 a 2 b 2 ( c + a ) 2 b 2 c 2 c 2 ( a + b ) 2 ∣ = ∣ ( b + c ) 2 a 2 0 b 2 ( c + a ) 2 ( a + b + c ) ( b − a − c ) c 2 c 2 ( a + b + c ) ( a + b − c ) ∣ . \left| \begin{array}{ccc} (b + c)^2 & a^2 & a^2 \\ b^2 & (c + a)^2 & b^2 \\ c^2 & c^2 & (a + b)^2 \end{array} \right| = \left| \begin{array}{ccc} (b + c)^2 & a^2 & 0 \\ b^2 & (c + a)^2 & (a + b + c)(b - a - c) \\ c^2 & c^2 & (a + b + c)(a + b - c) \end{array} \right|. ∣ ∣ ( b + c ) 2 b 2 c 2 a 2 ( c + a ) 2 c 2 a 2 b 2 ( a + b ) 2 ∣ ∣ = ∣ ∣ ( b + c ) 2 b 2 c 2 a 2 ( c + a ) 2 c 2 0 ( a + b + c ) ( b − a − c ) ( a + b + c ) ( a + b − c ) ∣ ∣ .
Further, multiply the first column by − 1 -1 − 1
∣ ( b + c ) 2 a 2 0 b 2 ( c + a ) 2 ( a + b + c ) ( b − a − c ) c 2 c 2 ( a + b + c ) ( a + b − c ) ∣ = ∣ ( b + c ) 2 ( a + b + c ) ( a − b − c ) 0 b 2 ( a + b + c ) ( a + c − b ) ( a + b + c ) ( b − a − c ) c 2 0 ( a + b + c ) ( a + b − c ) ∣ . \left| \begin{array}{ccc} (b + c)^2 & a^2 & 0 \\ b^2 & (c + a)^2 & (a + b + c)(b - a - c) \\ c^2 & c^2 & (a + b + c)(a + b - c) \end{array} \right| = \left| \begin{array}{ccc} (b + c)^2 & (a + b + c)(a - b - c) & 0 \\ b^2 & (a + b + c)(a + c - b) & (a + b + c)(b - a - c) \\ c^2 & 0 & (a + b + c)(a + b - c) \end{array} \right|. ∣ ∣ ( b + c ) 2 b 2 c 2 a 2 ( c + a ) 2 c 2 0 ( a + b + c ) ( b − a − c ) ( a + b + c ) ( a + b − c ) ∣ ∣ = ∣ ∣ ( b + c ) 2 b 2 c 2 ( a + b + c ) ( a − b − c ) ( a + b + c ) ( a + c − b ) 0 0 ( a + b + c ) ( b − a − c ) ( a + b + c ) ( a + b − c ) ∣ ∣ .
Take the common factor a + b + c a + b + c a + b + c
∣ ( b + c ) 2 ( a + b + c ) ( a − b − c ) 0 b 2 ( a + b + c ) ( a + c − b ) ( a + b + c ) ( b − a − c ) c 2 0 ( a + b + c ) ( a + b − c ) ∣ = ( a + b + c ) 2 ∣ ( b + c ) 2 a − b − c 0 b 2 a + c − b b − a − c c 2 0 a + b − c ∣ . \left| \begin{array}{ccc} (b + c)^2 & (a + b + c)(a - b - c) & 0 \\ b^2 & (a + b + c)(a + c - b) & (a + b + c)(b - a - c) \\ c^2 & 0 & (a + b + c)(a + b - c) \end{array} \right| = (a + b + c)^2 \left| \begin{array}{ccc} (b + c)^2 & a - b - c & 0 \\ b^2 & a + c - b & b - a - c \\ c^2 & 0 & a + b - c \end{array} \right|. ∣ ∣ ( b + c ) 2 b 2 c 2 ( a + b + c ) ( a − b − c ) ( a + b + c ) ( a + c − b ) 0 0 ( a + b + c ) ( b − a − c ) ( a + b + c ) ( a + b − c ) ∣ ∣ = ( a + b + c ) 2 ∣ ∣ ( b + c ) 2 b 2 c 2 a − b − c a + c − b 0 0 b − a − c a + b − c ∣ ∣ .
In the last determinant, add the third column to the second column:
( a + b + c ) 2 ∣ ( b + c ) 2 a − b − c 0 b 2 a + c − b b − a − c c 2 0 a + b − c ∣ = ( a + b + c ) 2 ∣ ( b + c ) 2 a − b − c 0 b 2 0 b − a − c c 2 a + b − c a + b − c ∣ . (a + b + c)^2 \left| \begin{array}{ccc} (b + c)^2 & a - b - c & 0 \\ b^2 & a + c - b & b - a - c \\ c^2 & 0 & a + b - c \end{array} \right| = (a + b + c)^2 \left| \begin{array}{ccc} (b + c)^2 & a - b - c & 0 \\ b^2 & 0 & b - a - c \\ c^2 & a + b - c & a + b - c \end{array} \right|. ( a + b + c ) 2 ∣ ∣ ( b + c ) 2 b 2 c 2 a − b − c a + c − b 0 0 b − a − c a + b − c ∣ ∣ = ( a + b + c ) 2 ∣ ∣ ( b + c ) 2 b 2 c 2 a − b − c 0 a + b − c 0 b − a − c a + b − c ∣ ∣ .
Further, add the second row to the third row:
( a + b + c ) 2 ∣ ( b + c ) 2 a − b − c 0 b 2 0 b − a − c c 2 a + b − c a + b − c ∣ = . (a + b + c)^2 \left| \begin{array}{ccc} (b + c)^2 & a - b - c & 0 \\ b^2 & 0 & b - a - c \\ c^2 & a + b - c & a + b - c \end{array} \right| =. ( a + b + c ) 2 ∣ ∣ ( b + c ) 2 b 2 c 2 a − b − c 0 a + b − c 0 b − a − c a + b − c ∣ ∣ = . = ( a + b + c ) 2 ∣ ( b + c ) 2 a − b − c 0 b 2 0 b − a − c b 2 + c 2 a + b − c 2 b − 2 c ∣ . Multiply the third row by − 1 and add it to the first row. Further take the common factor 2 from the first row: = (a + b + c) ^ {2} \left| \begin{array}{c c c} (b + c) ^ {2} & a - b - c & 0 \\ b ^ {2} & 0 & b - a - c \\ b ^ {2} + c ^ {2} & a + b - c & 2 b - 2 c \end{array} \right|. \text{Multiply the third row by } -1 \text{ and add it to the first row. Further take the common factor 2 from the first row:} = ( a + b + c ) 2 ∣ ∣ ( b + c ) 2 b 2 b 2 + c 2 a − b − c 0 a + b − c 0 b − a − c 2 b − 2 c ∣ ∣ . Multiply the third row by − 1 and add it to the first row. Further take the common factor 2 from the first row: ( a + b + c ) 2 ∣ ( b + c ) 2 a − b − c 0 b 2 0 b − a − c b 2 + c 2 a + b − c 2 b − 2 c ∣ = = ( a + b + c ) 2 ∣ 2 b c − 2 b 2 c − 2 b b 2 0 b − a − c b 2 + c 2 a + b − c 2 b − 2 c ∣ = \begin{array}{l} (a + b + c) ^ {2} \left| \begin{array}{c c c} (b + c) ^ {2} & a - b - c & 0 \\ b ^ {2} & 0 & b - a - c \\ b ^ {2} + c ^ {2} & a + b - c & 2 b - 2 c \end{array} \right| = \\ = (a + b + c) ^ {2} \left| \begin{array}{c c c} 2 b c & - 2 b & 2 c - 2 b \\ b ^ {2} & 0 & b - a - c \\ b ^ {2} + c ^ {2} & a + b - c & 2 b - 2 c \end{array} \right| = \\ \end{array} ( a + b + c ) 2 ∣ ∣ ( b + c ) 2 b 2 b 2 + c 2 a − b − c 0 a + b − c 0 b − a − c 2 b − 2 c ∣ ∣ = = ( a + b + c ) 2 ∣ ∣ 2 b c b 2 b 2 + c 2 − 2 b 0 a + b − c 2 c − 2 b b − a − c 2 b − 2 c ∣ ∣ = = 2 ( a + b + c ) 2 ∣ b c − b c − b b 2 0 b − a − c b 2 + c 2 a + b − c 2 b − 2 c ∣ . Multiply the second row by − 1 and add it to the third row: = 2 (a + b + c) ^ {2} \left| \begin{array}{c c c} b c & - b & c - b \\ b ^ {2} & 0 & b - a - c \\ b ^ {2} + c ^ {2} & a + b - c & 2 b - 2 c \end{array} \right|. \text{Multiply the second row by } -1 \text{ and add it to the third row:} = 2 ( a + b + c ) 2 ∣ ∣ b c b 2 b 2 + c 2 − b 0 a + b − c c − b b − a − c 2 b − 2 c ∣ ∣ . Multiply the second row by − 1 and add it to the third row: 2 ( a + b + c ) 2 ∣ b c − b c − b b 2 0 b − a − c b 2 + c 2 a + b − c 2 b − 2 c ∣ = 2 (a + b + c) ^ {2} \left| \begin{array}{c c c} b c & - b & c - b \\ b ^ {2} & 0 & b - a - c \\ b ^ {2} + c ^ {2} & a + b - c & 2 b - 2 c \end{array} \right| = 2 ( a + b + c ) 2 ∣ ∣ b c b 2 b 2 + c 2 − b 0 a + b − c c − b b − a − c 2 b − 2 c ∣ ∣ = = 2 ( a + b + c ) 2 ∣ b c − b c − b b 2 0 b − a − c c 2 a + b − c a + b − c ∣ . Multiply the second column by − 1 and add it to the third column: = 2 (a + b + c) ^ {2} \left| \begin{array}{c c c} b c & - b & c - b \\ b ^ {2} & 0 & b - a - c \\ c ^ {2} & a + b - c & a + b - c \end{array} \right|. \text{Multiply the second column by } -1 \text{ and add it to the third column:} = 2 ( a + b + c ) 2 ∣ ∣ b c b 2 c 2 − b 0 a + b − c c − b b − a − c a + b − c ∣ ∣ . Multiply the second column by − 1 and add it to the third column: 2 ( a + b + c ) 2 ∣ b c − b c − b b 2 0 b − a − c c 2 a + b − c a + b − c ∣ = 2 (a + b + c) ^ {2} \left| \begin{array}{c c c} b c & - b & c - b \\ b ^ {2} & 0 & b - a - c \\ c ^ {2} & a + b - c & a + b - c \end{array} \right| = 2 ( a + b + c ) 2 ∣ ∣ b c b 2 c 2 − b 0 a + b − c c − b b − a − c a + b − c ∣ ∣ = = 2 ( a + b + c ) 2 ∣ b c − b c b 2 0 b − a − c c 2 a + b − c 0 ∣ . We shall evaluate the last determinant using the rule of triangle: = 2 (a + b + c) ^ {2} \left| \begin{array}{c c c} b c & - b & c \\ b ^ {2} & 0 & b - a - c \\ c ^ {2} & a + b - c & 0 \end{array} \right|. \text{We shall evaluate the last determinant using the rule of triangle:} = 2 ( a + b + c ) 2 ∣ ∣ b c b 2 c 2 − b 0 a + b − c c b − a − c 0 ∣ ∣ . We shall evaluate the last determinant using the rule of triangle: 2 ( a + b + c ) 2 ∣ b c − b c b 2 0 b − a − c c 2 a + b − c 0 ∣ = 2 (a + b + c) ^ {2} \left| \begin{array}{c c c} b c & - b & c \\ b ^ {2} & 0 & b - a - c \\ c ^ {2} & a + b - c & 0 \end{array} \right| = 2 ( a + b + c ) 2 ∣ ∣ b c b 2 c 2 − b 0 a + b − c c b − a − c 0 ∣ ∣ = = 2 ( a + b + c ) 2 [ − b c 2 ( b − a − c ) + b 2 c ( a + b − c ) − b c ( a + b − c ) ( b − a − c ) ] = = 2 b c ( a + b + c ) 2 [ − b c + a c + c 2 + a b + b 2 − b c − a b + a 2 + a c − b 2 + a b + b c + b c − \begin{array}{l} = 2 (a + b + c) ^ {2} \left[ - b c ^ {2} (b - a - c) + b ^ {2} c (a + b - c) - b c (a + b - c) (b - a - c) \right] = \\ = 2 b c (a + b + c) ^ {2} \left[ - b c + a c + c ^ {2} + a b + b ^ {2} - b c - a b + a ^ {2} + a c - b ^ {2} + a b + b c + b c - \right. \\ \end{array} = 2 ( a + b + c ) 2 [ − b c 2 ( b − a − c ) + b 2 c ( a + b − c ) − b c ( a + b − c ) ( b − a − c ) ] = = 2 b c ( a + b + c ) 2 [ − b c + a c + c 2 + ab + b 2 − b c − ab + a 2 + a c − b 2 + ab + b c + b c − − a c − c 2 ] = 2 b c ( a + b + c ) 2 [ a 2 + a b + a c ] = 2 a b c ( a + b + c ) 3 . - a c - c ^ {2} \quad ] = 2 b c (a + b + c) ^ {2} [ a ^ {2} + a b + a c ] = 2 a b c (a + b + c) ^ {3}. − a c − c 2 ] = 2 b c ( a + b + c ) 2 [ a 2 + ab + a c ] = 2 ab c ( a + b + c ) 3 .
Answer. ∣ ( b + c ) 2 a 2 a 2 b 2 ( c + a ) 2 b 2 c 2 c 2 ( a + b ) 2 ∣ = 2 a b c ( a + b + c ) 3 . \left| \begin{array}{ccc} (b + c)^2 & a^2 & a^2 \\ b^2 & (c + a)^2 & b^2 \\ c^2 & c^2 & (a + b)^2 \end{array} \right| = 2 a b c (a + b + c)^3. ∣ ∣ ( b + c ) 2 b 2 c 2 a 2 ( c + a ) 2 c 2 a 2 b 2 ( a + b ) 2 ∣ ∣ = 2 ab c ( a + b + c ) 3 .
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