Answer on Question #46092-Math-Linear Algebra
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}{c c c} (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
Let
Δ = ∣ ( b + c ) 2 a 2 a 2 b 2 ( c + a ) 2 b 2 c 2 c 2 ( a + b ) 2 ∣ . \Delta = \left| \begin{array}{c c c} (b + c) ^ {2} & a ^ {2} & a ^ {2} \\ b ^ {2} & (c + a) ^ {2} & b ^ {2} \\ c ^ {2} & c ^ {2} & (a + b) ^ {2} \end{array} \right|. Δ = ∣ ∣ ( b + c ) 2 b 2 c 2 a 2 ( c + a ) 2 c 2 a 2 b 2 ( a + b ) 2 ∣ ∣ .
Applying C 1 → C 1 − C 3 C_1 \to C_1 - C_3 C 1 → C 1 − C 3 C 2 → C 2 − C 3 C_2 \to C_2 - C_3 C 2 → C 2 − C 3
Δ = ∣ ( b + c ) 2 − a 2 0 a 2 0 ( c + a ) 2 − b 2 b 2 c 2 − ( a + b ) 2 c 2 − ( a + b ) 2 ( a + b ) 2 ∣ = ( a + b + c ) 2 ∣ b + c − a 0 a 2 0 c + a − b b 2 c − a − b c − a − b ( a + b ) 2 ∣ \Delta = \left| \begin{array}{c c c} (b + c) ^ {2} - a ^ {2} & 0 & a ^ {2} \\ 0 & (c + a) ^ {2} - b ^ {2} & b ^ {2} \\ c ^ {2} - (a + b) ^ {2} & c ^ {2} - (a + b) ^ {2} & (a + b) ^ {2} \end{array} \right| = (a + b + c) ^ {2} \left| \begin{array}{c c c} b + c - a & 0 & a ^ {2} \\ 0 & c + a - b & b ^ {2} \\ c - a - b & c - a - b & (a + b) ^ {2} \end{array} \right| Δ = ∣ ∣ ( b + c ) 2 − a 2 0 c 2 − ( a + b ) 2 0 ( c + a ) 2 − b 2 c 2 − ( a + b ) 2 a 2 b 2 ( a + b ) 2 ∣ ∣ = ( a + b + c ) 2 ∣ ∣ b + c − a 0 c − a − b 0 c + a − b c − a − b a 2 b 2 ( a + b ) 2 ∣ ∣
here, take common ( a + b + c ) (a + b + c) ( a + b + c ) C 1 C_1 C 1 C 2 C_2 C 2
Now, applying R 3 → R 3 − ( R 1 + R 2 ) R_3 \to R_3 - (R_1 + R_2) R 3 → R 3 − ( R 1 + R 2 )
Δ = ( a + b + c ) 2 ∣ b + c − a 0 a 2 0 c + a − b b 2 − 2 b − 2 a 2 a b ∣ = ( a + b + c ) 2 a b ∣ a b + a c − a 2 0 a 2 0 c + a − b b 2 − 2 a b − 2 a b 2 a b ∣ , \Delta = (a + b + c) ^ {2} \left| \begin{array}{c c c} b + c - a & 0 & a ^ {2} \\ 0 & c + a - b & b ^ {2} \\ - 2 b & - 2 a & 2 a b \end{array} \right| = \frac {(a + b + c) ^ {2}}{a b} \left| \begin{array}{c c c} a b + a c - a ^ {2} & 0 & a ^ {2} \\ 0 & c + a - b & b ^ {2} \\ - 2 a b & - 2 a b & 2 a b \end{array} \right|, Δ = ( a + b + c ) 2 ∣ ∣ b + c − a 0 − 2 b 0 c + a − b − 2 a a 2 b 2 2 ab ∣ ∣ = ab ( a + b + c ) 2 ∣ ∣ ab + a c − a 2 0 − 2 ab 0 c + a − b − 2 ab a 2 b 2 2 ab ∣ ∣ ,
applying C 1 → a C 1 C_1 \to aC_1 C 1 → a C 1 C 2 → b C 2 C_2 \to bC_2 C 2 → b C 2
Δ = ( a + b + c ) 2 a b ∣ a b + a c a 2 a 2 b 2 b c + b a b 2 0 0 2 a b ∣ , \Delta = \frac {(a + b + c) ^ {2}}{a b} \left| \begin{array}{c c c} a b + a c & a ^ {2} & a ^ {2} \\ b ^ {2} & b c + b a & b ^ {2} \\ 0 & 0 & 2 a b \end{array} \right|, Δ = ab ( a + b + c ) 2 ∣ ∣ ab + a c b 2 0 a 2 b c + ba 0 a 2 b 2 2 ab ∣ ∣ ,
applying C 1 → C 1 + C 3 C_1 \to C_1 + C_3 C 1 → C 1 + C 3 C 2 → C 2 + C 3 C_2 \to C_2 + C_3 C 2 → C 2 + C 3
Δ = ( a + b + c ) 2 a b a b ⋅ 2 a b ∣ b + c a a b c + a b 0 0 1 ∣ , \Delta = \frac {(a + b + c) ^ {2}}{a b} a b \cdot 2 a b \left| \begin{array}{c c c} b + c & a & a \\ b & c + a & b \\ 0 & 0 & 1 \end{array} \right|, Δ = ab ( a + b + c ) 2 ab ⋅ 2 ab ∣ ∣ b + c b 0 a c + a 0 a b 1 ∣ ∣ ,
here, take a , b , c a, b, c a , b , c R 1 , R 2 R_1, R_2 R 1 , R 2 R 3 R_3 R 3
Δ = 2 a b ( a + b + c ) 2 ∣ b + c a b c + a ∣ , \Delta = 2 a b (a + b + c) ^ {2} \left| \begin{array}{c c} b + c & a \\ b & c + a \end{array} \right|, Δ = 2 ab ( a + b + c ) 2 ∣ ∣ b + c b a c + a ∣ ∣ ,
expand along R 3 R_3 R 3
Δ = 2 a b ( a + b + c ) 2 [ ( b + c ) ( c + a ) − a b ] = 2 a b ( a + b + c ) 2 [ a b + a c + b c + c 2 − a b ] = 2 a b c ( a + b + c ) 2 ( a + b + c ) = 2 a b c ( a + b + c ) 3 . \begin{array}{l}
\Delta = 2 a b (a + b + c) ^ {2} \left[ (b + c) (c + a) - a b \right] = 2 a b (a + b + c) ^ {2} [ a b + a c + b c + c ^ {2} - a b ] \\
= 2 a b c (a + b + c) ^ {2} (a + b + c) = 2 a b c (a + b + c) ^ {3}.
\end{array} Δ = 2 ab ( a + b + c ) 2 [ ( b + c ) ( c + a ) − ab ] = 2 ab ( a + b + c ) 2 [ ab + a c + b c + c 2 − ab ] = 2 ab c ( a + b + c ) 2 ( a + b + c ) = 2 ab c ( a + b + c ) 3 .
Answer: 2 a b c ( a + b + c ) 3 2a b c(a + b + c)^{3} 2 ab c ( a + b + c ) 3
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#339914 on Dec 2023