Prove that: (b+c)^2&a^2&bc c+a)^2&b^2&ca a+b)^2&c^2&ab =(a-b)(b-c)(c-b)(a+b+c)(a^2+b^2+c^2)

L.H.S= (b+c)^2&a^2&bc c+a)^2&b^2&ca a+b)^2&c^2&ab = (b+c)^2-(c+a)^2&a^2-b^2&bc-ca (c+a)^2-(a+b)^2&b^2-c^2&ca-ab (a+b)^2&c^2&ab [Applying R_1 R_1-R_2 and R_2 R_2-R_3] = (b-a)(b+2c+a)&(a+b)(a-b)&c(b-a) c-a)(b+2a+c)&(b-c)(b+c)&a(c-b) a+b)^2&c^2&ab =(a-b)(b-c) -(b+2c+a)&a+b&-c -(b+2a+c)&b+c&-a a+b)^2&c^2&ab [Applying x_2 - y_2 = (x + y)(x - y) and taking out (a - b) common from R_1 and (b - c) from R_2] =(a-b)(b-c) -2(b+c+a)&a+b&-c -2(b+a+c)&b+c&-a a+b)^2-c^2&c^2&ab [Applying C_1 C_1-C_2] =(a-b)(b-c) -2(b+c+a)&a+b&-c -2(b+a+c)&b+c&-a a+b+c)(a+b-c)&c^2&ab [Applying x^2 - y^2 = (x + y)(x - y) in C_1] =(a-b)(b-c)(a+b+c) -2&a+b&-c -2&b+c&-a a+b-c)&c^2&ab [Taking out (a + b + c) common from C_1] =(a-b)(b-c)(a+b+c) -2&a+b&-c 0&c-a&c-a a+b-c)&c^2&ab [Applying R_2 → R_2 - R_1] =(a-b)(b-c)(a+b+c)(c-a) -2&a+b&-c 0&1&1 a+b-c)&c^2&ab [Taking out (c - a) common from R_2] =(a-b)(b-c)(a+b+c)(c-a) -2&a+b+c&-c 0&0&1 a+b-c)&c^2-ab&ab [Applying C_2 → C_2 - C_3] =(a-b)(b-c)(a+b+c)(c-a) (-1) -2&a+b+c&(a+b-c)&c^2-ab [Expanding along R_2] =-(a-b)(b-c)(a+b+c)(c-a) -2c^2+2ab-a^2-b^2-2ab+c^2 =-(a-b)(b-c)(a+b+c)(c-a)(-a^2-b^2-c^2) =(a-b)(b-c)(c-b)(a+b+c)(a^2+b^2+c^2) =R.H.S

Prove that: (b+c)^2&a^2&bc c+a)^2&b^2&ca a+b)^2&c^2&ab =(a-b)(b-c…

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Prove that: $\begin{vmatrix}(\text{b}+\text{c})^2&\text{a}^2&\text{bc}\\(\text{c}+\text{a})^2&\text{b}^2&\text{ca}\\(\text{a}+\text{b})^2&\text{c}^2&\text{ab}\end{vmatrix}$ $=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2)$

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Answer

$\text{L.H.S}=\begin{vmatrix}(\text{b}+\text{c})^2&\text{a}^2&\text{bc}\\(\text{c}+\text{a})^2&\text{b}^2&\text{ca}\\(\text{a}+\text{b})^2&\text{c}^2&\text{ab}\end{vmatrix}$
$=\begin{vmatrix}(\text{b}+\text{c})^2-(\text{c}+\text{a})^2&\text{a}^2-\text{b}^2&\text{bc}-\text{ca} \\ (\text{c}+\text{a})^2-(\text{a}+\text{b})^2&\text{b}^2-\text{c}^2&\text{ca}-\text{ab} \\ (\text{a}+\text{b})^2&\text{c}^2&\text{ab}\end{vmatrix}$ [Applying $\mathrm{R}_1 \rightarrow \mathrm{R}_1-\mathrm{R}_2$ and $\mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_3]$
$=\begin{vmatrix}(\text{b}-\text{a})(\text{b}+2\text{c}+\text{a})&(\text{a}+\text{b})(\text{a}-\text{b})&\text{c}(\text{b}-\text{a})\\(\text{c}-\text{a})(\text{b}+2\text{a}+\text{c})&(\text{b}-\text{c})(\text{b}+\text{c})&\text{a}(\text{c}-\text{b})\\(\text{a}+\text{b})^2&\text{c}^2&\text{ab}\end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}-(\text{b}+2\text{c}+\text{a})&\text{a}+\text{b}&-\text{c}\\-(\text{b}+2\text{a}+\text{c})&\text{b}+\text{c}&-\text{a}\\(\text{a}+\text{b})^2&\text{c}^2&\text{ab}\end{vmatrix}$ [Applying x$_2$ - y$_2$ = (x + y)(x - y) and taking out (a - b) common from R$_1$ and (b - c) from R$_2$]
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}-2(\text{b}+\text{c}+\text{a})&\text{a}+\text{b}&-\text{c}\\-2(\text{b}+\text{a}+\text{c})&\text{b}+\text{c}&-\text{a}\\(\text{a}+\text{b})^2-\text{c}^2&\text{c}^2&\text{ab}\end{vmatrix}$ [Applying $\mathrm{C}_1 \rightarrow \mathrm{C}_1-\mathrm{C}_2$]
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}-2(\text{b}+\text{c}+\text{a})&\text{a}+\text{b}&-\text{c}\\-2(\text{b}+\text{a}+\text{c})&\text{b}+\text{c}&-\text{a}\\(\text{a}+\text{b}+\text{c})(\text{a}+\text{b}-\text{c})&\text{c}^2&\text{ab}\end{vmatrix}$ [Applying x$^2$ - y$^2$ = (x + y)(x - y) in C$_1$]
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})\begin{vmatrix}-2&\text{a}+\text{b}&-\text{c}\\-2&\text{b}+\text{c}&-\text{a}\\(\text{a}+\text{b}-\text{c})&\text{c}^2&\text{ab}\end{vmatrix}$ [Taking out (a + b + c) common from C$_1$]
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})\begin{vmatrix}-2&\text{a}+\text{b}&-\text{c}\\0&\text{c}-\text{a}&\text{c}-\text{a}\\(\text{a}+\text{b}-\text{c})&\text{c}^2&\text{ab}\end{vmatrix}$ [Applying R$_2$ → R$_2$ - R$_1$]
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a})\begin{vmatrix}-2&\text{a}+\text{b}&-\text{c}\\0&1&1\\(\text{a}+\text{b}-\text{c})&\text{c}^2&\text{ab}\end{vmatrix}$ [Taking out (c - a) common from R$_2$]
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a})\begin{vmatrix}-2&\text{a}+\text{b}+\text{c}&-\text{c}\\0&0&1\\(\text{a}+\text{b}-\text{c})&\text{c}^2-\text{ab}&\text{ab}\end{vmatrix}$ [Applying C$_2$ → C$_2$ - C$_3$]
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a}) \left\{(-1)\begin{vmatrix}-2&\text{a}+\text{b}+\text{c}&(\text{a}+\text{b}-\text{c})&\text{c}^2-\text{ab}\end{vmatrix}\right\}$ [Expanding along R$_2$]
$=-(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a})\{-2\text{c}^2+2\text{ab}-\text{a}^2-\text{b}^2-2\text{ab}+\text{c}^2\}$
$=-(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a})(-\text{a}^2-\text{b}^2-\text{c}^2)$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2)$ $=\text{R.H.S}$

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