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

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 R1 → R1 - R2 and R2 → R2 - R1]
$=\begin{vmatrix}(\text{b}+\text{c})(\text{b}+2\text{c}+\text{a})&(\text{a}+\text{b})(\text{a}-\text{b})&\text{c}(\text{b}-\text{a})\$\text{c}-\text{a})(\text{b}+2\text{a}+\text{c})&(\text{b}-\text{c})(\text{b}+\text{c})&\text{a}(\text{c}-\text{b})\$\text{a}+\text{b})^2&\text{c}^2&\text{ab}\end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}-(\text{b}+2\text{c}+\text{a})&\text{a}+\text{b}&-\text{c}\\-(\text{b}+2\text{a}+\text{c})&\text{b}+\text{c}&-\text{a}\$\text{a}+\text{b})^2&\text{c}^2&\text{ab}\end{vmatrix}$
[Applying x2 - y2 = (x + y)(x - y) and taking out (a - b) common from R1 and (b - c) from R2]
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}-2(\text{b}+\text{c}+\text{a})&\text{a}+\text{b}&-\text{c}\\-2(\text{b}+\text{a}+\text{c})&\text{b}+\text{c}&-\text{a}\$\text{a}+\text{b})^2-\text{c}^2&\text{c}^2&\text{ab}\end{vmatrix}$ [Applying C1 → C1 - C2]
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}-2(\text{b}+\text{c}+\text{a})&\text{a}+\text{b}&-\text{c}\\-2(\text{b}+\text{a}+\text{c})&\text{b}+\text{c}&-\text{a}\$\text{a}+\text{b}+\text{c})(\text{a}+\text{b}-\text{c})&\text{c}^2&\text{ab}\end{vmatrix}$
[Applying x2 - y2 = (x + y)(x - y) in C1]
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})\begin{vmatrix}-2&\text{a}+\text{b}&-\text{c}\\-2&\text{b}+\text{c}&-\text{a}\$\text{a}+\text{b}-\text{c})&\text{c}^2&\text{ab}\end{vmatrix}$
[Taking out (a + b + c) common from C1]
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})\begin{vmatrix}-2&\text{a}+\text{b}&-\text{c}\\0&\text{c}-\text{a}&\text{c}-\text{a}\$\text{a}+\text{b}-\text{c})&\text{c}^2&\text{ab}\end{vmatrix}$
[Applying R2 → R2 - R1]
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a})\begin{vmatrix}-2&\text{a}+\text{b}&-\text{c}\\0&1&1\$\text{a}+\text{b}-\text{c})&\text{c}^2&\text{ab}\end{vmatrix}$
[Taking out (c - a) common from R2]
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a})\begin{vmatrix}-2&\text{a}+\text{b}+\text{c}&-\text{c}\\0&0&1\$\text{a}+\text{b}-\text{c})&\text{c}^2-\text{ab}&\text{ab}\end{vmatrix}$
[Applying C2 → C2 - C3]
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a}) \left\{(-1)\begin{vmatrix}-2&\text{a}+\text{b}+\text{c}&\$\text{a}+\text{b}-\text{c})&\text{c}^2-\text{ab}\end{vmatrix}\right\}$ [Expanding along R2]
$=-(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a})\{-2\text{c}^2+2\text{ab}-\text{a}^2-\text{b}^2-2\text{ab}+\text{c}^2\}$
$=-(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a})(-\text{a}^2-\text{b}^2-\text{c}^2)$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2)$
$=\text{R.H.S}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Find the distance of the point (3, 3, 3) from the plane $\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})+9=0$
If $\text{x}=2\cos\text{t}-\cos2\text{t},\text{y}=2\sin\text{t}-\sin2\text{t},$ find $\frac{\text{d}^2\text{y}}{\text{dx}^2}\ \text{at}\ \text{t}=\frac{\pi}{2}.$
$\text{if y } = \sqrt{\frac{(x - 3)(x^{2} + 4)}{3x^{2} + 4x + 5}}, \text{find} \frac{dy}{dx}$
Find the shortest distance between the following pairs of lines whose vector equation are:
$\vec{\text{r}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}\big)$ and $\vec{\text{r}}=\big(2\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}\big)+\mu\big(3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}\big)$
Evaluate the following integrals:
$\int\limits^1_{-1}|\text{x}\cos\pi\text{x}|\text{dx}$
A shopkeeper has 3 varieties of pens 'A', 'B' and 'C'. Meenu purchased 1 pen of each variety for a total of Rs 21. Jeevan purchased 4 pens of 'A' variety 3 pens of 'B' variety and 2 pens of 'C' variety for Rs 60. While Shikha purchased 6 pens of 'A' variety, 2 pens of 'B' variety and 3 pens of 'C' variety for Rs 70. Using matrix method, find cost of each variety of pen.
If $\overrightarrow{\text{a}}\text{ and } \overrightarrow{\text{b}}$are two vectors such that $|\overrightarrow{\text{a}} + \overrightarrow{\text{b}}| = | \overrightarrow{\text{a}}|,$ then prove that vector 2 $\overrightarrow{\text{a}} + \overrightarrow{\text{b}}$is perpendicular to vector$\overrightarrow{\text{b}}.$
Find the area of the region in the first quadrant enclosed by x-axis, line $\text{x}=\sqrt3\text{ y}$ and the circle x2 + y2 = 4.
Differentiate the following functions with respect to x:
$\sqrt{\frac{1+\sin\text{x}}{1-\sin\text{x}}}$
Write the points where f(x) = |loge x| is not differentiable.