Class 12: Determinants – Exercise 6.2

$\displaystyle \textbf{Question 1: }\text{Evaluate the following determinants:}$
$\displaystyle \text{(i) }\begin{vmatrix}1&3&5\\2&6&10\\31&11&38\end{vmatrix}\qquad \text{(ii) }\begin{vmatrix}67&19&21\\39&13&14\\81&24&26\end{vmatrix}$
$\displaystyle \text{(iii) }\begin{vmatrix}a&h&g\\h&b&f\\g&f&c\end{vmatrix}\qquad \text{(iv) }\begin{vmatrix}1&-3&2\\4&-1&2\\3&5&2\end{vmatrix}$
$\displaystyle \text{(v) }\begin{vmatrix}1&4&9\\4&9&16\\9&16&25\end{vmatrix}\qquad \text{(vi) }\begin{vmatrix}6&-3&2\\2&-1&2\\-10&5&2\end{vmatrix}$
$\displaystyle \text{(vii) }\begin{vmatrix}1&3&9&27\\3&9&27&1\\9&27&1&3\\27&1&3&9\end{vmatrix}$
$\displaystyle \text{(viii) }\begin{vmatrix}102&18&36\\1&3&4\\17&3&6\end{vmatrix}\hfill\text{[CBSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i)}$
$\displaystyle \Delta=\begin{vmatrix}1&3&5\\2&6&10\\31&11&38\end{vmatrix}$
$\displaystyle \text{Here, }R_2=2R_1.$
$\displaystyle \therefore \Delta=0$
$\displaystyle \\$

$\displaystyle \text{(ii)}$
$\displaystyle \Delta=\begin{vmatrix}67&19&21\\39&13&14\\81&24&26\end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \Delta=67\begin{vmatrix}13&14\\24&26\end{vmatrix} -19\begin{vmatrix}39&14\\81&26\end{vmatrix} +21\begin{vmatrix}39&13\\81&24\end{vmatrix}$
$\displaystyle =67(338-336)-19(1014-1134)+21(936-1053)$
$\displaystyle =67(2)-19(-120)+21(-117)$
$\displaystyle =134+2280-2457$
$\displaystyle =-43$
$\displaystyle \\$

$\displaystyle \text{(iii)}$
$\displaystyle \Delta=\begin{vmatrix}a&h&g\\h&b&f\\g&f&c\end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \Delta=a\begin{vmatrix}b&f\\f&c\end{vmatrix} -h\begin{vmatrix}h&f\\g&c\end{vmatrix} +g\begin{vmatrix}h&b\\g&f\end{vmatrix}$
$\displaystyle =a(bc-f^2)-h(hc-fg)+g(hf-bg)$
$\displaystyle =abc-af^2-ch^2+fgh+fgh-bg^2$
$\displaystyle =abc+2fgh-af^2-bg^2-ch^2$
$\displaystyle \\$

$\displaystyle \text{(iv)}$
$\displaystyle \Delta=\begin{vmatrix}1&-3&2\\4&-1&2\\3&5&2\end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \Delta=\begin{vmatrix}-1&2\\5&2\end{vmatrix} +3\begin{vmatrix}4&2\\3&2\end{vmatrix} +2\begin{vmatrix}4&-1\\3&5\end{vmatrix}$
$\displaystyle =(-2-10)+3(8-6)+2(20+3)$
$\displaystyle =-12+6+46$
$\displaystyle =40$
$\displaystyle \\$

$\displaystyle \text{(v)}$
$\displaystyle \Delta=\begin{vmatrix}1&4&9\\4&9&16\\9&16&25\end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \Delta=\begin{vmatrix}9&16\\16&25\end{vmatrix} -4\begin{vmatrix}4&16\\9&25\end{vmatrix} +9\begin{vmatrix}4&9\\9&16\end{vmatrix}$
$\displaystyle =(225-256)-4(100-144)+9(64-81)$
$\displaystyle =-31+176-153$
$\displaystyle =-8$
$\displaystyle \\$

$\displaystyle \text{(vi)}$
$\displaystyle \Delta=\begin{vmatrix}6&-3&2\\2&-1&2\\-10&5&2\end{vmatrix}$
$\displaystyle \text{Here, }C_1=-2C_2.$
$\displaystyle \therefore \Delta=0$
$\displaystyle \\$

$\displaystyle \text{(vii)}$
$\displaystyle D=\begin{vmatrix}1&3&9&27\\3&9&27&1\\9&27&1&3\\27&1&3&9\end{vmatrix}$
$\displaystyle \text{Applying }C_2\to C_2-3C_1,\quad C_3\to C_3-9C_1,$
$\displaystyle \text{and }C_4\to C_4-27C_1,$
$\displaystyle D=\begin{vmatrix}1&0&0&0\\3&0&0&-80\\9&0&-80&-240\\27&-80&-240&-720\end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle D=\begin{vmatrix}0&0&-80\\0&-80&-240\\-80&-240&-720\end{vmatrix}$
$\displaystyle \text{Expanding along the first row,}$
$\displaystyle D=(-80)\begin{vmatrix}0&-80\\-80&-240\end{vmatrix}$
$\displaystyle =-80\left[0-(-80)(-80)\right]$
$\displaystyle =-80(-6400)$
$\displaystyle =512000$
$\displaystyle \\$

$\displaystyle \text{(viii)}$
$\displaystyle \Delta=\begin{vmatrix}102&18&36\\1&3&4\\17&3&6\end{vmatrix}$
$\displaystyle \text{Here, }R_1=6R_3.$
$\displaystyle \therefore \Delta=0$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Without expanding, show that the value of each of the following}$
$\displaystyle \text{determinants is zero:}$
$\displaystyle \text{(i) }\begin{vmatrix}8&2&7\\12&3&5\\16&4&3\end{vmatrix}\qquad \text{(ii) }\begin{vmatrix}6&-3&2\\2&-1&2\\-10&5&2\end{vmatrix}$
$\displaystyle \text{(iii) }\begin{vmatrix}2&3&7\\13&17&5\\15&20&12\end{vmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i)}$
$\displaystyle \Delta=\begin{vmatrix}8&2&7\\12&3&5\\16&4&3\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1-4C_2,$
$\displaystyle \Delta=\begin{vmatrix}0&2&7\\0&3&5\\0&4&3\end{vmatrix}$
$\displaystyle \text{Since all the elements of }C_1\text{ are zero,}$
$\displaystyle \therefore \Delta=0$

$\displaystyle \text{(ii)}$
$\displaystyle \Delta=\begin{vmatrix}6&-3&2\\2&-1&2\\-10&5&2\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1+2C_2,$
$\displaystyle \Delta=\begin{vmatrix}0&-3&2\\0&-1&2\\0&5&2\end{vmatrix}$
$\displaystyle \text{Since all the elements of }C_1\text{ are zero,}$
$\displaystyle \therefore \Delta=0$

$\displaystyle \text{(iii)}$
$\displaystyle \Delta=\begin{vmatrix}2&3&7\\13&17&5\\15&20&12\end{vmatrix}$
$\displaystyle \text{Applying }R_3\to R_3-R_1,$
$\displaystyle \Delta=\begin{vmatrix}2&3&7\\13&17&5\\13&17&5\end{vmatrix}$
$\displaystyle \text{Since }R_2=R_3,$
$\displaystyle \therefore \Delta=0$

$\displaystyle \textbf{Question 2: }\text{Without expanding, show that the value of each of the following}$
$\displaystyle \text{determinants is zero:}$
$\displaystyle \text{(iv) }\begin{vmatrix}\frac{1}{a}&a^2&bc\\\frac{1}{b}&b^2&ac\\\frac{1}{c}&c^2&ab\end{vmatrix}$
$\displaystyle \text{(v) }\begin{vmatrix}a+b&2a+b&3a+b\\2a+b&3a+b&4a+b\\4a+b&5a+b&6a+b\end{vmatrix}$
$\displaystyle \text{(vi) }\begin{vmatrix}1&a&a^2-bc\\1&b&b^2-ac\\1&c&c^2-ab\end{vmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(iv)}$
$\displaystyle \Delta=\begin{vmatrix}\frac{1}{a}&a^2&bc\\\frac{1}{b}&b^2&ac\\\frac{1}{c}&c^2&ab\end{vmatrix}$
$\displaystyle \text{Here, }C_3=abc\,C_1.$
$\displaystyle \text{Thus, }C_1\text{ and }C_3\text{ are proportional.}$
$\displaystyle \therefore \Delta=0$

$\displaystyle \text{(v)}$
$\displaystyle \Delta=\begin{vmatrix}a+b&2a+b&3a+b\\2a+b&3a+b&4a+b\\4a+b&5a+b&6a+b\end{vmatrix}$
$\displaystyle \text{Applying }C_3\to C_3-C_2\text{ and }C_2\to C_2-C_1,$
$\displaystyle \Delta=\begin{vmatrix}a+b&a&a\\2a+b&a&a\\4a+b&a&a\end{vmatrix}$
$\displaystyle \text{Since }C_2=C_3,$
$\displaystyle \therefore \Delta=0$

$\displaystyle \text{(vi)}$
$\displaystyle \Delta=\begin{vmatrix}1&a&a^2-bc\\1&b&b^2-ac\\1&c&c^2-ab\end{vmatrix}$
$\displaystyle \text{Applying }R_1\to R_1-R_2\text{ and }R_2\to R_2-R_3,$
$\displaystyle \Delta=\begin{vmatrix}0&a-b&a^2-b^2+ac-bc\\0&b-c&b^2-c^2+ab-ac\\1&c&c^2-ab\end{vmatrix}$
$\displaystyle =\begin{vmatrix}0&a-b&(a-b)(a+b+c)\\0&b-c&(b-c)(a+b+c)\\1&c&c^2-ab\end{vmatrix}$
$\displaystyle =(a-b)(b-c)\begin{vmatrix}0&1&a+b+c\\0&1&a+b+c\\1&c&c^2-ab\end{vmatrix}$
$\displaystyle \text{Since }R_1=R_2,$
$\displaystyle \therefore \Delta=0$

$\displaystyle \textbf{Question 2: }\text{Without expanding, show that the value of each of the following}$
$\displaystyle \text{determinants is zero:}$
$\displaystyle \text{(vii) }\begin{vmatrix}49&1&6\\39&7&4\\26&2&3\end{vmatrix}\qquad \text{(viii) }\begin{vmatrix}0&x&y\\-x&0&z\\-y&-z&0\end{vmatrix}$
$\displaystyle \text{(ix) }\begin{vmatrix}1&43&6\\7&35&4\\3&17&2\end{vmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(vii)}$
$\displaystyle \Delta=\begin{vmatrix}49&1&6\\39&7&4\\26&2&3\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1-8C_3,$
$\displaystyle \Delta=\begin{vmatrix}1&1&6\\7&7&4\\2&2&3\end{vmatrix}$
$\displaystyle \text{Since }C_1=C_2,$
$\displaystyle \therefore \Delta=0$

$\displaystyle \text{(viii)}$
$\displaystyle \Delta=\begin{vmatrix}0&x&y\\-x&0&z\\-y&-z&0\end{vmatrix}$
$\displaystyle \text{By interchanging rows and columns,}$
$\displaystyle \Delta=\begin{vmatrix}0&-x&-y\\x&0&-z\\y&z&0\end{vmatrix}$
$\displaystyle =(-1)^3\begin{vmatrix}0&x&y\\-x&0&z\\-y&-z&0\end{vmatrix}$
$\displaystyle =-\Delta$
$\displaystyle \Rightarrow 2\Delta=0$
$\displaystyle \therefore \Delta=0$

$\displaystyle \text{(ix)}$
$\displaystyle \Delta=\begin{vmatrix}1&43&6\\7&35&4\\3&17&2\end{vmatrix}$
$\displaystyle \text{Applying }C_2\to C_2-7C_3,$
$\displaystyle \Delta=\begin{vmatrix}1&1&6\\7&7&4\\3&3&2\end{vmatrix}$
$\displaystyle \text{Since }C_1=C_2,$
$\displaystyle \therefore \Delta=0$

$\displaystyle \textbf{Question 2: }\text{Without expanding, show that the value of each of the following}$
$\displaystyle \text{determinants is zero:}$
$\displaystyle \text{(x) }\begin{vmatrix}1^2&2^2&3^2&4^2\\2^2&3^2&4^2&5^2\\3^2&4^2&5^2&6^2\\4^2&5^2&6^2&7^2\end{vmatrix}$
$\displaystyle \text{(xi) }\begin{vmatrix}a&b&c\\a+2x&b+2y&c+2z\\x&y&z\end{vmatrix}$
$\displaystyle \text{(xii) }\begin{vmatrix}(2^x+2^{-x})^2&(2^x-2^{-x})^2&1\\(3^x+3^{-x})^2&(3^x-3^{-x})^2&1\\(4^x+4^{-x})^2&(4^x-4^{-x})^2&1\end{vmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(x)}$
$\displaystyle \Delta=\begin{vmatrix}1&4&9&16\\4&9&16&25\\9&16&25&36\\16&25&36&49\end{vmatrix}$
$\displaystyle \text{Applying }R_4\to R_4-R_3,\quad R_3\to R_3-R_2,\quad R_2\to R_2-R_1,$
$\displaystyle \Delta=\begin{vmatrix}1&4&9&16\\3&5&7&9\\5&7&9&11\\7&9&11&13\end{vmatrix}$
$\displaystyle \text{Now applying }R_4\to R_4-R_3\text{ and }R_3\to R_3-R_2,$
$\displaystyle \Delta=\begin{vmatrix}1&4&9&16\\3&5&7&9\\2&2&2&2\\2&2&2&2\end{vmatrix}$
$\displaystyle \text{Since }R_3=R_4,$
$\displaystyle \therefore \Delta=0$

$\displaystyle \text{(xi)}$
$\displaystyle \Delta=\begin{vmatrix}a&b&c\\a+2x&b+2y&c+2z\\x&y&z\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1-2R_3,$
$\displaystyle \Delta=\begin{vmatrix}a&b&c\\0&0&0\\x&y&z\end{vmatrix}$
$\displaystyle \text{Since all the elements of }R_2\text{ are zero,}$
$\displaystyle \therefore \Delta=0$

$\displaystyle \text{(xii)}$
$\displaystyle \Delta=\begin{vmatrix}(2^x+2^{-x})^2&(2^x-2^{-x})^2&1\\(3^x+3^{-x})^2&(3^x-3^{-x})^2&1\\(4^x+4^{-x})^2&(4^x-4^{-x})^2&1\end{vmatrix}$
$\displaystyle =\begin{vmatrix}2^{2x}+2^{-2x}+2&2^{2x}+2^{-2x}-2&1\\3^{2x}+3^{-2x}+2&3^{2x}+3^{-2x}-2&1\\4^{2x}+4^{-2x}+2&4^{2x}+4^{-2x}-2&1\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1-C_2,$
$\displaystyle \Delta=\begin{vmatrix}4&2^{2x}+2^{-2x}-2&1\\4&3^{2x}+3^{-2x}-2&1\\4&4^{2x}+4^{-2x}-2&1\end{vmatrix}$
$\displaystyle \text{Here, }C_1=4C_3.$
$\displaystyle \text{Thus, }C_1\text{ and }C_3\text{ are proportional.}$
$\displaystyle \therefore \Delta=0$

$\displaystyle \textbf{Question 2: }\text{Without expanding, show that the value of each of the following}$
$\displaystyle \text{determinants is zero:}$
$\displaystyle \text{(xiii) }\begin{vmatrix}\sin\alpha&\cos\alpha&\cos(\alpha+\delta)\\\sin\beta&\cos\beta&\cos(\beta+\delta)\\\sin\gamma&\cos\gamma&\cos(\gamma+\delta)\end{vmatrix}$
$\displaystyle \text{(xiv) }\begin{vmatrix}\sin^2 23^\circ&\sin^2 67^\circ&\cos180^\circ\\-\sin^2 67^\circ&-\sin^2 23^\circ&\cos^2 180^\circ\\\cos180^\circ&\sin^2 23^\circ&\sin^2 67^\circ\end{vmatrix}$
$\displaystyle \text{(xv) }\begin{vmatrix}\cos(x+y)&-\sin(x+y)&\cos2y\\\sin x&\cos x&\sin y\\-\cos x&\sin x&-\cos y\end{vmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(xiii)}$
$\displaystyle \Delta=\begin{vmatrix}\sin\alpha&\cos\alpha&\cos(\alpha+\delta)\\\sin\beta&\cos\beta&\cos(\beta+\delta)\\\sin\gamma&\cos\gamma&\cos(\gamma+\delta)\end{vmatrix}$
$\displaystyle \text{Using }\cos(A+B)=\cos A\cos B-\sin A\sin B,$
$\displaystyle C_3=\cos\delta\,C_2-\sin\delta\,C_1.$
$\displaystyle \text{Thus, the columns are linearly dependent.}$
$\displaystyle \therefore \Delta=0$

$\displaystyle \text{(xiv)}$
$\displaystyle \Delta=\begin{vmatrix}\sin^2 23^\circ&\sin^2 67^\circ&\cos180^\circ\\-\sin^2 67^\circ&-\sin^2 23^\circ&\cos^2 180^\circ\\\cos180^\circ&\sin^2 23^\circ&\sin^2 67^\circ\end{vmatrix}$
$\displaystyle \text{Since }\sin67^\circ=\cos23^\circ,\quad \cos180^\circ=-1,$
$\displaystyle \text{and }\cos^2 180^\circ=1,$
$\displaystyle \Delta=\begin{vmatrix}\sin^2 23^\circ&\cos^2 23^\circ&-1\\-\cos^2 23^\circ&-\sin^2 23^\circ&1\\-1&\sin^2 23^\circ&\cos^2 23^\circ\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1+C_2+C_3,$
$\displaystyle \Delta=\begin{vmatrix}0&\cos^2 23^\circ&-1\\0&-\sin^2 23^\circ&1\\0&\sin^2 23^\circ&\cos^2 23^\circ\end{vmatrix}$
$\displaystyle \text{since }\sin^2 23^\circ+\cos^2 23^\circ=1.$
$\displaystyle \text{As all the elements of }C_1\text{ are zero,}$
$\displaystyle \therefore \Delta=0$

$\displaystyle \text{(xv)}$
$\displaystyle \Delta=\begin{vmatrix}\cos(x+y)&-\sin(x+y)&\cos2y\\\sin x&\cos x&\sin y\\-\cos x&\sin x&-\cos y\end{vmatrix}$
$\displaystyle \text{Applying }R_1\to R_1+\sin y\,R_2+\cos y\,R_3,$
$\displaystyle \Delta=\begin{vmatrix}\cos(x+y)+\sin x\sin y-\cos x\cos y&-\sin(x+y)+\cos x\sin y+\sin x\cos y&\cos2y+\sin^2y-\cos^2y\\\sin x&\cos x&\sin y\\-\cos x&\sin x&-\cos y\end{vmatrix}$
$\displaystyle \text{Using }\cos(x+y)=\cos x\cos y-\sin x\sin y,$
$\displaystyle \sin(x+y)=\sin x\cos y+\cos x\sin y$
$\displaystyle \text{and }\cos2y=\cos^2y-\sin^2y,$
$\displaystyle \Delta=\begin{vmatrix}0&0&0\\\sin x&\cos x&\sin y\\-\cos x&\sin x&-\cos y\end{vmatrix}$
$\displaystyle \therefore \Delta=0$

$\displaystyle \textbf{Question 2: }\text{Without expanding, show that the value of each of the following}$
$\displaystyle \text{determinants is zero:}$
$\displaystyle \text{(xvi) }\begin{vmatrix}\sqrt{23}+\sqrt3&\sqrt5&\sqrt5\\\sqrt{15}+\sqrt{46}&5&\sqrt{10}\\3+\sqrt{115}&\sqrt{15}&5\end{vmatrix}$
$\displaystyle \text{(xvii) }\begin{vmatrix}\sin^2A&\cot A&1\\\sin^2B&\cot B&1\\\sin^2C&\cot C&1\end{vmatrix},\quad \text{where }A,B,C\text{ are the angles of }\triangle ABC.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(xvi)}$
$\displaystyle \Delta=\begin{vmatrix}\sqrt{23}+\sqrt3&\sqrt5&\sqrt5\\\sqrt{15}+\sqrt{46}&5&\sqrt{10}\\3+\sqrt{115}&\sqrt{15}&5\end{vmatrix}$
$\displaystyle \text{Using the linearity of the determinant with respect to }C_1,$
$\displaystyle \Delta=\begin{vmatrix}\sqrt3&\sqrt5&\sqrt5\\\sqrt{15}&5&\sqrt{10}\\3&\sqrt{15}&5\end{vmatrix} +\begin{vmatrix}\sqrt{23}&\sqrt5&\sqrt5\\\sqrt{46}&5&\sqrt{10}\\\sqrt{115}&\sqrt{15}&5\end{vmatrix}$
$\displaystyle =\sqrt3\sqrt5\begin{vmatrix}1&1&\sqrt5\\\sqrt5&\sqrt5&\sqrt{10}\\\sqrt3&\sqrt3&5\end{vmatrix} +\sqrt{23}\sqrt5\begin{vmatrix}1&\sqrt5&1\\\sqrt2&5&\sqrt2\\\sqrt5&\sqrt{15}&\sqrt5\end{vmatrix}$
$\displaystyle \text{In the first determinant, }C_1=C_2,\text{ and in the second determinant, }C_1=C_3.$
$\displaystyle \therefore \Delta=0+0=0$

$\displaystyle \text{(xvii)}$
$\displaystyle \Delta=\begin{vmatrix}\sin^2A&\cot A&1\\\sin^2B&\cot B&1\\\sin^2C&\cot C&1\end{vmatrix}$
$\displaystyle \text{Applying }R_1\to R_1-R_2\text{ and }R_3\to R_3-R_2,$
$\displaystyle \Delta=\begin{vmatrix}\sin^2A-\sin^2B&\cot A-\cot B&0\\\sin^2B&\cot B&1\\\sin^2C-\sin^2B&\cot C-\cot B&0\end{vmatrix}$
$\displaystyle \text{Since }A+B+C=\pi,$
$\displaystyle \sin^2A-\sin^2B=\sin(A+B)\sin(A-B)=\sin C\sin(A-B)$
$\displaystyle \cot A-\cot B=-\frac{\sin(A-B)}{\sin A\sin B}$
$\displaystyle \sin^2C-\sin^2B=\sin(C+B)\sin(C-B)=\sin A\sin(C-B)$
$\displaystyle \cot C-\cot B=-\frac{\sin(C-B)}{\sin B\sin C}$
$\displaystyle \therefore R_1=\sin(A-B)\left[\sin C,\,-\frac{1}{\sin A\sin B},\,0\right]$
$\displaystyle \text{and }R_3=\sin(C-B)\left[\sin A,\,-\frac{1}{\sin B\sin C},\,0\right].$
$\displaystyle \text{Also, }\left[\sin C,\,-\frac{1}{\sin A\sin B},\,0\right] =\frac{\sin C}{\sin A}\left[\sin A,\,-\frac{1}{\sin B\sin C},\,0\right].$
$\displaystyle \text{Thus, }R_1\text{ and }R_3\text{ are proportional.}$
$\displaystyle \therefore \Delta=0$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Evaluate:}$
$\displaystyle \begin{vmatrix}a&b+c&a^2\\b&c+a&b^2\\c&a+b&c^2\end{vmatrix} \hfill\text{[CBSE 2006]}$
$\displaystyle \text{Answer:}$
$\displaystyle \Delta=\begin{vmatrix}a&b+c&a^2\\b&c+a&b^2\\c&a+b&c^2\end{vmatrix}$
$\displaystyle \text{Applying }C_2\to C_1+C_2,$
$\displaystyle \Delta=\begin{vmatrix}a&a+b+c&a^2\\b&a+b+c&b^2\\c&a+b+c&c^2\end{vmatrix}$
$\displaystyle =(a+b+c)\begin{vmatrix}a&1&a^2\\b&1&b^2\\c&1&c^2\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \Delta=(a+b+c)\begin{vmatrix}a&1&a^2\\b-a&0&b^2-a^2\\c-a&0&c^2-a^2\end{vmatrix}$
$\displaystyle =(a+b+c)\begin{vmatrix}a&1&a^2\\b-a&0&(b-a)(a+b)\\c-a&0&(c-a)(a+c)\end{vmatrix}$
$\displaystyle \text{Expanding along }C_2,$
$\displaystyle \Delta=-(a+b+c)\begin{vmatrix}b-a&(b-a)(a+b)\\c-a&(c-a)(a+c)\end{vmatrix}$
$\displaystyle =-(a+b+c)(b-a)(c-a)\begin{vmatrix}1&a+b\\1&a+c\end{vmatrix}$
$\displaystyle =-(a+b+c)(b-a)(c-a)(c-b)$
$\displaystyle =(a+b+c)(a-b)(a-c)(b-c)$
$\displaystyle \therefore \Delta=(a+b+c)(a-b)(a-c)(b-c)$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Evaluate:}$
$\displaystyle \begin{vmatrix}1&a&bc\\1&b&ca\\1&c&ab\end{vmatrix} \hfill\text{[CBSE 2006]}$
$\displaystyle \text{Answer:}$
$\displaystyle \Delta=\begin{vmatrix}1&a&bc\\1&b&ca\\1&c&ab\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \Delta=\begin{vmatrix}1&a&bc\\0&b-a&ca-bc\\0&c-a&ab-bc\end{vmatrix}$
$\displaystyle =\begin{vmatrix}1&a&bc\\0&b-a&c(a-b)\\0&c-a&b(a-c)\end{vmatrix}$
$\displaystyle \text{Expanding along }C_1,$
$\displaystyle \Delta=\begin{vmatrix}b-a&c(a-b)\\c-a&b(a-c)\end{vmatrix}$
$\displaystyle =(b-a)b(a-c)-c(a-b)(c-a)$
$\displaystyle =-(a-b)b(a-c)+c(a-b)(a-c)$
$\displaystyle =(a-b)(a-c)(c-b)$
$\displaystyle =(a-b)(b-c)(c-a)$
$\displaystyle \therefore \Delta=(a-b)(b-c)(c-a)$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Evaluate:}$
$\displaystyle \begin{vmatrix}x+\lambda&x&x\\x&x+\lambda&x\\x&x&x+\lambda\end{vmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \Delta=\begin{vmatrix}x+\lambda&x&x\\x&x+\lambda&x\\x&x&x+\lambda\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1+C_2+C_3,$
$\displaystyle \Delta=\begin{vmatrix}3x+\lambda&x&x\\3x+\lambda&x+\lambda&x\\3x+\lambda&x&x+\lambda\end{vmatrix}$
$\displaystyle =(3x+\lambda)\begin{vmatrix}1&x&x\\1&x+\lambda&x\\1&x&x+\lambda\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \Delta=(3x+\lambda)\begin{vmatrix}1&x&x\\0&\lambda&0\\0&0&\lambda\end{vmatrix}$
$\displaystyle =(3x+\lambda)(1)(\lambda)(\lambda)$
$\displaystyle \therefore \Delta=\lambda^2(3x+\lambda)$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Evaluate:}$
$\displaystyle \begin{vmatrix}a&b&c\\c&a&b\\b&c&a\end{vmatrix} \hfill\text{[CBSE 2004]}$
$\displaystyle \text{Answer:}$
$\displaystyle \Delta=\begin{vmatrix}a&b&c\\c&a&b\\b&c&a\end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \Delta=a(a^2-bc)-b(ac-b^2)+c(c^2-ab)$
$\displaystyle =a^3-abc-abc+b^3+c^3-abc$
$\displaystyle =a^3+b^3+c^3-3abc$
$\displaystyle \therefore \Delta=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Evaluate:}$
$\displaystyle \begin{vmatrix}x&1&1\\1&x&1\\1&1&x\end{vmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \Delta=\begin{vmatrix}x&1&1\\1&x&1\\1&1&x\end{vmatrix}$
$\displaystyle \text{Applying }R_1\to R_1-R_2\text{ and }R_3\to R_3-R_2,$
$\displaystyle \Delta=\begin{vmatrix}x-1&1-x&0\\1&x&1\\0&1-x&x-1\end{vmatrix}$
$\displaystyle =(x-1)^2\begin{vmatrix}1&-1&0\\1&x&1\\0&-1&1\end{vmatrix}$
$\displaystyle \text{Applying }C_2\to C_2+C_3,$
$\displaystyle \Delta=(x-1)^2\begin{vmatrix}1&-1&0\\1&x+1&1\\0&0&1\end{vmatrix}$
$\displaystyle \text{Expanding along }R_3,$
$\displaystyle \Delta=(x-1)^2\begin{vmatrix}1&-1\\1&x+1\end{vmatrix}$
$\displaystyle =(x-1)^2(x+2)$
$\displaystyle \therefore \Delta=(x-1)^2(x+2)$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Evaluate:}$
$\displaystyle \begin{vmatrix}0&xy^2&xz^2\\x^2y&0&yz^2\\x^2z&zy^2&0\end{vmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \Delta=\begin{vmatrix}0&xy^2&xz^2\\x^2y&0&yz^2\\x^2z&zy^2&0\end{vmatrix}$
$\displaystyle =x^2y^2z^2\begin{vmatrix}0&x&x\\y&0&y\\z&z&0\end{vmatrix}$
$\displaystyle \text{Taking }x^2,\ y^2,\ z^2\text{ common from }C_1,\ C_2,\ C_3\text{ respectively.}$
$\displaystyle =x^3y^3z^3\begin{vmatrix}0&1&1\\1&0&1\\1&1&0\end{vmatrix}$
$\displaystyle \text{Taking }x,\ y,\ z\text{ common from }R_1,\ R_2,\ R_3\text{ respectively.}$
$\displaystyle \text{Applying }C_2\to C_2-C_3,$
$\displaystyle \Delta=x^3y^3z^3\begin{vmatrix}0&0&1\\1&-1&1\\1&1&0\end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \Delta=x^3y^3z^3\begin{vmatrix}1&-1\\1&1\end{vmatrix}$
$\displaystyle =x^3y^3z^3(1+1)$
$\displaystyle \therefore \Delta=2x^3y^3z^3$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{Evaluate:}$
$\displaystyle \begin{vmatrix}a+x&y&z\\x&a+y&z\\x&y&a+z\end{vmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \Delta=\begin{vmatrix}a+x&y&z\\x&a+y&z\\x&y&a+z\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1+C_2+C_3,$
$\displaystyle \Delta=\begin{vmatrix}a+x+y+z&y&z\\a+x+y+z&a+y&z\\a+x+y+z&y&a+z\end{vmatrix}$
$\displaystyle =(a+x+y+z)\begin{vmatrix}1&y&z\\1&a+y&z\\1&y&a+z\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \Delta=(a+x+y+z)\begin{vmatrix}1&y&z\\0&a&0\\0&0&a\end{vmatrix}$
$\displaystyle =(a+x+y+z)(1)(a)(a)$
$\displaystyle \therefore \Delta=a^2(a+x+y+z)$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{If}$
$\displaystyle \Delta=\begin{vmatrix}1&x&x^2\\1&y&y^2\\1&z&z^2\end{vmatrix}, \qquad \Delta_1=\begin{vmatrix}1&1&1\\yz&zx&xy\\x&y&z\end{vmatrix},$
$\displaystyle \text{then prove that }\Delta+\Delta_1=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \Delta+\Delta_1= \begin{vmatrix}1&x&x^2\\1&y&y^2\\1&z&z^2\end{vmatrix} +\begin{vmatrix}1&1&1\\yz&zx&xy\\x&y&z\end{vmatrix}$
$\displaystyle \text{Taking the transpose of }\Delta_1,$
$\displaystyle \Delta+\Delta_1= \begin{vmatrix}1&x&x^2\\1&y&y^2\\1&z&z^2\end{vmatrix} +\begin{vmatrix}1&yz&x\\1&zx&y\\1&xy&z\end{vmatrix}$
$\displaystyle \text{Interchanging }C_2\text{ and }C_3\text{ in the second determinant,}$
$\displaystyle \Delta+\Delta_1= \begin{vmatrix}1&x&x^2\\1&y&y^2\\1&z&z^2\end{vmatrix} -\begin{vmatrix}1&x&yz\\1&y&zx\\1&z&xy\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1\text{ in both determinants,}$
$\displaystyle \Delta+\Delta_1= \begin{vmatrix}1&x&x^2\\0&y-x&y^2-x^2\\0&z-x&z^2-x^2\end{vmatrix} -\begin{vmatrix}1&x&yz\\0&y-x&zx-yz\\0&z-x&xy-yz\end{vmatrix}$
$\displaystyle =(y-x)(z-x) \begin{vmatrix}1&x&x^2\\0&1&x+y\\0&1&x+z\end{vmatrix} -(y-x)(z-x) \begin{vmatrix}1&x&yz\\0&1&-z\\0&1&-y\end{vmatrix}$
$\displaystyle \text{Expanding both determinants along }C_1,$
$\displaystyle \Delta+\Delta_1=(y-x)(z-x) \begin{vmatrix}1&x+y\\1&x+z\end{vmatrix} -(y-x)(z-x) \begin{vmatrix}1&-z\\1&-y\end{vmatrix}$
$\displaystyle =(y-x)(z-x)(z-y)-(y-x)(z-x)(z-y)$
$\displaystyle =0$
$\displaystyle \therefore \Delta+\Delta_1=0$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}a&b&c\\a-b&b-c&c-a\\b+c&c+a&a+b\end{vmatrix} =a^3+b^3+c^3-3abc \hfill\text{[CBSE 2009]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}a&b&c\\a-b&b-c&c-a\\b+c&c+a&a+b\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1+C_2+C_3,$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}a+b+c&b&c\\0&b-c&c-a\\2(a+b+c)&c+a&a+b\end{vmatrix}$
$\displaystyle =(a+b+c) \begin{vmatrix}1&b&c\\0&b-c&c-a\\2&c+a&a+b\end{vmatrix}$
$\displaystyle \text{Applying }R_3\to R_3-2R_1,$
$\displaystyle \text{L.H.S.}=(a+b+c) \begin{vmatrix}1&b&c\\0&b-c&c-a\\0&c+a-2b&a+b-2c\end{vmatrix}$
$\displaystyle \text{Expanding along }C_1,$
$\displaystyle \text{L.H.S.}=(a+b+c) \begin{vmatrix}b-c&c-a\\c+a-2b&a+b-2c\end{vmatrix}$
$\displaystyle =(a+b+c)\big[(b-c)(a+b-2c)-(c-a)(c+a-2b)\big]$
$\displaystyle =(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$\displaystyle =a^3+b^3+c^3-3abc$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}b+c&a-b&a\\c+a&b-c&b\\a+b&c-a&c\end{vmatrix} =3abc-a^3-b^3-c^3$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}b+c&a-b&a\\c+a&b-c&b\\a+b&c-a&c\end{vmatrix}$
$\displaystyle \text{Applying }R_1\to R_1+R_2+R_3,$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}2(a+b+c)&0&a+b+c\\c+a&b-c&b\\a+b&c-a&c\end{vmatrix}$
$\displaystyle =(a+b+c) \begin{vmatrix}2&0&1\\c+a&b-c&b\\a+b&c-a&c\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1-2C_3,$
$\displaystyle \text{L.H.S.}=(a+b+c) \begin{vmatrix}0&0&1\\a+c-2b&b-c&b\\a+b-2c&c-a&c\end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \text{L.H.S.}=(a+b+c) \begin{vmatrix}a+c-2b&b-c\\a+b-2c&c-a\end{vmatrix}$
$\displaystyle =(a+b+c)\big[(a+c-2b)(c-a)-(b-c)(a+b-2c)\big]$
$\displaystyle =(a+b+c)(-a^2-b^2-c^2+ab+bc+ca)$
$\displaystyle =-(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$\displaystyle =-(a^3+b^3+c^3-3abc)$
$\displaystyle =3abc-a^3-b^3-c^3$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}a+b&b+c&c+a\\b+c&c+a&a+b\\c+a&a+b&b+c\end{vmatrix} =2\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix} \hfill\text{[CBSE 2004, 2006, 2007]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}a+b&b+c&c+a\\b+c&c+a&a+b\\c+a&a+b&b+c\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1-C_2+C_3,$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}2a&b+c&c+a\\2b&c+a&a+b\\2c&a+b&b+c\end{vmatrix}$
$\displaystyle =2\begin{vmatrix}a&b+c&c+a\\b&c+a&a+b\\c&a+b&b+c\end{vmatrix}$
$\displaystyle \text{Applying }C_2\to C_2-C_3+C_1,$
$\displaystyle \text{L.H.S.}=2 \begin{vmatrix}a&b&c+a\\b&c&a+b\\c&a&b+c\end{vmatrix}$
$\displaystyle \text{Applying }C_3\to C_3-C_1,$
$\displaystyle \text{L.H.S.}=2 \begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix}$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}a+b+2c&a&b\\c&b+c+2a&b\\c&a&c+a+2b\end{vmatrix} =2(a+b+c)^3\hfill\text{[CBSE 2006, 2008, 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}a+b+2c&a&b\\c&b+c+2a&b\\c&a&c+a+2b\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1+C_2+C_3,$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}2(a+b+c)&a&b\\2(a+b+c)&b+c+2a&b\\2(a+b+c)&a&c+a+2b\end{vmatrix}$
$\displaystyle =2(a+b+c) \begin{vmatrix}1&a&b\\1&b+c+2a&b\\1&a&c+a+2b\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \text{L.H.S.}=2(a+b+c) \begin{vmatrix}1&a&b\\0&a+b+c&0\\0&0&a+b+c\end{vmatrix}$
$\displaystyle =2(a+b+c)(1)(a+b+c)(a+b+c)$
$\displaystyle =2(a+b+c)^3$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}a-b-c&2a&2a\\2b&b-c-a&2b\\2c&2c&c-a-b\end{vmatrix} =(a+b+c)^3$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}a-b-c&2a&2a\\2b&b-c-a&2b\\2c&2c&c-a-b\end{vmatrix}$
$\displaystyle \text{Applying }R_1\to R_1+R_2+R_3,$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}a+b+c&a+b+c&a+b+c\\2b&b-c-a&2b\\2c&2c&c-a-b\end{vmatrix}$
$\displaystyle =(a+b+c) \begin{vmatrix}1&1&1\\2b&b-c-a&2b\\2c&2c&c-a-b\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1-C_2,$
$\displaystyle \text{L.H.S.}=(a+b+c) \begin{vmatrix}0&1&1\\a+b+c&b-c-a&2b\\0&2c&c-a-b\end{vmatrix}$
$\displaystyle \text{Expanding along }C_1,$
$\displaystyle \text{L.H.S.}=-(a+b+c)^2 \begin{vmatrix}1&1\\2c&c-a-b\end{vmatrix}$
$\displaystyle =-(a+b+c)^2\big[(c-a-b)-2c\big]$
$\displaystyle =-(a+b+c)^2[-(a+b+c)]$
$\displaystyle =(a+b+c)^3$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}1&b+c&b^2+c^2\\1&c+a&c^2+a^2\\1&a+b&a^2+b^2\end{vmatrix} =(a-b)(b-c)(c-a)\hfill\text{[CBSE 2002]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}1&b+c&b^2+c^2\\1&c+a&c^2+a^2\\1&a+b&a^2+b^2\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_3\text{ and then }R_1\to R_1-R_2,$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}0&b-a&b^2-a^2\\0&c-b&c^2-b^2\\1&a+b&a^2+b^2\end{vmatrix}$
$\displaystyle = \begin{vmatrix}0&a-b&a^2-b^2\\0&b-c&b^2-c^2\\1&a+b&a^2+b^2\end{vmatrix}$
$\displaystyle \text{Taking }-1\text{ common from each of }R_1\text{ and }R_2,$
$\displaystyle \text{L.H.S.}=(a-b)(b-c) \begin{vmatrix}0&1&a+b\\0&1&b+c\\1&a+b&a^2+b^2\end{vmatrix}$
$\displaystyle \text{Expanding along }C_1,$
$\displaystyle \text{L.H.S.}=(a-b)(b-c) \begin{vmatrix}1&a+b\\1&b+c\end{vmatrix}$
$\displaystyle =(a-b)(b-c)\big[(b+c)-(a+b)\big]$
$\displaystyle =(a-b)(b-c)(c-a)$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}a&a+b&a+2b\\a+2b&a&a+b\\a+b&a+2b&a\end{vmatrix} =9(a+b)b^2\hfill\text{[CBSE 2002, 2013, 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}a&a+b&a+2b\\a+2b&a&a+b\\a+b&a+2b&a\end{vmatrix}$
$\displaystyle \text{Applying }R_1\to R_1+R_2+R_3,$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}3a+3b&3a+3b&3a+3b\\a+2b&a&a+b\\a+b&a+2b&a\end{vmatrix}$
$\displaystyle =3(a+b) \begin{vmatrix}1&1&1\\a+2b&a&a+b\\a+b&a+2b&a\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1-C_2\text{ and }C_2\to C_2-C_3,$
$\displaystyle \text{L.H.S.}=3(a+b) \begin{vmatrix}0&0&1\\2b&-b&a+b\\-b&2b&a\end{vmatrix}$
$\displaystyle =3(a+b)b^2 \begin{vmatrix}0&0&1\\2&-1&a+b\\-1&2&a\end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \text{L.H.S.}=3(a+b)b^2 \begin{vmatrix}2&-1\\-1&2\end{vmatrix}$
$\displaystyle =3(a+b)b^2(4-1)$
$\displaystyle =9(a+b)b^2$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}1&a&bc\\1&b&ca\\1&c&ab\end{vmatrix} =\begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let L.H.S.}=\Delta= \begin{vmatrix}1&a&bc\\1&b&ca\\1&c&ab\end{vmatrix}$
$\displaystyle \text{Multiplying }R_1,\ R_2,\ R_3\text{ by }a,\ b,\ c\text{ respectively,}$
$\displaystyle abc\,\Delta= \begin{vmatrix}a&a^2&abc\\b&b^2&abc\\c&c^2&abc\end{vmatrix}$
$\displaystyle \text{Taking }abc\text{ common from }C_3,$
$\displaystyle abc\,\Delta =abc\begin{vmatrix}a&a^2&1\\b&b^2&1\\c&c^2&1\end{vmatrix}$
$\displaystyle \text{Interchanging }C_1\text{ and }C_3,$
$\displaystyle abc\,\Delta =-abc\begin{vmatrix}1&a^2&a\\1&b^2&b\\1&c^2&c\end{vmatrix}$
$\displaystyle \text{Interchanging }C_2\text{ and }C_3,$
$\displaystyle abc\,\Delta =abc\begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatrix}$
$\displaystyle \therefore \begin{vmatrix}1&a&bc\\1&b&ca\\1&c&ab\end{vmatrix} =\begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatrix}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$ $\displaystyle \textbf{Question 19: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}z&x&y\\z^2&x^2&y^2\\z^4&x^4&y^4\end{vmatrix} =\begin{vmatrix}x&y&z\\x^2&y^2&z^2\\x^4&y^4&z^4\end{vmatrix} =\begin{vmatrix}x^2&y^2&z^2\\x^4&y^4&z^4\\x&y&z\end{vmatrix}$
$\displaystyle =xyz(x-y)(y-z)(z-x)(x+y+z).$
$\displaystyle \text{Answer:}$
$\displaystyle \Delta_1=\begin{vmatrix}z&x&y\\z^2&x^2&y^2\\z^4&x^4&y^4\end{vmatrix}, \quad \Delta_2=\begin{vmatrix}x&y&z\\x^2&y^2&z^2\\x^4&y^4&z^4\end{vmatrix}$
$\displaystyle \text{and }\Delta_3= \begin{vmatrix}x^2&y^2&z^2\\x^4&y^4&z^4\\x&y&z\end{vmatrix}.$
$\displaystyle \text{Interchanging }C_1\text{ and }C_2,\text{ and then }C_2\text{ and }C_3\text{ in }\Delta_1,$
$\displaystyle \Delta_1=(-1)^2 \begin{vmatrix}x&y&z\\x^2&y^2&z^2\\x^4&y^4&z^4\end{vmatrix} =\Delta_2$
$\displaystyle \text{Interchanging }R_1\text{ and }R_2,\text{ and then }R_2\text{ and }R_3\text{ in }\Delta_2,$
$\displaystyle \Delta_2=(-1)^2 \begin{vmatrix}x^2&y^2&z^2\\x^4&y^4&z^4\\x&y&z\end{vmatrix} =\Delta_3$
$\displaystyle \therefore \Delta_1=\Delta_2=\Delta_3.$
$\displaystyle \text{Now,}$
$\displaystyle \Delta_2= \begin{vmatrix}x&y&z\\x^2&y^2&z^2\\x^4&y^4&z^4\end{vmatrix}$
$\displaystyle =xyz \begin{vmatrix}1&1&1\\x&y&z\\x^3&y^3&z^3\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1-C_2\text{ and }C_2\to C_2-C_3,$
$\displaystyle \Delta_2=xyz \begin{vmatrix}0&0&1\\x-y&y-z&z\\x^3-y^3&y^3-z^3&z^3\end{vmatrix}$
$\displaystyle =xyz(x-y)(y-z) \begin{vmatrix}0&0&1\\1&1&z\\x^2+xy+y^2&y^2+yz+z^2&z^3\end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \Delta_2=xyz(x-y)(y-z) \begin{vmatrix}1&1\\x^2+xy+y^2&y^2+yz+z^2\end{vmatrix}$
$\displaystyle =xyz(x-y)(y-z)\big[y^2+yz+z^2-x^2-xy-y^2\big]$
$\displaystyle =xyz(x-y)(y-z)\big[yz+z^2-x^2-xy\big]$
$\displaystyle =xyz(x-y)(y-z)\big[y(z-x)+(z-x)(z+x)\big]$
$\displaystyle =xyz(x-y)(y-z)(z-x)(x+y+z)$
$\displaystyle \therefore \Delta_1=\Delta_2=\Delta_3 =xyz(x-y)(y-z)(z-x)(x+y+z).$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}(b+c)^2&a^2&bc\\(c+a)^2&b^2&ca\\(a+b)^2&c^2&ab\end{vmatrix}$
$\displaystyle =(a-b)(b-c)(c-a)(a+b+c)(a^2+b^2+c^2)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}(b+c)^2&a^2&bc\\(c+a)^2&b^2&ca\\(a+b)^2&c^2&ab\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_3\text{ and then }R_1\to R_1-R_2,$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}(b+c)^2-(c+a)^2&a^2-b^2&bc-ca\\ (c+a)^2-(a+b)^2&b^2-c^2&ca-ab\\ (a+b)^2&c^2&ab\end{vmatrix}$
$\displaystyle = \begin{vmatrix}(b-a)(a+b+2c)&(a-b)(a+b)&c(b-a)\\ (c-b)(2a+b+c)&(b-c)(b+c)&a(c-b)\\ (a+b)^2&c^2&ab\end{vmatrix}$
$\displaystyle =(a-b)(b-c) \begin{vmatrix}-(a+b+2c)&a+b&-c\\ -(2a+b+c)&b+c&-a\\ (a+b)^2&c^2&ab\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1-C_2,$
$\displaystyle \text{L.H.S.}=(a-b)(b-c) \begin{vmatrix}-2(a+b+c)&a+b&-c\\ -2(a+b+c)&b+c&-a\\ (a+b)^2-c^2&c^2&ab\end{vmatrix}$
$\displaystyle =(a-b)(b-c)(a+b+c) \begin{vmatrix}-2&a+b&-c\\ -2&b+c&-a\\ a+b-c&c^2&ab\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1,$
$\displaystyle \text{L.H.S.}=(a-b)(b-c)(a+b+c) \begin{vmatrix}-2&a+b&-c\\ 0&c-a&c-a\\ a+b-c&c^2&ab\end{vmatrix}$
$\displaystyle =(a-b)(b-c)(c-a)(a+b+c) \begin{vmatrix}-2&a+b&-c\\ 0&1&1\\ a+b-c&c^2&ab\end{vmatrix}$
$\displaystyle \text{Applying }C_2\to C_2-C_3,$
$\displaystyle \text{L.H.S.}=(a-b)(b-c)(c-a)(a+b+c) \begin{vmatrix}-2&a+b+c&-c\\ 0&0&1\\ a+b-c&c^2-ab&ab\end{vmatrix}$
$\displaystyle \text{Expanding along }R_2,$
$\displaystyle \text{L.H.S.}=-(a-b)(b-c)(c-a)(a+b+c) \begin{vmatrix}-2&a+b+c\\a+b-c&c^2-ab\end{vmatrix}$
$\displaystyle =(a-b)(b-c)(c-a)(a+b+c)$
$\displaystyle \qquad\times\big[2(c^2-ab)+(a+b+c)(a+b-c)\big]$
$\displaystyle =(a-b)(b-c)(c-a)(a+b+c)$
$\displaystyle \qquad\times\big[2c^2-2ab+(a+b)^2-c^2\big]$
$\displaystyle =(a-b)(b-c)(c-a)(a+b+c)(a^2+b^2+c^2)$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}(a+1)(a+2)&a+2&1\\(a+2)(a+3)&a+3&1\\(a+3)(a+4)&a+4&1\end{vmatrix}=-2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}(a+1)(a+2)&a+2&1\\(a+2)(a+3)&a+3&1\\(a+3)(a+4)&a+4&1\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1-C_2\text{ and }C_2\to C_2-C_3,$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}(a+1)(a+2)-(a+2)(a+3)&(a+2)-(a+3)&0\\ (a+2)(a+3)-(a+3)(a+4)&(a+3)-(a+4)&0\\ (a+3)(a+4)&a+4&1 \end{vmatrix}$
$\displaystyle = \begin{vmatrix} -2(a+2)&-1&0\\ -2(a+3)&-1&0\\ (a+3)(a+4)&a+4&1 \end{vmatrix}$
$\displaystyle \text{Expanding along }C_3,$
$\displaystyle \text{L.H.S.}= \begin{vmatrix} -2(a+2)&-1\\ -2(a+3)&-1 \end{vmatrix}$
$\displaystyle =2(a+2)-2(a+3)$
$\displaystyle =-2$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}a^2&a^2-(b-c)^2&bc\\b^2&b^2-(c-a)^2&ca\\c^2&c^2-(a-b)^2&ab\end{vmatrix}$
$\displaystyle =(a-b)(b-c)(c-a)(a+b+c)(a^2+b^2+c^2) \hfill\text{[CBSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}a^2&a^2-(b-c)^2&bc\\b^2&b^2-(c-a)^2&ca\\c^2&c^2-(a-b)^2&ab\end{vmatrix}$
$\displaystyle \text{Applying }C_2\to C_2-C_1,$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}a^2&-(b-c)^2&bc\\b^2&-(c-a)^2&ca\\c^2&-(a-b)^2&ab\end{vmatrix}$
$\displaystyle =- \begin{vmatrix}a^2&(b-c)^2&bc\\b^2&(c-a)^2&ca\\c^2&(a-b)^2&ab\end{vmatrix}$
$\displaystyle \text{Applying }C_2\to C_2+2C_3,$
$\displaystyle \text{L.H.S.}=- \begin{vmatrix}a^2&b^2+c^2&bc\\b^2&c^2+a^2&ca\\c^2&a^2+b^2&ab\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1+C_2,$
$\displaystyle \text{L.H.S.}=- \begin{vmatrix}a^2+b^2+c^2&b^2+c^2&bc\\ a^2+b^2+c^2&c^2+a^2&ca\\ a^2+b^2+c^2&a^2+b^2&ab \end{vmatrix}$
$\displaystyle =-(a^2+b^2+c^2) \begin{vmatrix}1&b^2+c^2&bc\\1&c^2+a^2&ca\\1&a^2+b^2&ab\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \text{L.H.S.}=-(a^2+b^2+c^2) \begin{vmatrix}1&b^2+c^2&bc\\ 0&a^2-b^2&c(a-b)\\ 0&a^2-c^2&b(a-c) \end{vmatrix}$
$\displaystyle =-(a^2+b^2+c^2)(a-b)(a-c) \begin{vmatrix}1&b^2+c^2&bc\\0&a+b&c\\0&a+c&b\end{vmatrix}$
$\displaystyle \text{Expanding along }C_1,$
$\displaystyle \text{L.H.S.}=-(a^2+b^2+c^2)(a-b)(a-c) \begin{vmatrix}a+b&c\\a+c&b\end{vmatrix}$
$\displaystyle =-(a^2+b^2+c^2)(a-b)(a-c) \big[ab+b^2-ac-c^2\big]$
$\displaystyle =-(a^2+b^2+c^2)(a-b)(a-c) \big[a(b-c)+(b+c)(b-c)\big]$
$\displaystyle =-(a^2+b^2+c^2)(a-b)(a-c)(b-c)(a+b+c)$
$\displaystyle =(a-b)(b-c)(c-a)(a+b+c)(a^2+b^2+c^2)$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}1&a^2+bc&a^3\\1&b^2+ca&b^3\\1&c^2+ab&c^3\end{vmatrix} =-(a-b)(b-c)(c-a)(a^2+b^2+c^2)$
$\displaystyle \hfill\text{[CBSE 2008]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}1&a^2+bc&a^3\\1&b^2+ca&b^3\\1&c^2+ab&c^3\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_3\text{ and then }R_1\to R_1-R_2,$
$\displaystyle \text{L.H.S.}= \begin{vmatrix} 0&(a^2+bc)-(b^2+ca)&a^3-b^3\\ 0&(b^2+ca)-(c^2+ab)&b^3-c^3\\ 1&c^2+ab&c^3 \end{vmatrix}$
$\displaystyle = \begin{vmatrix} 0&(a-b)(a+b-c)&(a-b)(a^2+ab+b^2)\\ 0&(b-c)(b+c-a)&(b-c)(b^2+bc+c^2)\\ 1&c^2+ab&c^3 \end{vmatrix}$
$\displaystyle =(a-b)(b-c) \begin{vmatrix} 0&a+b-c&a^2+ab+b^2\\ 0&b+c-a&b^2+bc+c^2\\ 1&c^2+ab&c^3 \end{vmatrix}$
$\displaystyle \text{Expanding along }C_1,$
$\displaystyle \text{L.H.S.}=(a-b)(b-c) \begin{vmatrix} a+b-c&a^2+ab+b^2\\ b+c-a&b^2+bc+c^2 \end{vmatrix}$
$\displaystyle =(a-b)(b-c)\big[(a+b-c)(b^2+bc+c^2)$
$\displaystyle \qquad -(b+c-a)(a^2+ab+b^2)\big]$
$\displaystyle =(a-b)(b-c)\big[a^3-a^2c+ab^2+ac^2-b^2c-c^3\big]$
$\displaystyle =(a-b)(b-c)\big[(a^3-c^3)-ac(a-c)+b^2(a-c)\big]$
$\displaystyle =(a-b)(b-c)(a-c)(a^2+b^2+c^2)$
$\displaystyle =-(a-b)(b-c)(c-a)(a^2+b^2+c^2)$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}a^2&bc&ac+c^2\\a^2+ab&b^2&ac\\ab&b^2+bc&c^2\end{vmatrix} =4a^2b^2c^2\hfill\text{[CBSE 2014, 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}a^2&bc&ac+c^2\\a^2+ab&b^2&ac\\ab&b^2+bc&c^2\end{vmatrix}$
$\displaystyle =abc \begin{vmatrix}a&c&a+c\\a+b&b&a\\b&b+c&c\end{vmatrix}$
$\displaystyle \text{Applying }C_3\to C_3-C_2-C_1,$
$\displaystyle \text{L.H.S.}=abc \begin{vmatrix}a&c&0\\a+b&b&-2b\\b&b+c&-2b\end{vmatrix}$
$\displaystyle =-2ab^2c \begin{vmatrix}a&c&0\\a+b&b&1\\b&b+c&1\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_3,$
$\displaystyle \text{L.H.S.}=-2ab^2c \begin{vmatrix}a&c&0\\a&-c&0\\b&b+c&1\end{vmatrix}$
$\displaystyle \text{Expanding along }C_3,$
$\displaystyle \text{L.H.S.}=-2ab^2c \begin{vmatrix}a&c\\a&-c\end{vmatrix}$
$\displaystyle =-2ab^2c\big[-ac-ac\big]$
$\displaystyle =-2ab^2c(-2ac)$
$\displaystyle =4a^2b^2c^2$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}x+4&x&x\\x&x+4&x\\x&x&x+4\end{vmatrix} =16(3x+4)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}x+4&x&x\\x&x+4&x\\x&x&x+4\end{vmatrix}$
$\displaystyle \text{Applying }R_1\to R_1+R_2+R_3,$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}3x+4&3x+4&3x+4\\x&x+4&x\\x&x&x+4\end{vmatrix}$
$\displaystyle =(3x+4) \begin{vmatrix}1&1&1\\x&x+4&x\\x&x&x+4\end{vmatrix}$
$\displaystyle \text{Applying }C_2\to C_2-C_1\text{ and }C_3\to C_3-C_1,$
$\displaystyle \text{L.H.S.}=(3x+4) \begin{vmatrix}1&0&0\\x&4&0\\x&0&4\end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \text{L.H.S.}=(3x+4) \begin{vmatrix}4&0\\0&4\end{vmatrix}$
$\displaystyle =(3x+4)(16)$
$\displaystyle =16(3x+4)$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 26: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}1&1+p&1+p+q\\2&3+2p&4+3p+2q\\3&6+3p&10+6p+3q\end{vmatrix}=1$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}1&1+p&1+p+q\\2&3+2p&4+3p+2q\\3&6+3p&10+6p+3q\end{vmatrix}$
$\displaystyle \text{Applying }C_2\to C_2-pC_1,$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}1&1&1+p+q\\2&3&4+3p+2q\\3&6&10+6p+3q\end{vmatrix}$
$\displaystyle \text{Applying }C_3\to C_3-pC_2-qC_1,$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}1&1&1\\2&3&4\\3&6&10\end{vmatrix}$
$\displaystyle \text{Applying }C_2\to C_2-C_1\text{ and }C_3\to C_3-C_1,$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}1&0&0\\2&1&2\\3&3&7\end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}1&2\\3&7\end{vmatrix}$
$\displaystyle =7-6$
$\displaystyle =1$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 27: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}a&b-c&c-b\\a-c&b&c-a\\a-b&b-a&c\end{vmatrix} =(a+b-c)(b+c-a)(c+a-b)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}a&b-c&c-b\\a-c&b&c-a\\a-b&b-a&c\end{vmatrix}$
$\displaystyle \text{Applying }C_2\to C_2+C_3\text{ and }C_3\to C_1+C_3,$
$\displaystyle \text{L.H.S.}= \begin{vmatrix} a&0&a+c-b\\ a-c&b+c-a&0\\ a-b&b+c-a&a+c-b \end{vmatrix}$
$\displaystyle =(b+c-a)(c+a-b) \begin{vmatrix} a&0&1\\ a-c&1&0\\ a-b&1&1 \end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \text{L.H.S.}=(b+c-a)(c+a-b) \left[ a\begin{vmatrix}1&0\\1&1\end{vmatrix} +\begin{vmatrix}a-c&1\\a-b&1\end{vmatrix} \right]$
$\displaystyle =(b+c-a)(c+a-b)\big[a+(a-c-a+b)\big]$
$\displaystyle =(b+c-a)(c+a-b)(a+b-c)$
$\displaystyle =(a+b-c)(b+c-a)(c+a-b)$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 28: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}a^2&2ab&b^2\\b^2&a^2&2ab\\2ab&b^2&a^2\end{vmatrix} =(a^3+b^3)^2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}a^2&2ab&b^2\\b^2&a^2&2ab\\2ab&b^2&a^2\end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \text{L.H.S.} =a^2\begin{vmatrix}a^2&2ab\\b^2&a^2\end{vmatrix} -2ab\begin{vmatrix}b^2&2ab\\2ab&a^2\end{vmatrix} +b^2\begin{vmatrix}b^2&a^2\\2ab&b^2\end{vmatrix}$
$\displaystyle =a^2(a^4-2ab^3) -2ab(a^2b^2-4a^2b^2) +b^2(b^4-2a^3b)$
$\displaystyle =a^6-2a^3b^3+6a^3b^3+b^6-2a^3b^3$
$\displaystyle =a^6+2a^3b^3+b^6$
$\displaystyle =(a^3+b^3)^2$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 29: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}a^2+1&ab&ac\\ab&b^2+1&bc\\ca&cb&c^2+1\end{vmatrix} =1+a^2+b^2+c^2\hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}a^2+1&ab&ac\\ab&b^2+1&bc\\ac&bc&c^2+1\end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \text{L.H.S.}=(a^2+1) \begin{vmatrix}b^2+1&bc\\bc&c^2+1\end{vmatrix} -ab\begin{vmatrix}ab&bc\\ac&c^2+1\end{vmatrix}$
$\displaystyle \qquad +ac\begin{vmatrix}ab&b^2+1\\ac&bc\end{vmatrix}$
$\displaystyle =(a^2+1)\big[(b^2+1)(c^2+1)-b^2c^2\big]$
$\displaystyle \qquad -ab\big[ab(c^2+1)-abc^2\big] +ac\big[ab^2c-ac(b^2+1)\big]$
$\displaystyle =(a^2+1)(b^2+c^2+1)-a^2b^2-a^2c^2$
$\displaystyle =a^2b^2+a^2c^2+a^2+b^2+c^2+1-a^2b^2-a^2c^2$
$\displaystyle =1+a^2+b^2+c^2$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 30: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}1&a&a^2\\a^2&1&a\\a&a^2&1\end{vmatrix} =(a^3-1)^2\hfill\text{[CBSE 2013, 2014, 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}1&a&a^2\\a^2&1&a\\a&a^2&1\end{vmatrix}$
$\displaystyle \text{Applying }R_1\to R_1+R_2+R_3,$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}1+a+a^2&1+a+a^2&1+a+a^2\\a^2&1&a\\a&a^2&1\end{vmatrix}$
$\displaystyle =(1+a+a^2) \begin{vmatrix}1&1&1\\a^2&1&a\\a&a^2&1\end{vmatrix}$
$\displaystyle \text{Applying }C_2\to C_2-C_1\text{ and }C_3\to C_3-C_1,$
$\displaystyle \text{L.H.S.}=(1+a+a^2) \begin{vmatrix}1&0&0\\a^2&1-a^2&a-a^2\\a&a^2-a&1-a\end{vmatrix}$
$\displaystyle =(1+a+a^2)(a-1)^2 \begin{vmatrix}1&0&0\\a^2&-(a+1)&-a\\a&a&-1\end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \text{L.H.S.}=(1+a+a^2)(a-1)^2 \begin{vmatrix}-(a+1)&-a\\a&-1\end{vmatrix}$
$\displaystyle =(1+a+a^2)(a-1)^2\big[(a+1)+a^2\big]$
$\displaystyle =(a-1)^2(1+a+a^2)^2$
$\displaystyle =\big[(a-1)(1+a+a^2)\big]^2$
$\displaystyle =(a^3-1)^2$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 31: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}a+b+c&-c&-b\\-c&a+b+c&-a\\-b&-a&a+b+c\end{vmatrix} =2(a+b)(b+c)(c+a)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}a+b+c&-c&-b\\-c&a+b+c&-a\\-b&-a&a+b+c\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1+C_2+C_3,$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}a&-c&-b\\b&a+b+c&-a\\c&-a&a+b+c\end{vmatrix}$
$\displaystyle \text{Applying }R_1\to R_1+R_2\text{ and }R_2\to R_2+R_3,$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}a+b&a+b&-(a+b)\\b+c&b+c&b+c\\c&-a&a+b+c\end{vmatrix}$
$\displaystyle =(a+b)(b+c) \begin{vmatrix}1&1&-1\\1&1&1\\c&-a&a+b+c\end{vmatrix}$
$\displaystyle \text{Applying }R_1\to R_1-R_2,$
$\displaystyle \text{L.H.S.}=(a+b)(b+c) \begin{vmatrix}0&0&-2\\1&1&1\\c&-a&a+b+c\end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \text{L.H.S.}=(a+b)(b+c)(-2) \begin{vmatrix}1&1\\c&-a\end{vmatrix}$
$\displaystyle =(a+b)(b+c)(-2)(-a-c)$
$\displaystyle =2(a+b)(b+c)(a+c)$
$\displaystyle =2(a+b)(b+c)(c+a)$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 32: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}b+c&a&a\\b&c+a&b\\c&c&a+b\end{vmatrix} =4abc\hfill\text{[CBSE 2006]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}b+c&a&a\\b&c+a&b\\c&c&a+b\end{vmatrix}$
$\displaystyle \text{Applying }R_1\to R_1-R_2-R_3,$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}0&-2c&-2b\\b&c+a&b\\c&c&a+b\end{vmatrix}$
$\displaystyle \text{Applying }C_2\to C_2-C_1\text{ and }C_3\to C_3-C_1,$
$\displaystyle \text{L.H.S.}= \begin{vmatrix}0&-2c&-2b\\b&c+a-b&0\\c&0&a+b-c\end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \text{L.H.S.} =2c\begin{vmatrix}b&0\\c&a+b-c\end{vmatrix} -2b\begin{vmatrix}b&c+a-b\\c&0\end{vmatrix}$
$\displaystyle =2bc(a+b-c)+2bc(c+a-b)$
$\displaystyle =2bc\big[(a+b-c)+(c+a-b)\big]$
$\displaystyle =2bc(2a)$
$\displaystyle =4abc$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 33: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix} b^2+c^2&ab&ac\\ ab&c^2+a^2&bc\\ ac&bc&a^2+b^2 \end{vmatrix} =4a^2b^2c^2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}= \begin{vmatrix} b^2+c^2&ab&ac\\ ab&c^2+a^2&bc\\ ac&bc&a^2+b^2 \end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \text{L.H.S.}=(b^2+c^2) \begin{vmatrix} c^2+a^2&bc\\ bc&a^2+b^2 \end{vmatrix}$
$\displaystyle \qquad -ab\begin{vmatrix} ab&bc\\ ac&a^2+b^2 \end{vmatrix} +ac\begin{vmatrix} ab&c^2+a^2\\ ac&bc \end{vmatrix}$
$\displaystyle =(b^2+c^2) \big[(c^2+a^2)(a^2+b^2)-b^2c^2\big]$
$\displaystyle \qquad -ab\big[ab(a^2+b^2)-abc^2\big] +ac\big[ab^2c-ac(c^2+a^2)\big]$
$\displaystyle =a^2(b^2+c^2)(a^2+b^2+c^2) -a^2b^2(a^2+b^2-c^2)$
$\displaystyle \qquad +a^2c^2(b^2-c^2-a^2)$
$\displaystyle =a^2\big[(b^2+c^2)(a^2+b^2+c^2) -b^2(a^2+b^2-c^2)$
$\displaystyle \qquad +c^2(b^2-c^2-a^2)\big]$
$\displaystyle =a^2(4b^2c^2)$
$\displaystyle =4a^2b^2c^2$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 34: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix} 0&b^2a&c^2a\\ a^2b&0&c^2b\\ a^2c&b^2c&0 \end{vmatrix} =2a^3b^3c^3\hfill\text{[CBSE 2003]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}= \begin{vmatrix} 0&b^2a&c^2a\\ a^2b&0&c^2b\\ a^2c&b^2c&0 \end{vmatrix}$
$\displaystyle \text{Taking }a,b,c\text{ common from }R_1,R_2,R_3,\text{ respectively,}$
$\displaystyle \text{L.H.S.}=abc \begin{vmatrix} 0&b^2&c^2\\ a^2&0&c^2\\ a^2&b^2&0 \end{vmatrix}$
$\displaystyle \text{Taking }a^2,b^2,c^2\text{ common from }C_1,C_2,C_3,\text{ respectively,}$
$\displaystyle \text{L.H.S.}=a^3b^3c^3 \begin{vmatrix} 0&1&1\\ 1&0&1\\ 1&1&0 \end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \text{L.H.S.}=a^3b^3c^3 \left[ -\begin{vmatrix}1&1\\1&0\end{vmatrix} +\begin{vmatrix}1&0\\1&1\end{vmatrix} \right]$
$\displaystyle =a^3b^3c^3[-(-1)+1]$
$\displaystyle =2a^3b^3c^3$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 35: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix} \dfrac{a^2+b^2}{c}&c&c\\ a&\dfrac{b^2+c^2}{a}&a\\ b&b&\dfrac{c^2+a^2}{b} \end{vmatrix} =4abc$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Here, }a\ne0,\ b\ne0\text{ and }c\ne0.$
$\displaystyle \text{L.H.S.}= \begin{vmatrix} \dfrac{a^2+b^2}{c}&c&c\\ a&\dfrac{b^2+c^2}{a}&a\\ b&b&\dfrac{c^2+a^2}{b} \end{vmatrix}$
$\displaystyle =\frac1{abc} \begin{vmatrix} a^2+b^2&c^2&c^2\\ a^2&b^2+c^2&a^2\\ b^2&b^2&c^2+a^2 \end{vmatrix}$
$\displaystyle \text{Applying }C_2\to C_2-C_1\text{ and }C_3\to C_3-C_1,$
$\displaystyle \text{L.H.S.}=\frac1{abc} \begin{vmatrix} a^2+b^2&c^2-a^2-b^2&c^2-a^2-b^2\\ a^2&b^2+c^2-a^2&0\\ b^2&0&c^2+a^2-b^2 \end{vmatrix}$
$\displaystyle \text{Applying }R_1\to R_1-R_2-R_3,$
$\displaystyle \text{L.H.S.}=\frac1{abc} \begin{vmatrix} 0&-2b^2&-2a^2\\ a^2&b^2+c^2-a^2&0\\ b^2&0&c^2+a^2-b^2 \end{vmatrix}$
$\displaystyle \text{Expanding along }C_1,$
$\displaystyle \text{L.H.S.}=\frac1{abc}\left[ -a^2\begin{vmatrix} -2b^2&-2a^2\\ 0&c^2+a^2-b^2 \end{vmatrix} +b^2\begin{vmatrix} -2b^2&-2a^2\\ b^2+c^2-a^2&0 \end{vmatrix} \right]$
$\displaystyle =\frac1{abc}\left[ 2a^2b^2(c^2+a^2-b^2) +2a^2b^2(b^2+c^2-a^2) \right]$
$\displaystyle =\frac{2a^2b^2}{abc} \big[(c^2+a^2-b^2)+(b^2+c^2-a^2)\big]$
$\displaystyle =\frac{2a^2b^2}{abc}(2c^2)$
$\displaystyle =4abc$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 36: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix} -bc&b^2+bc&c^2+bc\\ a^2+ac&-ac&c^2+ac\\ a^2+ab&b^2+ab&-ab \end{vmatrix} =(ab+bc+ca)^3$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }S=ab+bc+ca.$
$\displaystyle \text{L.H.S.}= \begin{vmatrix} -bc&b^2+bc&c^2+bc\\ a^2+ac&-ac&c^2+ac\\ a^2+ab&b^2+ab&-ab \end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \text{L.H.S.} =-bc\begin{vmatrix}-ac&c^2+ac\\b^2+ab&-ab\end{vmatrix} -(b^2+bc)\begin{vmatrix}a^2+ac&c^2+ac\\a^2+ab&-ab\end{vmatrix}$
$\displaystyle \qquad +(c^2+bc)\begin{vmatrix}a^2+ac&-ac\\a^2+ab&b^2+ab\end{vmatrix}$
$\displaystyle \begin{vmatrix}-ac&c^2+ac\\b^2+ab&-ab\end{vmatrix} =a^2bc-c(a+c)b(a+b)$
$\displaystyle =bc\big[a^2-(a+c)(a+b)\big]$
$\displaystyle =-bc(ab+bc+ca)=-bcS$
$\displaystyle \begin{vmatrix}a^2+ac&c^2+ac\\a^2+ab&-ab\end{vmatrix} =-a^2b(a+c)-ac(a+c)(a+b)$
$\displaystyle =-a(a+c)\big[ab+ac+bc\big]=-a(a+c)S$
$\displaystyle \begin{vmatrix}a^2+ac&-ac\\a^2+ab&b^2+ab\end{vmatrix} =ab(a+c)(a+b)+a^2c(a+b)$
$\displaystyle =a(a+b)\big[ab+ac+bc\big]=a(a+b)S$
$\displaystyle \therefore \text{L.H.S.} =b^2c^2S+ab(b+c)(a+c)S+ac(b+c)(a+b)S$
$\displaystyle =S\big[b^2c^2+ab(a+c)(b+c)+ac(a+b)(b+c)\big]$
$\displaystyle =S\big[b^2c^2+a(b+c)\{b(a+c)+c(a+b)\}\big]$
$\displaystyle =S\big[b^2c^2+a(b+c)(ab+ac+2bc)\big]$
$\displaystyle =S\big[b^2c^2+a(b+c)(S+bc)\big]$
$\displaystyle =S\big[b^2c^2+a(b+c)S+abc(b+c)\big]$
$\displaystyle =S\big[S\{bc+a(b+c)\}\big]$
$\displaystyle =S\cdot S^2=S^3$
$\displaystyle =(ab+bc+ca)^3$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 37: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}x+\lambda&2x&2x\\2x&x+\lambda&2x\\2x&2x&x+\lambda\end{vmatrix}=(5x+\lambda)(\lambda-x)^2\hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}=\begin{vmatrix}x+\lambda&2x&2x\\2x&x+\lambda&2x\\2x&2x&x+\lambda\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \text{L.H.S.}=\begin{vmatrix}x+\lambda&2x&2x\\-(\lambda-x)&\lambda-x&0\\-(\lambda-x)&0&\lambda-x\end{vmatrix}$
$\displaystyle =(\lambda-x)^2\begin{vmatrix}x+\lambda&2x&2x\\-1&1&0\\-1&0&1\end{vmatrix}$
$\displaystyle \text{Expanding along }R_3,$
$\displaystyle \text{L.H.S.}=(\lambda-x)^2\left[-\begin{vmatrix}2x&2x\\1&0\end{vmatrix}+\begin{vmatrix}x+\lambda&2x\\-1&1\end{vmatrix}\right]$
$\displaystyle =(\lambda-x)^2\left[2x+(x+\lambda+2x)\right]$
$\displaystyle =(\lambda-x)^2(\lambda+5x)$
$\displaystyle =(5x+\lambda)(\lambda-x)^2$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 38: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}x+4&2x&2x\\2x&x+4&2x\\2x&2x&x+4\end{vmatrix}=(5x+4)(4-x)^2\hfill\text{[CBSE 2007, 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}=\begin{vmatrix}x+4&2x&2x\\2x&x+4&2x\\2x&2x&x+4\end{vmatrix}$
$\displaystyle \text{Applying }R_1\to R_1+R_2+R_3,$
$\displaystyle \text{L.H.S.}=\begin{vmatrix}5x+4&5x+4&5x+4\\2x&x+4&2x\\2x&2x&x+4\end{vmatrix}$
$\displaystyle =(5x+4)\begin{vmatrix}1&1&1\\2x&x+4&2x\\2x&2x&x+4\end{vmatrix}$
$\displaystyle \text{Applying }C_2\to C_2-C_1\text{ and }C_3\to C_3-C_1,$
$\displaystyle \text{L.H.S.}=(5x+4)\begin{vmatrix}1&0&0\\2x&4-x&0\\2x&0&4-x\end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \text{L.H.S.}=(5x+4)\begin{vmatrix}4-x&0\\0&4-x\end{vmatrix}$
$\displaystyle =(5x+4)(4-x)^2$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 39: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}y+z&z&y\\z&z+x&x\\y&x&x+y\end{vmatrix}=4xyz$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}=\begin{vmatrix}y+z&z&y\\z&z+x&x\\y&x&x+y\end{vmatrix}$
$\displaystyle \text{Applying }R_1\to R_1-R_2-R_3,$
$\displaystyle \text{L.H.S.}=\begin{vmatrix}0&-2x&-2x\\z&z+x&x\\y&x&x+y\end{vmatrix}$
$\displaystyle =-2x\begin{vmatrix}0&1&1\\z&z+x&x\\y&x&x+y\end{vmatrix}$
$\displaystyle \text{Applying }C_2\to C_2-C_3,$
$\displaystyle \text{L.H.S.}=-2x\begin{vmatrix}0&0&1\\z&z&x\\y&-y&x+y\end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \text{L.H.S.}=-2x\begin{vmatrix}z&z\\y&-y\end{vmatrix}$
$\displaystyle =-2x(-zy-zy)$
$\displaystyle =-2x(-2yz)$
$\displaystyle =4xyz$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 40: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}-a(b^2+c^2-a^2)&2b^3&2c^3\\2a^3&-b(c^2+a^2-b^2)&2c^3\\2a^3&2b^3&-c(a^2+b^2-c^2)\end{vmatrix}=abc(a^2+b^2+c^2)^3$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}=\begin{vmatrix}-a(b^2+c^2-a^2)&2b^3&2c^3\\2a^3&-b(c^2+a^2-b^2)&2c^3\\2a^3&2b^3&-c(a^2+b^2-c^2)\end{vmatrix}$
$\displaystyle =abc\begin{vmatrix}a^2-b^2-c^2&2b^2&2c^2\\2a^2&b^2-c^2-a^2&2c^2\\2a^2&2b^2&c^2-a^2-b^2\end{vmatrix}\quad\text{Taking }a,b,c\text{ common from }C_1,C_2,C_3$
$\displaystyle =abc\begin{vmatrix}a^2+b^2+c^2&2b^2&2c^2\\a^2+b^2+c^2&b^2-a^2-c^2&2c^2\\a^2+b^2+c^2&2b^2&c^2-a^2-b^2\end{vmatrix}\quad\text{Applying }C_1\to C_1+C_2+C_3$
$\displaystyle =abc(a^2+b^2+c^2)\begin{vmatrix}1&2b^2&2c^2\\1&b^2-a^2-c^2&2c^2\\1&2b^2&c^2-a^2-b^2\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle =abc(a^2+b^2+c^2)\begin{vmatrix}1&2b^2&2c^2\\0&-(a^2+b^2+c^2)&0\\0&0&-(a^2+b^2+c^2)\end{vmatrix}$
$\displaystyle =abc(a^2+b^2+c^2)^3$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 41: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}1+a&1&1\\1&1+a&1\\1&1&1+a\end{vmatrix}=a^3+3a^2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}=\begin{vmatrix}1+a&1&1\\1&1+a&1\\1&1&1+a\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \text{L.H.S.}=\begin{vmatrix}1+a&1&1\\-a&a&0\\-a&0&a\end{vmatrix}$
$\displaystyle =a^2\begin{vmatrix}1+a&1&1\\-1&1&0\\-1&0&1\end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \text{L.H.S.}=a^2\left[(1+a)\begin{vmatrix}1&0\\0&1\end{vmatrix}-\begin{vmatrix}-1&0\\-1&1\end{vmatrix}+\begin{vmatrix}-1&1\\-1&0\end{vmatrix}\right]$
$\displaystyle =a^2[(1+a)+1+1]$
$\displaystyle =a^2(a+3)$
$\displaystyle =a^3+3a^2$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 42: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}2y&y-z-x&2y\\2z&2z&z-x-y\\x-y-z&2x&2x\end{vmatrix}=(x+y+z)^3\hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}=\begin{vmatrix}2y&y-z-x&2y\\2z&2z&z-x-y\\x-y-z&2x&2x\end{vmatrix}$
$\displaystyle \text{Applying }R_1\to R_1+R_2+R_3,$
$\displaystyle \text{L.H.S.}=\begin{vmatrix}x+y+z&x+y+z&x+y+z\\2z&2z&z-x-y\\x-y-z&2x&2x\end{vmatrix}$
$\displaystyle =(x+y+z)\begin{vmatrix}1&1&1\\2z&2z&z-x-y\\x-y-z&2x&2x\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1-C_2,$
$\displaystyle \text{L.H.S.}=(x+y+z)\begin{vmatrix}0&1&1\\0&2z&z-x-y\\-(x+y+z)&2x&2x\end{vmatrix}$
$\displaystyle =(x+y+z)^2\begin{vmatrix}0&1&1\\0&2z&z-x-y\\-1&2x&2x\end{vmatrix}$
$\displaystyle \text{Expanding along }C_1,$
$\displaystyle \text{L.H.S.}=(x+y+z)^2(-1)\begin{vmatrix}1&1\\2z&z-x-y\end{vmatrix}$
$\displaystyle =(x+y+z)^2(-1)\big[(z-x-y)-2z\big]$
$\displaystyle =(x+y+z)^2(-1)(-x-y-z)$
$\displaystyle =(x+y+z)^3$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 43: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}y+z&x&y\\z+x&z&x\\x+y&y&z\end{vmatrix}=(x+y+z)(x-z)^2\hfill\text{[CBSE 2007]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}=\begin{vmatrix}y+z&x&y\\z+x&z&x\\x+y&y&z\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1-C_2-C_3,$
$\displaystyle \text{L.H.S.}=\begin{vmatrix}z-x&x&y\\0&z&x\\x-z&y&z\end{vmatrix}$
$\displaystyle =(x-z)\begin{vmatrix}-1&x&y\\0&z&x\\1&y&z\end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \text{L.H.S.}=(x-z)\left[-\begin{vmatrix}z&x\\y&z\end{vmatrix}-x\begin{vmatrix}0&x\\1&z\end{vmatrix}+y\begin{vmatrix}0&z\\1&y\end{vmatrix}\right]$
$\displaystyle =(x-z)\left[-(z^2-xy)+x^2-yz\right]$
$\displaystyle =(x-z)\left[x^2-z^2+y(x-z)\right]$
$\displaystyle =(x-z)\left[(x-z)(x+z)+y(x-z)\right]$
$\displaystyle =(x-z)^2(x+y+z)$
$\displaystyle =(x+y+z)(x-z)^2$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 44: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}a+x&y&z\\x&a+y&z\\x&y&a+z\end{vmatrix}=a^2(a+x+y+z)\hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}=\begin{vmatrix}a+x&y&z\\x&a+y&z\\x&y&a+z\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1+C_2+C_3,$
$\displaystyle \text{L.H.S.}=\begin{vmatrix}a+x+y+z&y&z\\a+x+y+z&a+y&z\\a+x+y+z&y&a+z\end{vmatrix}$
$\displaystyle =(a+x+y+z)\begin{vmatrix}1&y&z\\1&a+y&z\\1&y&a+z\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \text{L.H.S.}=(a+x+y+z)\begin{vmatrix}1&y&z\\0&a&0\\0&0&a\end{vmatrix}$
$\displaystyle \text{Expanding along }C_1,$
$\displaystyle \text{L.H.S.}=(a+x+y+z)\begin{vmatrix}a&0\\0&a\end{vmatrix}$
$\displaystyle =a^2(a+x+y+z)$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 45: }\text{Prove the following identity:}$
$\displaystyle \begin{vmatrix}a^3&2&a\\b^3&2&b\\c^3&2&c\end{vmatrix}=2(a-b)(b-c)(c-a)(a+b+c)\hfill\text{[CBSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{L.H.S.}=\begin{vmatrix}a^3&2&a\\b^3&2&b\\c^3&2&c\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \text{L.H.S.}=\begin{vmatrix}a^3&2&a\\b^3-a^3&0&b-a\\c^3-a^3&0&c-a\end{vmatrix}$
$\displaystyle =-(a-b)(c-a)\begin{vmatrix}a^3&2&a\\a^2+ab+b^2&0&1\\a^2+ac+c^2&0&1\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_3,$
$\displaystyle \text{L.H.S.}=-(a-b)(c-a)\begin{vmatrix}a^3&2&a\\(b^2-c^2)+a(b-c)&0&0\\a^2+ac+c^2&0&1\end{vmatrix}$
$\displaystyle =-(a-b)(c-a)(b-c)(a+b+c)\begin{vmatrix}a^3&2&a\\1&0&0\\a^2+ac+c^2&0&1\end{vmatrix}$
$\displaystyle \text{Expanding along }C_2,$
$\displaystyle \text{L.H.S.}=-(a-b)(c-a)(b-c)(a+b+c)(-2)$
$\displaystyle =2(a-b)(b-c)(c-a)(a+b+c)$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 46: }\text{Without expanding, prove that}$
$\displaystyle \begin{vmatrix}a&b&c\\x&y&z\\p&q&r\end{vmatrix}=\begin{vmatrix}x&y&z\\p&q&r\\a&b&c\end{vmatrix}=\begin{vmatrix}y&b&q\\x&a&p\\z&c&r\end{vmatrix}$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{vmatrix}x&y&z\\p&q&r\\a&b&c\end{vmatrix}=-\begin{vmatrix}x&y&z\\a&b&c\\p&q&r\end{vmatrix}\quad[\text{Interchanging }R_2\text{ and }R_3]$
$\displaystyle =\begin{vmatrix}a&b&c\\x&y&z\\p&q&r\end{vmatrix}\quad[\text{Interchanging }R_1\text{ and }R_2]$
$\displaystyle \therefore \begin{vmatrix}a&b&c\\x&y&z\\p&q&r\end{vmatrix}=\begin{vmatrix}x&y&z\\p&q&r\\a&b&c\end{vmatrix}$
$\displaystyle \begin{vmatrix}y&b&q\\x&a&p\\z&c&r\end{vmatrix}=\begin{vmatrix}y&x&z\\b&a&c\\q&p&r\end{vmatrix}\quad[\text{Transposing the determinant}]$
$\displaystyle =-\begin{vmatrix}x&y&z\\b&a&c\\p&q&r\end{vmatrix}\quad[\text{Interchanging }C_1\text{ and }C_2]$
$\displaystyle =\begin{vmatrix}x&y&z\\a&b&c\\p&q&r\end{vmatrix}\quad[\text{Interchanging }R_2\text{ and }R_3]$
$\displaystyle =\begin{vmatrix}a&b&c\\x&y&z\\p&q&r\end{vmatrix}\quad[\text{Interchanging }R_1\text{ and }R_2]$
$\displaystyle \therefore \begin{vmatrix}a&b&c\\x&y&z\\p&q&r\end{vmatrix}=\begin{vmatrix}y&b&q\\x&a&p\\z&c&r\end{vmatrix}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 47: }\text{Show that}$
$\displaystyle \begin{vmatrix}x+1&x+2&x+a\\x+2&x+3&x+b\\x+3&x+4&x+c\end{vmatrix}=0,\quad\text{where }a,b,c\text{ are in A.P.}\hfill\text{[CBSE 2005]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }a,b,c\text{ are in A.P.,}$
$\displaystyle 2b=a+c$
$\displaystyle \text{Let }\Delta=\begin{vmatrix}x+1&x+2&x+a\\x+2&x+3&x+b\\x+3&x+4&x+c\end{vmatrix}$
$\displaystyle 2R_2=\bigl(2x+4,\;2x+6,\;2x+2b\bigr)$
$\displaystyle =\bigl(2x+4,\;2x+6,\;2x+a+c\bigr)$
$\displaystyle =R_1+R_3$
$\displaystyle \therefore R_1-2R_2+R_3=O$
$\displaystyle \text{Thus, the rows of }\Delta\text{ are linearly dependent.}$
$\displaystyle \therefore \Delta=0$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 48: }\text{Show that}$
$\displaystyle \begin{vmatrix}x-3&x-4&x-\alpha\\x-2&x-3&x-\beta\\x-1&x-2&x-\gamma\end{vmatrix}=0,\quad\text{where }\alpha,\beta,\gamma\text{ are in A.P.}\hfill\text{[CBSE 2007]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }\alpha,\beta,\gamma\text{ are in A.P.,}$
$\displaystyle 2\beta=\alpha+\gamma$
$\displaystyle \text{Let }\Delta=\begin{vmatrix}x-3&x-4&x-\alpha\\x-2&x-3&x-\beta\\x-1&x-2&x-\gamma\end{vmatrix}$
$\displaystyle 2R_2=\bigl(2x-4,\;2x-6,\;2x-2\beta\bigr)$
$\displaystyle =\bigl(2x-4,\;2x-6,\;2x-\alpha-\gamma\bigr)$
$\displaystyle =R_1+R_3$
$\displaystyle \therefore R_1-2R_2+R_3=O$
$\displaystyle \text{Thus, the rows of }\Delta\text{ are linearly dependent.}$
$\displaystyle \therefore \Delta=0$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 49: }\text{If }a,b,c\text{ are real numbers such that}$
$\displaystyle \begin{vmatrix}b+c&c+a&a+b\\c+a&a+b&b+c\\a+b&b+c&c+a\end{vmatrix}=0,\quad\text{then show that either }a+b+c=0\text{ or }a=b=c.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\Delta=\begin{vmatrix}b+c&c+a&a+b\\c+a&a+b&b+c\\a+b&b+c&c+a\end{vmatrix}$
$\displaystyle \text{Applying }R_1\to R_1+R_2+R_3,$
$\displaystyle \Delta=\begin{vmatrix}2(a+b+c)&2(a+b+c)&2(a+b+c)\\c+a&a+b&b+c\\a+b&b+c&c+a\end{vmatrix}$
$\displaystyle =2(a+b+c)\begin{vmatrix}1&1&1\\c+a&a+b&b+c\\a+b&b+c&c+a\end{vmatrix}$
$\displaystyle \text{Applying }C_2\to C_2-C_1\text{ and }C_3\to C_3-C_1,$
$\displaystyle \Delta=2(a+b+c)\begin{vmatrix}1&0&0\\c+a&b-c&b-a\\a+b&c-a&c-b\end{vmatrix}$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle \Delta=2(a+b+c)\begin{vmatrix}b-c&b-a\\c-a&c-b\end{vmatrix}$
$\displaystyle =2(a+b+c)\left[(b-c)(c-b)-(b-a)(c-a)\right]$
$\displaystyle =-2(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$\displaystyle =-(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$
$\displaystyle \text{But }\Delta=0.$
$\displaystyle \therefore (a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right]=0$
$\displaystyle \therefore a+b+c=0\text{ or }(a-b)^2+(b-c)^2+(c-a)^2=0$
$\displaystyle \text{Since }a,b,c\text{ are real numbers,}$
$\displaystyle (a-b)^2+(b-c)^2+(c-a)^2=0\Rightarrow a-b=b-c=c-a=0$
$\displaystyle \Rightarrow a=b=c$
$\displaystyle \therefore \text{Either }a+b+c=0\text{ or }a=b=c.$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 50: }\text{If}$
$\displaystyle \begin{vmatrix}p&b&c\\a&q&c\\a&b&r\end{vmatrix}=0,\quad\text{find }\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c},\quad p\ne a,\ q\ne b,\ r\ne c.\hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\Delta=\begin{vmatrix}p&b&c\\a&q&c\\a&b&r\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_3,$
$\displaystyle \Delta=\begin{vmatrix}p&b&c\\0&q-b&c-r\\a&b&r\end{vmatrix}$
$\displaystyle \text{Expanding along }C_1,$
$\displaystyle \Delta=p\begin{vmatrix}q-b&c-r\\b&r\end{vmatrix}+a\begin{vmatrix}b&c\\q-b&c-r\end{vmatrix}$
$\displaystyle =p\left[r(q-b)-b(c-r)\right]+a\left[b(c-r)-c(q-b)\right]$
$\displaystyle =pr(q-b)+pb(r-c)-ab(r-c)-ac(q-b)$
$\displaystyle =(pr-ac)(q-b)+b(p-a)(r-c)$
$\displaystyle \text{Since }\Delta=0,$
$\displaystyle (pr-ac)(q-b)+b(p-a)(r-c)=0$
$\displaystyle \frac{pr-ac}{(p-a)(r-c)}+\frac{b}{q-b}=0$
$\displaystyle \text{Since }pr-ac=r(p-a)+a(r-c),$
$\displaystyle \frac{r}{r-c}+\frac{a}{p-a}+\frac{b}{q-b}=0$
$\displaystyle \therefore \frac{a}{p-a}+\frac{b}{q-b}+\frac{r}{r-c}=0$
$\displaystyle \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$
$\displaystyle =\frac{p-a}{p-a}+\frac{q-b}{q-b}+\frac{a}{p-a}+\frac{b}{q-b}+\frac{r}{r-c}$
$\displaystyle =1+1+0$
$\displaystyle =2$
$\displaystyle \therefore \text{The required value is }2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 51: }\text{Show that }x=2\text{ is a root of the equation}$
$\displaystyle \begin{vmatrix}x&-6&-1\\2&-3x&x-3\\-3&2x&x+2\end{vmatrix}=0\text{ and solve it completely.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\Delta=\begin{vmatrix}x&-6&-1\\2&-3x&x-3\\-3&2x&x+2\end{vmatrix}$
$\displaystyle \text{Applying }R_3\to R_3-R_1,$
$\displaystyle \Delta=\begin{vmatrix}x&-6&-1\\2&-3x&x-3\\-x-3&2x+6&x+3\end{vmatrix}$
$\displaystyle =(x+3)\begin{vmatrix}x&-6&-1\\2&-3x&x-3\\-1&2&1\end{vmatrix}$
$\displaystyle \text{Applying }R_1\to R_1-R_2,$
$\displaystyle \Delta=(x+3)\begin{vmatrix}x-2&3x-6&2-x\\2&-3x&x-3\\-1&2&1\end{vmatrix}$
$\displaystyle =(x+3)(x-2)\begin{vmatrix}1&3&-1\\2&-3x&x-3\\-1&2&1\end{vmatrix}$
$\displaystyle \text{Applying }C_3\to C_3+C_1,$
$\displaystyle \Delta=(x+3)(x-2)\begin{vmatrix}1&3&0\\2&-3x&x-1\\-1&2&0\end{vmatrix}$
$\displaystyle =(x+3)(x-2)(x-1)\begin{vmatrix}1&3&0\\2&-3x&1\\-1&2&0\end{vmatrix}$
$\displaystyle \text{Expanding along }C_3,$
$\displaystyle \Delta=-(x+3)(x-2)(x-1)\begin{vmatrix}1&3\\-1&2\end{vmatrix}$
$\displaystyle =-(x+3)(x-2)(x-1)(2+3)$
$\displaystyle =-5(x+3)(x-2)(x-1)$
$\displaystyle \text{Since }\Delta=0,$
$\displaystyle -5(x+3)(x-2)(x-1)=0$
$\displaystyle \Rightarrow x+3=0,\quad x-2=0\quad\text{or}\quad x-1=0$
$\displaystyle \therefore x=-3,\;2,\;1$
$\displaystyle \text{Hence, }x=2\text{ is a root and all the roots are }-3,\;1,\;2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 52: }\text{Solve the following determinant equations:}$
$\displaystyle \text{(i) }\begin{vmatrix}x+a&b&c\\a&x+b&c\\a&b&x+c\end{vmatrix}=0\hfill\text{[CBSE 2006]}$
$\displaystyle \text{(ii) }\begin{vmatrix}x+a&x&x\\x&x+a&x\\x&x&x+a\end{vmatrix}=0,\quad a\ne0\hfill\text{[CBSE 2011]}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i)}$
$\displaystyle \text{Let }\Delta=\begin{vmatrix}x+a&b&c\\a&x+b&c\\a&b&x+c\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1+C_2+C_3,$
$\displaystyle \Delta=\begin{vmatrix}x+a+b+c&b&c\\x+a+b+c&x+b&c\\x+a+b+c&b&x+c\end{vmatrix}$
$\displaystyle =(x+a+b+c)\begin{vmatrix}1&b&c\\1&x+b&c\\1&b&x+c\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \Delta=(x+a+b+c)\begin{vmatrix}1&b&c\\0&x&0\\0&0&x\end{vmatrix}$
$\displaystyle =x^2(x+a+b+c)$
$\displaystyle \text{Since }\Delta=0,$
$\displaystyle x^2(x+a+b+c)=0$
$\displaystyle \Rightarrow x^2=0\quad\text{or}\quad x+a+b+c=0$
$\displaystyle \therefore x=0\quad\text{or}\quad x=-(a+b+c)$

$\displaystyle \text{(ii)}$
$\displaystyle \text{Let }\Delta=\begin{vmatrix}x+a&x&x\\x&x+a&x\\x&x&x+a\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1+C_2+C_3,$
$\displaystyle \Delta=\begin{vmatrix}3x+a&x&x\\3x+a&x+a&x\\3x+a&x&x+a\end{vmatrix}$
$\displaystyle =(3x+a)\begin{vmatrix}1&x&x\\1&x+a&x\\1&x&x+a\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \Delta=(3x+a)\begin{vmatrix}1&x&x\\0&a&0\\0&0&a\end{vmatrix}$
$\displaystyle =a^2(3x+a)$
$\displaystyle \text{Since }\Delta=0,$
$\displaystyle a^2(3x+a)=0$
$\displaystyle \text{But }a\ne0\Rightarrow a^2\ne0.$
$\displaystyle \therefore 3x+a=0$
$\displaystyle \Rightarrow x=-\frac{a}{3}$

$\displaystyle \textbf{Question 52: }\text{Solve the following determinant equations:}$
$\displaystyle \text{(iii) }\begin{vmatrix}3x-8&3&3\\3&3x-8&3\\3&3&3x-8\end{vmatrix}=0\hfill\text{[CBSE 2008]}$
$\displaystyle \text{(iv) }\begin{vmatrix}1&x&x^2\\1&a&a^2\\1&b&b^2\end{vmatrix}=0,\quad a\ne b$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(iii)}$
$\displaystyle \text{Let }\Delta=\begin{vmatrix}3x-8&3&3\\3&3x-8&3\\3&3&3x-8\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1+C_2+C_3,$
$\displaystyle \Delta=\begin{vmatrix}3x-2&3&3\\3x-2&3x-8&3\\3x-2&3&3x-8\end{vmatrix}$
$\displaystyle =(3x-2)\begin{vmatrix}1&3&3\\1&3x-8&3\\1&3&3x-8\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \Delta=(3x-2)\begin{vmatrix}1&3&3\\0&3x-11&0\\0&0&3x-11\end{vmatrix}$
$\displaystyle =(3x-2)(3x-11)^2$
$\displaystyle \text{Since }\Delta=0,$
$\displaystyle (3x-2)(3x-11)^2=0$
$\displaystyle \therefore x=\frac23\text{ or }x=\frac{11}{3}\;(\text{repeated root})$

$\displaystyle \text{(iv)}$
$\displaystyle \text{Let }\Delta=\begin{vmatrix}1&x&x^2\\1&a&a^2\\1&b&b^2\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \Delta=\begin{vmatrix}1&x&x^2\\0&a-x&a^2-x^2\\0&b-x&b^2-x^2\end{vmatrix}$
$\displaystyle =(a-x)(b-x)\begin{vmatrix}1&x&x^2\\0&1&a+x\\0&1&b+x\end{vmatrix}$
$\displaystyle \text{Applying }R_3\to R_3-R_2,$
$\displaystyle \Delta=(a-x)(b-x)\begin{vmatrix}1&x&x^2\\0&1&a+x\\0&0&b-a\end{vmatrix}$
$\displaystyle =(a-x)(b-x)(b-a)$
$\displaystyle \text{Since }a\ne b,\;b-a\ne0.$
$\displaystyle \text{Also, }\Delta=0.$
$\displaystyle \therefore (a-x)(b-x)=0$
$\displaystyle \therefore x=a\text{ or }x=b$
$\displaystyle \\$

$\displaystyle \textbf{Question 52: }\text{Solve the following determinant equations:}$
$\displaystyle \text{(v) }\begin{vmatrix}x+1&3&5\\2&x+2&5\\2&3&x+4\end{vmatrix}=0$
$\displaystyle \text{(vi) }\begin{vmatrix}1&x&x^3\\1&b&b^3\\1&c&c^3\end{vmatrix}=0,\quad b\ne c$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(v)}$
$\displaystyle \text{Let }\Delta=\begin{vmatrix}x+1&3&5\\2&x+2&5\\2&3&x+4\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1+C_2+C_3,$
$\displaystyle \Delta=\begin{vmatrix}x+9&3&5\\x+9&x+2&5\\x+9&3&x+4\end{vmatrix}$
$\displaystyle =(x+9)\begin{vmatrix}1&3&5\\1&x+2&5\\1&3&x+4\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \Delta=(x+9)\begin{vmatrix}1&3&5\\0&x-1&0\\0&0&x-1\end{vmatrix}$
$\displaystyle =(x+9)(x-1)^2$
$\displaystyle \text{Since }\Delta=0,$
$\displaystyle (x+9)(x-1)^2=0$
$\displaystyle \therefore x=-9\text{ or }x=1\;(\text{repeated root})$

$\displaystyle \text{(vi)}$
$\displaystyle \text{Let }\Delta=\begin{vmatrix}1&x&x^3\\1&b&b^3\\1&c&c^3\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,$
$\displaystyle \Delta=\begin{vmatrix}1&x&x^3\\0&b-x&b^3-x^3\\0&c-x&c^3-x^3\end{vmatrix}$
$\displaystyle =(b-x)(c-x)\begin{vmatrix}1&x&x^3\\0&1&b^2+bx+x^2\\0&1&c^2+cx+x^2\end{vmatrix}$
$\displaystyle \text{Applying }R_3\to R_3-R_2,$
$\displaystyle \Delta=(b-x)(c-x)\begin{vmatrix}1&x&x^3\\0&1&b^2+bx+x^2\\0&0&c^2-b^2+x(c-b)\end{vmatrix}$
$\displaystyle =(b-x)(c-x)(c-b)(x+b+c)$
$\displaystyle \text{Since }\Delta=0,$
$\displaystyle (b-x)(c-x)(c-b)(x+b+c)=0$
$\displaystyle \text{But }b\ne c\Rightarrow c-b\ne0.$
$\displaystyle \therefore b-x=0,\quad c-x=0\quad\text{or}\quad x+b+c=0$
$\displaystyle \therefore x=b,\quad x=c\quad\text{or}\quad x=-(b+c)$

$\displaystyle \textbf{Question 52: }\text{Solve the following determinant equations:}$
$\displaystyle \text{(vii) }\begin{vmatrix}15-2x&11-3x&7-x\\11&17&14\\10&16&13\end{vmatrix}=0$
$\displaystyle \text{(viii) }\begin{vmatrix}1&1&x\\p+1&p+1&p+x\\3&x+1&x+2\end{vmatrix}=0$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(vii)}$
$\displaystyle \text{Let }\Delta=\begin{vmatrix}15-2x&11-3x&7-x\\11&17&14\\10&16&13\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1-2C_3,$
$\displaystyle \Delta=\begin{vmatrix}1&11-3x&7-x\\-17&17&14\\-16&16&13\end{vmatrix}$
$\displaystyle \text{Using the original columns, apply }C_1\to C_1+C_2\text{ and }C_2\to C_2-C_3.$
$\displaystyle \Delta=\begin{vmatrix}12-3x&4-2x&7-x\\0&3&14\\0&3&13\end{vmatrix}$
$\displaystyle \text{Expanding along }C_1,$
$\displaystyle \Delta=(12-3x)\begin{vmatrix}3&14\\3&13\end{vmatrix}$
$\displaystyle =(12-3x)(39-42)$
$\displaystyle =-3(12-3x)$
$\displaystyle \text{Since }\Delta=0,$
$\displaystyle -3(12-3x)=0$
$\displaystyle \Rightarrow 12-3x=0$
$\displaystyle \therefore x=4$

$\displaystyle \text{(viii)}$
$\displaystyle \text{Let }\Delta=\begin{vmatrix}1&1&x\\p+1&p+1&p+x\\3&x+1&x+2\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-(p+1)R_1,$
$\displaystyle \Delta=\begin{vmatrix}1&1&x\\0&0&p(1-x)\\3&x+1&x+2\end{vmatrix}$
$\displaystyle \text{Expanding along }R_2,$
$\displaystyle \Delta=-p(1-x)\begin{vmatrix}1&1\\3&x+1\end{vmatrix}$
$\displaystyle =-p(1-x)(x+1-3)$
$\displaystyle =-p(1-x)(x-2)$
$\displaystyle =p(x-1)(x-2)$
$\displaystyle \text{Since }\Delta=0,$
$\displaystyle p(x-1)(x-2)=0$
$\displaystyle \text{If }p\ne0,\text{ then }(x-1)(x-2)=0.$
$\displaystyle \therefore x=1\text{ or }x=2$
$\displaystyle \text{If }p=0,\text{ the equation is satisfied for every value of }x.$
$\displaystyle \\$

$\displaystyle \textbf{Question 52: }\text{Solve the following determinant equation:}$
$\displaystyle \text{(ix) }\begin{vmatrix}3&-2&\sin(3\theta)\\-7&8&\cos(2\theta)\\-11&14&2\end{vmatrix}=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(ix)}$
$\displaystyle \text{Let }\Delta=\begin{vmatrix}3&-2&\sin(3\theta)\\-7&8&\cos(2\theta)\\-11&14&2\end{vmatrix}$
$\displaystyle \text{Applying }C_1\to C_1+C_2,$
$\displaystyle \Delta=\begin{vmatrix}1&-2&\sin(3\theta)\\1&8&\cos(2\theta)\\3&14&2\end{vmatrix}$
$\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-3R_1,$
$\displaystyle \Delta=\begin{vmatrix}1&-2&\sin(3\theta)\\0&10&\cos(2\theta)-\sin(3\theta)\\0&20&2-3\sin(3\theta)\end{vmatrix}$
$\displaystyle \text{Expanding along }C_1,$
$\displaystyle \Delta=10\bigl(2-3\sin(3\theta)\bigr)-20\bigl(\cos(2\theta)-\sin(3\theta)\bigr)$
$\displaystyle =20-30\sin(3\theta)-20\cos(2\theta)+20\sin(3\theta)$
$\displaystyle =20-10\sin(3\theta)-20\cos(2\theta)$
$\displaystyle \text{Since }\Delta=0,$
$\displaystyle \sin(3\theta)+2\cos(2\theta)-2=0$
$\displaystyle \text{Using }\sin(3\theta)=3\sin\theta-4\sin^3\theta\text{ and }\cos(2\theta)=1-2\sin^2\theta,$
$\displaystyle 3\sin\theta-4\sin^3\theta+2(1-2\sin^2\theta)-2=0$
$\displaystyle \Rightarrow 3\sin\theta-4\sin^2\theta-4\sin^3\theta=0$
$\displaystyle \Rightarrow \sin\theta\left(3-4\sin\theta-4\sin^2\theta\right)=0$
$\displaystyle \Rightarrow \sin\theta\left(2\sin\theta+3\right)\left(2\sin\theta-1\right)=0$
$\displaystyle \Rightarrow \sin\theta=0\text{ or }\sin\theta=-\frac32\text{ or }\sin\theta=\frac12$
$\displaystyle \text{Since }\sin\theta=-\frac32\text{ is impossible,}$
$\displaystyle \sin\theta=0\text{ or }\sin\theta=\frac12$
$\displaystyle \therefore \theta=n\pi\text{ or }\theta=n\pi+(-1)^n\frac{\pi}{6},\quad n\in\mathbb Z$
$\displaystyle \\$

$\displaystyle \textbf{Question 53: }\text{If }a,b\text{ and }c\text{ are all non-zero and}$
$\displaystyle \begin{vmatrix}1+a&1&1\\1&1+b&1\\1&1&1+c\end{vmatrix}=0,\text{ then prove that}$
$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1=0.\hfill\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have}$
$\displaystyle \begin{vmatrix}1+a&1&1\\1&1+b&1\\1&1&1+c\end{vmatrix}=0$
$\displaystyle \text{Applying }C_1\to C_1-C_2,$
$\displaystyle \begin{vmatrix}a&1&1\\-b&1+b&1\\0&1&1+c\end{vmatrix}=0$
$\displaystyle \text{Applying }C_2\to C_2-C_3,$
$\displaystyle \begin{vmatrix}a&0&1\\-b&b&1\\0&-c&1+c\end{vmatrix}=0$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle a\begin{vmatrix}b&1\\-c&1+c\end{vmatrix}+\begin{vmatrix}-b&b\\0&-c\end{vmatrix}=0$
$\displaystyle \Rightarrow a\{b(1+c)+c\}+bc=0$
$\displaystyle \Rightarrow ab+abc+ac+bc=0$
$\displaystyle \text{Since }a,b,c\ne0,\text{ dividing by }abc,\text{ we get}$
$\displaystyle \frac{1}{c}+1+\frac{1}{b}+\frac{1}{a}=0$
$\displaystyle \therefore \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1=0$
$\displaystyle \\$

$\displaystyle \textbf{Question 54: }\text{If}$
$\displaystyle \begin{vmatrix}a&b-y&c-z\\a-x&b&c-z\\a-x&b-y&c\end{vmatrix}=0,\text{ then using properties of determinants, find the value of}$
$\displaystyle \frac{a}{x}+\frac{b}{y}+\frac{c}{z},\quad\text{where }x,y,z\ne0.\hfill\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{vmatrix}a&b-y&c-z\\a-x&b&c-z\\a-x&b-y&c\end{vmatrix}=0$
$\displaystyle \text{Applying }R_1\to R_1-R_2,$
$\displaystyle \begin{vmatrix}x&-y&0\\a-x&b&c-z\\a-x&b-y&c\end{vmatrix}=0$
$\displaystyle \text{Applying }R_2\to R_2-R_3,$
$\displaystyle \begin{vmatrix}x&-y&0\\0&y&-z\\a-x&b-y&c\end{vmatrix}=0$
$\displaystyle \text{Expanding along }R_1,$
$\displaystyle x\begin{vmatrix}y&-z\\b-y&c\end{vmatrix}+y\begin{vmatrix}0&-z\\a-x&c\end{vmatrix}=0$
$\displaystyle \Rightarrow x(yc+zb-zy)+y(-za+zx)=0$
$\displaystyle \Rightarrow xyc+xzb-xyz-azy+xyz=0$
$\displaystyle \Rightarrow xyc+xzb+ayz-2xyz=0$
$\displaystyle \text{Since }x,y,z\ne0,\text{ dividing by }xyz,\text{ we get}$
$\displaystyle \frac{c}{z}+\frac{b}{y}+\frac{a}{x}-2=0$
$\displaystyle \therefore \frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2$
$\displaystyle \\$