Class 12: Determinants – Exercise 6.2

$\displaystyle \textbf{Question 1: } \text{Evaluate the following determinant:}$

$\displaystyle \textbf{(i) } \begin{vmatrix} 1 & 3 & 5\\ 2 & 6 & 10\\ 31 & 11 & 38 \end{vmatrix}$ $\displaystyle \textbf{(ii) } \begin{vmatrix} 67 & 19 & 21\\ 39 & 13 & 14\\ 81 & 24 & 26 \end{vmatrix}$ $\displaystyle \textbf{(iii) } \begin{vmatrix} a & h & g\\ h & b & f\\ g & f & c \end{vmatrix}$ $\displaystyle \textbf{(iv) } \begin{vmatrix} 1 & -3 & 2\\ 4 & -1 & 2\\ 3 & 5 & 2 \end{vmatrix}$

$\displaystyle \textbf{(v) } \begin{vmatrix} 1 & 4 & 9\\ 4 & 9 & 16\\ 9 & 16 & 25 \end{vmatrix}$ $\displaystyle \textbf{(vi) } \begin{vmatrix} 6 & -3 & 2\\ 2 & -1 & 2\\ -10 & 5 & 2 \end{vmatrix}$ $\displaystyle \textbf{(vii) } \begin{vmatrix} 1 & 3 & 9 & 27\\ 3 & 9 & 27 & 1\\ 9 & 27 & 1 & 3\\ 27 & 1 & 3 & 9 \end{vmatrix}$

$\displaystyle \textbf{(viii) } \left| \begin{array}{ccc} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{array} \right| \hspace{2.0cm} \text{[CBSE 2012]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle \Delta=\begin{vmatrix} 1 & 3 & 5\\ 2 & 6 & 10\\ 31 & 11 & 38 \end{vmatrix}$

$\displaystyle =1\begin{vmatrix} 6 & 10\\ 11 & 38 \end{vmatrix}-3\begin{vmatrix} 2 & 10\\ 31 & 38 \end{vmatrix}+5\begin{vmatrix} 2 & 6\\ 31 & 11 \end{vmatrix}$

$\displaystyle =1(228-110)-3(76-310)+5(22-186)$

$\displaystyle =1(118)-3(-234)+5(-164)$

$\displaystyle =118+702-820$

$\displaystyle =0$

$\displaystyle \text{(ii) }$

$\displaystyle \Delta=\begin{vmatrix} 67 & 19 & 21\\ 39 & 13 & 14\\ 81 & 24 & 26 \end{vmatrix}$

$\displaystyle =67(338-336)-19(1014-1134)+21(936-1053)$

$\displaystyle =67(2)-19(-120)+21(-117)$

$\displaystyle =134+2280-2457$

$\displaystyle =-43$

$\displaystyle \text{(iii) }$

$\displaystyle \Delta=\begin{vmatrix} a & h & g\\ h & b & f\\ g & f & c \end{vmatrix}$

$\displaystyle = a\begin{vmatrix} b & f\\ f & c \end{vmatrix} - h\begin{vmatrix} h & f\\ g & c \end{vmatrix} + g\begin{vmatrix} h & b\\ g & f \end{vmatrix}$

$\displaystyle = a(bc-f^2)-h(hc-fg)+g(hf-gb)$

$\displaystyle = abc-af^2-h^2c+fgh+fgh-g^2b$

$\displaystyle = abc+2fgh-af^2-ch^2-bg^2$

$\displaystyle \text{(iv) }$

$\displaystyle \Delta=\begin{vmatrix} 1 & -3 & 2\\ 4 & -1 & 2\\ 3 & 5 & 2 \end{vmatrix}$

$\displaystyle = 1\begin{vmatrix} -1 & 2\\ 5 & 2 \end{vmatrix} -(-3)\begin{vmatrix} 4 & 2\\ 3 & 2 \end{vmatrix} +2\begin{vmatrix} 4 & -1\\ 3 & 5 \end{vmatrix}$

$\displaystyle = 1((-1)(2)-(2)(5)) + 3((4)(2)-(2)(3)) + 2((4)(5)-(-1)(3))$

$\displaystyle = 1(-2-10)+3(8-6)+2(20+3)$

$\displaystyle = -12+6+46$

$\displaystyle = 40$

$\displaystyle \text{(v) }$

$\displaystyle \Delta=\begin{vmatrix} 1 & 4 & 9\\ 4 & 9 & 16\\ 9 & 16 & 25 \end{vmatrix}$

$\displaystyle = 1\begin{vmatrix} 9 & 16\\ 16 & 25 \end{vmatrix} -4\begin{vmatrix} 4 & 16\\ 9 & 25 \end{vmatrix} +9\begin{vmatrix} 4 & 9\\ 9 & 16 \end{vmatrix}$

$\displaystyle = 1(9\cdot25-16\cdot16) -4(4\cdot25-16\cdot9) +9(4\cdot16-9\cdot9)$

$\displaystyle = 1(225-256)-4(100-144)+9(64-81)$

$\displaystyle = 1(-31)-4(-44)+9(-17)$

$\displaystyle = -31+176-153$

$\displaystyle = -8$

$\displaystyle \text{(vi) }$

$\displaystyle \Delta=\begin{vmatrix} 6 & -3 & 2\\ 2 & -1 & 2\\ -10 & 5 & 2 \end{vmatrix}$

$\displaystyle = 6\begin{vmatrix} -1 & 2\\ 5 & 2 \end{vmatrix} -(-3)\begin{vmatrix} 2 & 2\\ -10 & 2 \end{vmatrix} +2\begin{vmatrix} 2 & -1\\ -10 & 5 \end{vmatrix}$

$\displaystyle = 6(-1\cdot2-2\cdot5)-(-3)(2\cdot2-2\cdot(-10))+2(2\cdot5-(-1)\cdot(-10))$

$\displaystyle = 6(-2-10)-(-3)(4+20)+2(10-10)$

$\displaystyle = -72+72+0$

$\displaystyle = 0$

$\displaystyle \text{(vii) }$

$\displaystyle D=\begin{vmatrix} 1 & 3 & 9 & 27\\ 3 & 9 & 27 & 1\\ 9 & 27 & 1 & 3\\ 27 & 1 & 3 & 9 \end{vmatrix}$

$\displaystyle \text{Let's label the columns:}$

$\displaystyle C_1=\begin{bmatrix}1\\3\\9\\27\end{bmatrix},\; C_2=\begin{bmatrix}3\\9\\27\\1\end{bmatrix},\; C_3=\begin{bmatrix}9\\27\\1\\3\end{bmatrix},\; C_4=\begin{bmatrix}27\\1\\3\\9\end{bmatrix}$

$\displaystyle C_2'=C_2-3C_1,\; C_3'=C_3-9C_1,\; C_4'=C_4-27C_1$

$\displaystyle \text{Compute each:}$

$\displaystyle C_2-3C_1=[3,9,27,1]-3[1,3,9,27]=[0,0,0,1-81=-80]$

$\displaystyle C_3-9C_1=[9,27,1,3]-9[1,3,9,27]=[0,0,-80,3-243=-240]$

$\displaystyle C_4-27C_1=[27,1,3,9]-27[1,3,9,27]=[0,-80,-240,9-729=-720]$

$\displaystyle D=\begin{vmatrix} 1 & 0 & 0 & 0\\ 3 & 0 & 0 & -80\\ 9 & 0 & -80 & -240\\ 27 & -80 & -240 & -720 \end{vmatrix}$

$\displaystyle =1\cdot \begin{vmatrix} 0 & 0 & -80\\ 0 & -80 & -240\\ -80 & -240 & -720 \end{vmatrix}$

$\displaystyle \text{Evaluate the }3\times 3\text{ determinant:}$

$\displaystyle \begin{vmatrix} 0 & 0 & -80\\ 0 & -80 & -240\\ -80 & -240 & -720 \end{vmatrix}$

$\displaystyle \text{Only the third element of the first row is nonzero:}$

$\displaystyle =(-1)^{1+3}(-80) \begin{vmatrix} 0 & -80\\ -80 & -240 \end{vmatrix}$

$\displaystyle =-80\bigl(0-(-80)(-80)\bigr)$

$\displaystyle =-80(0-6400)$

$\displaystyle =-80(-6400)=512000$

$\displaystyle D=1\cdot512000=512000$

$\displaystyle \text{(viii) }$

$\displaystyle \left| \begin{array}{ccc} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{array} \right| = 102(3\cdot6-4\cdot3)-18(1\cdot6-4\cdot17)+36(1\cdot3-3\cdot17) \\[6pt] = 102(18-12)-18(6-68)+36(3-51) \\[6pt] = 102\cdot6-18(-62)+36(-48) \\[6pt] = 612+1116-1728 \\[6pt] =0$

$\\$

$\displaystyle \textbf{Question 2: } \text{Without expanding, show that the value of the following determinant is zero:}$

$\displaystyle \textbf{(i) } \begin{vmatrix} 8 & 2 & 7\\ 12 & 3 & 5\\ 16 & 4 & 3 \end{vmatrix}$ $\displaystyle \textbf{(ii) } \begin{vmatrix} 6 & -3 & 2\\ 2 & -1 & 2\\ -10 & 5 & 2 \end{vmatrix}$ $\displaystyle \textbf{(iii) } \begin{vmatrix} 2 & 3 & 7\\ 13 & 17 & 5\\ 15 & 20 & 12 \end{vmatrix}$ $\displaystyle \textbf{(iv) } \begin{vmatrix} \frac{1}{a} & a^{2} & bc\\ \frac{1}{b} & b^{2} & ac\\ \frac{1}{c} & c^{2} & ab \end{vmatrix}$

$\displaystyle \textbf{(v) } \begin{vmatrix} a+b & 2a+b & 3a+b\\ 2a+b & 3a+b & 4a+b\\ 4a+b & 5a+b & 6a+b \end{vmatrix}$ $\displaystyle \textbf{(vi) } \begin{vmatrix} 1 & a & a^{2}-bc\\ 1 & b & b^{2}-ac\\ 1 & c & c^{2}-ab \end{vmatrix}$ $\displaystyle \textbf{(vii) } \begin{vmatrix} 49 & 1 & 6\\ 39 & 7 & 4\\ 26 & 2 & 3 \end{vmatrix}$

$\displaystyle \textbf{(viii) } \begin{vmatrix} 0 & x & y\\ -x & 0 & z\\ -y & -z & 0 \end{vmatrix}$ $\displaystyle \textbf{(ix) } \begin{vmatrix} 1 & 43 & 6\\ 7 & 35 & 4\\ 3 & 17 & 2 \end{vmatrix}$

$\displaystyle \textbf{(x) } \begin{vmatrix} 1^{2} & 2^{2} & 3^{2} & 4^{2}\\ 2^{2} & 3^{2} & 4^{2} & 5^{2}\\ 3^{2} & 4^{2} & 5^{2} & 6^{2}\\ 4^{2} & 5^{2} & 6^{2} & 7^{2} \end{vmatrix}$ $\displaystyle \textbf{(xi) } \begin{vmatrix} a & b & c\\ a+2x & b+2y & c+2z\\ x & y & z \end{vmatrix}$

$\displaystyle \textbf{(xii) } \begin{vmatrix} (2^{x}+2^{-x})^{2} & (2^{x}-2^{-x})^{2} & 1\\ (3^{x}+3^{-x})^{2} & (3^{x}-3^{-x})^{2} & 1\\ (4^{x}+4^{-x})^{2} & (4^{x}-4^{-x})^{2} & 1 \end{vmatrix}$ $\displaystyle \textbf{(xiii) } \begin{vmatrix} \sin\alpha & \cos\alpha & \cos(\alpha+\delta)\\ \sin\beta & \cos\beta & \cos(\beta+\delta)\\ \sin\gamma & \cos\gamma & \cos(\gamma+\delta) \end{vmatrix}$

$\displaystyle \textbf{(xiv) } \begin{vmatrix} \sin^{2}23^\circ & \sin^{2}67^\circ & \cos180^\circ\\ -\sin^{2}67^\circ & -\sin^{2}23^\circ & \cos^{2}180^\circ\\ \cos180^\circ & \sin^{2}23^\circ & \sin^{2}67^\circ \end{vmatrix}$ $\displaystyle \textbf{(xv) } \begin{vmatrix} \cos(x+y) & -\sin(x+y) & \cos2y\\ \sin x & \cos x & \sin y\\ -\cos x & \sin x & -\cos y \end{vmatrix}$

$\displaystyle \textbf{(xvi) } \begin{vmatrix} \sqrt{23}+\sqrt{3} & \sqrt{5} & \sqrt{5}\\ \sqrt{15}+\sqrt{46} & 5 & \sqrt{10}\\ 3+\sqrt{115} & \sqrt{15} & 5 \end{vmatrix}$ $\displaystyle \textbf{(xvii) } \begin{vmatrix} \sin^{2}A & \cot A & 1\\ \sin^{2}B & \cot B & 1\\ \sin^{2}C & \cot C & 1 \end{vmatrix}, \text{where }A,B,C\text{ are the angles of }\triangle ABC.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle \Delta= \begin{vmatrix} 8 & 2 & 7\\ 12 & 3 & 5\\ 16 & 4 & 3 \end{vmatrix}$

$\displaystyle = \begin{vmatrix} 0 & 2 & 7\\ 0 & 3 & 5\\ 0 & 4 & 3 \end{vmatrix} \quad [\text{Applying } C_1 \rightarrow C_1 - 4C_2]$

$\displaystyle \Rightarrow \Delta = 0$

$\displaystyle \text{(ii) }$

$\displaystyle \Delta = \begin{vmatrix} 6 & -3 & 2\\ 2 & -1 & 2\\ -10 & 5 & 2 \end{vmatrix}$

$\displaystyle = \begin{vmatrix} 0 & -3 & 2\\ 0 & -1 & 2\\ 0 & 5 & 2 \end{vmatrix} \quad [\text{Applying } C_1 \rightarrow C_1 + 2C_2]$

$\displaystyle \Rightarrow \Delta = 0$

$\displaystyle \text{(iii) }$

$\displaystyle \Delta = \begin{vmatrix} 2 & 3 & 7\\ 13 & 17 & 5\\ 15 & 20 & 12 \end{vmatrix}$

$\displaystyle = \begin{vmatrix} 2 & 3 & 7\\ 13 & 17 & 5\\ 13 & 17 & 5 \end{vmatrix} \quad [\text{Applying } R_3 \rightarrow R_3 - R_1]$

$\displaystyle \Rightarrow \Delta = 0$

$\displaystyle \text{(iv) }$

$\displaystyle \Delta = \begin{vmatrix} \frac{1}{a} & a^{2} & bc\\ \frac{1}{b} & b^{2} & ac\\ \frac{1}{c} & c^{2} & ab \end{vmatrix}$

$\displaystyle = \begin{vmatrix} 1 & a^{3} & abc\\ 1 & b^{3} & abc\\ 1 & c^{3} & abc \end{vmatrix} \quad [\text{Applying } R_1 \rightarrow aR_1,\; R_2 \rightarrow bR_2,\; R_3 \rightarrow cR_3]$

$\displaystyle = abc \begin{vmatrix} 1 & a^{3} & 1\\ 1 & b^{3} & 1\\ 1 & c^{3} & 1 \end{vmatrix}$

$\displaystyle \Rightarrow \Delta = 0$

$\displaystyle \text{(v) }$

$\displaystyle \Delta = \begin{vmatrix} a+b & 2a+b & 3a+b\\ 2a+b & 3a+b & 4a+b\\ 4a+b & 5a+b & 6a+b \end{vmatrix}$

$\displaystyle = \begin{vmatrix} a & a & a\\ 2a & 2a & 2a\\ 4a+b & 5a+b & 6a+b \end{vmatrix} \quad [\text{Applying } R_1 \rightarrow R_2-R_1 \text{ and } R_2 \rightarrow R_3-R_2]$

$\displaystyle = 2\begin{vmatrix} a & a & a\\ a & a & a\\ 4a+b & 5a+b & 6a+b \end{vmatrix} =0$

$\displaystyle \text{(vi) }$

$\displaystyle \Delta = \begin{vmatrix} 1 & a & a^{2}-bc\\ 1 & b & b^{2}-ac\\ 1 & c & c^{2}-ab \end{vmatrix}$

$\displaystyle = \begin{vmatrix} 0 & a-b & a^{2}-bc-b^{2}+ac\\ 0 & b-c & b^{2}-ac-c^{2}+ab\\ 1 & c & c^{2}-ab \end{vmatrix} \quad [\text{Applying } R_1 \rightarrow R_1-R_2,\; R_2 \rightarrow R_2-R_3]$

$\displaystyle = \begin{vmatrix} 0 & a-b & (a-b)(a+b)+c(a-b)\\ 0 & b-c & (b-c)(b+c)+a(b-c)\\ 1 & c & c^{2}-ab \end{vmatrix}$

$\displaystyle = (a-b)(b-c) \begin{vmatrix} 0 & 1 & (a+b+c)\\ 0 & 1 & (a+b+c)\\ 1 & c & c^{2}-ab \end{vmatrix}$

$\displaystyle \Rightarrow \Delta = 0$

$\displaystyle \text{(vii) }$

$\displaystyle \Delta = \begin{vmatrix} 49 & 1 & 6\\ 39 & 7 & 4\\ 26 & 2 & 3 \end{vmatrix}$

$\displaystyle = \begin{vmatrix} 1 & 1 & 6\\ 7 & 7 & 4\\ 2 & 2 & 3 \end{vmatrix} \quad [\text{Applying } C_1 \rightarrow C_1 - 8C_3]$

$\displaystyle \Rightarrow \Delta = 0$

$\displaystyle \text{(viii) }$

$\displaystyle \Delta = \begin{vmatrix} 0 & x & y\\ -x & 0 & z\\ -y & -z & 0 \end{vmatrix}$

$\displaystyle = \frac{xyz}{xyz} \begin{vmatrix} 0 & x & y\\ -x & 0 & z\\ -y & -z & 0 \end{vmatrix}$

$\displaystyle = \frac{1}{xyz} \begin{vmatrix} 0 & xz & yz\\ -xy & 0 & xz\\ -yx & -zx & 0 \end{vmatrix}$

$\displaystyle = \frac{1}{xyz} \begin{vmatrix} -2xy & 0 & 2yz\\ -xy & 0 & zy\\ -yx & -zx & 0 \end{vmatrix} \quad [\text{Applying } R_1 \rightarrow R_1 + R_2 + R_3]$

$\displaystyle = \frac{1}{xyz} \begin{vmatrix} 0 & 0 & 0\\ -xy & 0 & zy\\ -yx & -zx & 0 \end{vmatrix} \quad [\text{Applying } R_1 \rightarrow R_1 - 2R_2]$

$\displaystyle = 0$

$\displaystyle \text{(ix) }$

$\displaystyle \Delta = \begin{vmatrix} 1 & 43 & 6\\ 7 & 35 & 4\\ 3 & 17 & 2 \end{vmatrix}$

$\displaystyle = \begin{vmatrix} 1 & 1 & 6\\ 7 & 7 & 4\\ 3 & 3 & 2 \end{vmatrix} \quad [\text{Applying } C_2 \rightarrow C_2 - 7C_3]$

$\displaystyle \Rightarrow \Delta = 0$

$\displaystyle \text{(x) }$

$\displaystyle \Delta= \begin{vmatrix} 1^2 & 2^2 & 3^2 & 4^2\\ 2^2 & 3^2 & 4^2 & 5^2\\ 3^2 & 4^2 & 5^2 & 6^2\\ 4^2 & 5^2 & 6^2 & 7^2 \end{vmatrix}$

$\displaystyle = \begin{vmatrix} 1 & 4 & 9 & 16\\ 4 & 9 & 16 & 25\\ 9 & 16 & 25 & 36\\ 16 & 25 & 36 & 49 \end{vmatrix}$

$\displaystyle = \begin{vmatrix} 1 & 4 & 9 & 16\\ 4 & 9 & 16 & 25\\ 5 & 7 & 9 & 11\\ 7 & 9 & 11 & 13 \end{vmatrix} \;[\text{Applying } R_3 \rightarrow R_3 - R_2 \text{ and } R_4 \rightarrow R_4 - R_3]$

$\displaystyle = \begin{vmatrix} 1 & 4 & 9 & 16\\ 4 & 9 & 16 & 25\\ 7 & 9 & 11 & 13\\ 7 & 9 & 11 & 13 \end{vmatrix} =0\;[\text{Applying } R_3 \rightarrow 2 + R_3]$

$\displaystyle \text{(xi) }$

$\displaystyle \Delta= \begin{vmatrix} a & b & c\\ a+2x & b+2y & c+2z\\ x & y & z \end{vmatrix}$

$\displaystyle = \begin{vmatrix} a+2x & b+2y & c+2z\\ a+2x & b+2y & c+2z\\ x & y & z \end{vmatrix} \;[\text{Applying } R_1 \rightarrow R_1 + 2R_3]$

$\displaystyle = \begin{vmatrix} 0 & 0 & 0\\ a+2x & b+2y & c+2z\\ x & y & z \end{vmatrix} = 0 \;[\text{Applying } R_1 \rightarrow R_1 - R_2]$

$\displaystyle \text{(xii) }$

$\displaystyle \begin{vmatrix} (2^x+2^{-x})^2 & (2^x-2^{-x})^2 & 1\\ (3^x+3^{-x})^2 & (3^x-3^{-x})^2 & 1\\ (4^x+4^{-x})^2 & (4^x-4^{-x})^2 & 1 \end{vmatrix}$

$\displaystyle = \begin{vmatrix} (2^{2x}+2^{-2x}+2) & (2^{2x}+2^{-2x}-2) & 1\\ (3^{2x}+3^{-2x}+2) & (3^{2x}+3^{-2x}-2) & 1\\ (4^{2x}+4^{-2x}+2) & (4^{2x}+4^{-2x}-2) & 1 \end{vmatrix}$

$\displaystyle = \begin{vmatrix} 4 & (2^{2x}+2^{-2x}-2) & 1\\ 4 & (3^{2x}+3^{-2x}-2) & 1\\ 4 & (4^{2x}+4^{-2x}-2) & 1 \end{vmatrix} \;[\text{Applying } C_1 \rightarrow C_1 - C_2]$

$\displaystyle = 4 \begin{vmatrix} 1 & (2^{2x}+2^{-2x}-2) & 1\\ 1 & (3^{2x}+3^{-2x}-2) & 1\\ 1 & (4^{2x}+4^{-2x}-2) & 1 \end{vmatrix}$

$\displaystyle = 0$

$\displaystyle \text{(xiii) }$

$\displaystyle \begin{vmatrix} \sin\alpha & \cos\alpha & \cos(\alpha+\delta)\\ \sin\beta & \cos\beta & \cos(\beta+\delta)\\ \sin\gamma & \cos\gamma & \cos(\gamma+\delta) \end{vmatrix}$

$\displaystyle = \begin{vmatrix} \sin\alpha\sin\delta & \cos\alpha\cos\delta & \cos(\alpha+\delta)\\ \sin\beta\sin\delta & \cos\beta\cos\delta & \cos(\beta+\delta)\\ \sin\gamma\sin\delta & \cos\gamma\cos\delta & \cos(\gamma+\delta) \end{vmatrix} \;[\text{Applying } C_1 \rightarrow \sin\delta\,C_1 \text{ and } C_2 \rightarrow \cos\delta\,C_2]$

$\displaystyle = \begin{vmatrix} \sin\alpha\sin\delta & \cos(\alpha+\delta) & \cos(\alpha+\delta)\\ \sin\beta\sin\delta & \cos(\beta+\delta) & \cos(\beta+\delta)\\ \sin\gamma\sin\delta & \cos(\gamma+\delta) & \cos(\gamma+\delta) \end{vmatrix} \;[\text{Applying } C_2 \rightarrow C_2 - C_1]$

$\displaystyle = 0$

$\displaystyle \text{(xiv) }$

$\displaystyle \begin{vmatrix} \sin^2 23^\circ & \sin^2 67^\circ & \cos 180^\circ\\ -\sin^2 67^\circ & -\sin^2 23^\circ & \cos^2 180^\circ\\ \cos 180^\circ & \sin^2 23^\circ & \sin^2 67^\circ \end{vmatrix}$

$\displaystyle = \begin{vmatrix} \sin^2 23^\circ & \sin^2(90^\circ-23^\circ) & -1\\ -\sin^2(90^\circ-23^\circ) & -\sin^2 23^\circ & 1\\ -1 & \sin^2 23^\circ & \sin^2(90^\circ-23^\circ) \end{vmatrix}$

$\displaystyle = \begin{vmatrix} \sin^2 23^\circ & \cos^2 23^\circ & -1\\ -\cos^2 23^\circ & -\sin^2 23^\circ & 1\\ -1 & \sin^2 23^\circ & \cos^2 23^\circ \end{vmatrix}$

$\displaystyle = \begin{vmatrix} \sin^2 23^\circ + \cos^2 23^\circ & \cos^2 23^\circ & -1\\ -\cos^2 23^\circ - \sin^2 23^\circ & -\sin^2 23^\circ & 1\\ -1 + \sin^2 23^\circ & \sin^2 23^\circ & \cos^2 23^\circ \end{vmatrix} \;[\text{Applying } C_1 \rightarrow C_1 + C_2]$

$\displaystyle = \begin{vmatrix} 1 & 1 & -1\\ -1 & -\sin^2 23^\circ & 1\\ -\cos^2 23^\circ & \sin^2 23^\circ & \cos^2 23^\circ \end{vmatrix}$

$\displaystyle = (-1) \begin{vmatrix} -1 & 1 & -1\\ 1 & -\sin^2 23^\circ & 1\\ \cos^2 23^\circ & \sin^2 23^\circ & \cos^2 23^\circ \end{vmatrix}$

$\displaystyle = 0$

$\displaystyle \text{(xv) }$

$\displaystyle \begin{vmatrix} \cos(x+y) & -\sin(x+y) & \cos 2y\\ \sin x & \cos x & \sin y\\ -\cos x & \sin x & -\cos y \end{vmatrix}$

$\displaystyle = \frac{1}{\sin y \cos y} \begin{vmatrix} \cos(x+y) & -\sin(x+y) & \cos 2y\\ \sin x \sin y & \cos x \sin y & \sin^2 y\\ -\cos x \cos y & \sin x \cos y & -\cos^2 y \end{vmatrix} \;[\text{Applying } R_2 \rightarrow \sin y\,R_2 \text{ and } R_3 \rightarrow \cos y\,R_3]$

$\displaystyle = \frac{1}{\sin y \cos y} \begin{vmatrix} \cos(x+y) & -\sin(x+y) & \cos 2y\\ \sin x \sin y - \cos x \cos y & \cos x \sin y + \sin x \cos y & \sin^2 y - \cos^2 y\\ -\cos x \cos y & \sin x \cos y & -\cos^2 y \end{vmatrix} \;[\text{Applying } R_2 \rightarrow R_2 + R_3]$

$\displaystyle = \frac{-1}{\sin y \cos y} \begin{vmatrix} \cos(x+y) & -\sin(x+y) & \cos 2y\\ \cos(x+y) & -\sin(x+y) & \cos 2y\\ -\cos x \cos y & \sin x \cos y & -\cos^2 y \end{vmatrix}$

$\displaystyle = 0$

$\displaystyle \text{(xvi) }$

$\displaystyle \begin{vmatrix} \sqrt{23}+\sqrt{3} & \sqrt{5} & \sqrt{5}\\ \sqrt{15}+\sqrt{46} & 5 & \sqrt{10}\\ 3+\sqrt{115} & \sqrt{15} & 5 \end{vmatrix}$

$\displaystyle = \begin{vmatrix} \sqrt{3} & \sqrt{5} & \sqrt{5}\\ \sqrt{15} & 5 & \sqrt{10}\\ 3 & \sqrt{15} & 5 \end{vmatrix} + \begin{vmatrix} \sqrt{23} & \sqrt{5} & \sqrt{5}\\ \sqrt{46} & 5 & \sqrt{10}\\ \sqrt{115} & \sqrt{15} & 5 \end{vmatrix}$

$\displaystyle = \sqrt{3} \begin{vmatrix} 1 & \sqrt{5} & \sqrt{5}\\ \sqrt{5} & 5 & \sqrt{10}\\ \sqrt{3} & \sqrt{15} & 5 \end{vmatrix} + \sqrt{23} \begin{vmatrix} 1 & \sqrt{5} & \sqrt{5}\\ \sqrt{2} & 5 & \sqrt{10}\\ \sqrt{5} & \sqrt{15} & 5 \end{vmatrix}$

$\displaystyle = \sqrt{3}\times\sqrt{5} \begin{vmatrix} 1 & 1 & \sqrt{5}\\ \sqrt{5} & \sqrt{5} & \sqrt{10}\\ \sqrt{3} & \sqrt{3} & 5 \end{vmatrix} + \sqrt{23}\times\sqrt{5} \begin{vmatrix} 1 & \sqrt{5} & 1\\ \sqrt{2} & 5 & \sqrt{2}\\ \sqrt{5} & \sqrt{15} & \sqrt{5} \end{vmatrix}$

$\displaystyle = 0 + 0$

$\displaystyle = 0$

$\displaystyle \text{(xvii) }$

$\displaystyle \begin{vmatrix} \sin^2 A & \cot A & 1\\ \sin^2 B & \cot B & 1\\ \sin^2 C & \cot C & 1 \end{vmatrix} = \begin{vmatrix} \sin^2 A-\sin^2 B & \cot A-\cot B & 0\\ \sin^2 B & \cot B & 1\\ \sin^2 C-\sin^2 B & \cot C-\cot B & 0 \end{vmatrix} \ \big[\text{Applying } R_1\to R_1-R_2 \text{ and } R_3\to R_3-R_2\big]$

$\displaystyle = \begin{vmatrix} \sin(A+B)\sin(A-B) & \dfrac{\cos A\sin B-\cos B\sin A}{\sin A\,\sin B} & 0\\ \sin^2 B & \cot B & 1\\ \sin(C+B)\sin(C-B) & \dfrac{\cos C\sin B-\cos B\sin C}{\sin B\,\sin C} & 0 \end{vmatrix}$

$\displaystyle = \begin{vmatrix} \sin(\pi-C)\sin(A-B) & -\dfrac{\sin(A-B)}{\sin A\,\sin B} & 0\\ \sin^2 B & \cot B & 1\\ \sin(\pi-A)\sin(C-B) & -\dfrac{\sin(C-B)}{\sin B\,\sin C} & 0 \end{vmatrix} \ \big[\because\ A+B+C=\pi\big]$

$\displaystyle = \begin{vmatrix} \sin C\,\sin(A-B) & -\dfrac{\sin(A-B)}{\sin A\,\sin B} & 0\\ \sin^2 B & \dfrac{\cos B}{\sin B} & 1\\ \sin A\,\sin(C-B) & -\dfrac{\sin(C-B)}{\sin B\,\sin C} & 0 \end{vmatrix}$

$\displaystyle = \frac{\sin(A-B)\sin(C-B)}{\sin B} \begin{vmatrix} \sin C & -\dfrac{1}{\sin A} & 0\\ \sin^2 B & \cos B & 1\\ \sin A & -\dfrac{1}{\sin C} & 0 \end{vmatrix}$

$\displaystyle = \frac{\sin(A-B)\sin(C-B)}{\sin B\,\sin A\,\sin C} \begin{vmatrix} \sin C\sin A & -1 & 0\\ \sin^2 B & \cos B & 1\\ \sin A\sin C & -1 & 0 \end{vmatrix} \ \big[\text{Applying } R_1\to \sin A\,R_1 \text{ and } R_3\to \sin C\,R_3\big]$

$\displaystyle = \frac{\sin(A-B)\sin(C-B)}{\sin B\,\sin A\,\sin C} \begin{vmatrix} 0 & 0 & 0\\ \sin^2 B & \cos B & 1\\ \sin A\sin C & -1 & 0 \end{vmatrix} \ \big[\text{Applying } R_1\to R_1-R_3\big]$

$\displaystyle = 0$

$\\$

$\displaystyle \textbf{Question 3: } \text{Evaluate:}$

$\displaystyle \begin{vmatrix} a & b+c & a^{2}\\ b & c+a & b^{2}\\ c & a+b & c^{2} \end{vmatrix} \hspace{2.0cm} \text{[CBSE 2006]}$

$\displaystyle \text{Answer:}$

$\displaystyle \Delta= \begin{vmatrix} a & b+c & a^2\\ b & c+a & b^2\\ c & a+b & c^2 \end{vmatrix}$

$\displaystyle \text{When } a=b,\ \text{the first two rows become identical. Hence, } a-b \text{ is a factor.}$

$\displaystyle \text{Similarly, when } b=c\ \text{the second and third rows become identical. So, } b-c \text{ is also a factor.}$

$\displaystyle \text{Also, when } c=a,\ \text{the third and first rows become identical. Hence, } c-a \text{ is also a factor.}$

$\displaystyle \text{The product of diagonal elements, } a(c+a)c^2 \text{ is } 4.\ \text{So, the other factor should be a linear in } a,b,c.$

$\displaystyle \text{It should also remain unaltered when any two letters are changed. Let this factor be } \lambda(a+b+c).$

$\displaystyle \text{Here, } \lambda \text{ is a constant. To find this, we have}$

$\displaystyle a=0,\ b=1,\ c=2$

$\displaystyle \begin{vmatrix} 0 & 3 & 2\\ 1 & 2 & 1\\ 2 & 1 & 4 \end{vmatrix} =\lambda(a-b)(b-c)(c-a)(a+b+c)$

$\displaystyle \begin{vmatrix} 0 & 3 & 2\\ 1 & 2 & 1\\ 2 & 1 & 4 \end{vmatrix} =\lambda(0-1)(1-2)(2-0)(0+1+2)$

$\displaystyle \Rightarrow -6=6\lambda$

$\displaystyle \Rightarrow \lambda=-1$

$\displaystyle \text{Thus,} \qquad \begin{vmatrix} a & b+c & a^2\\ b & c+a & b^2\\ c & a+b & c^2 \end{vmatrix} =-\big((a+b+c)\big)(a-b)(b-c)(c-a)$

$\\$

$\displaystyle \textbf{Question 4: } \text{Evaluate:}$

$\displaystyle \begin{vmatrix} 1 & a & bc\\ 1 & b & ca\\ 1 & c & ab \end{vmatrix} \hspace{2.0cm} \text{[CBSE 2006]}$

$\displaystyle \text{Answer:}$

$\displaystyle \Delta= \begin{vmatrix} 1 & a & bc\\ 1 & b & ca\\ 1 & c & ab \end{vmatrix}$

$\displaystyle \text{When } a=b,\ \text{the first two rows become identical. Hence, } a-b \text{ is a factor.}$

$\displaystyle \text{Similarly, when } b=c \text{ and } c=a,\ \text{the second and third and third and first rows become identical.}$

$\displaystyle \text{Hence, } b-c \text{ and } c-a \text{ are also factors.}$

$\displaystyle \text{The degree of the product of the diagonal elements is } 3.\ \text{Hence, there are no other factors.}$

$\displaystyle \begin{vmatrix} 1 & a & bc\\ 1 & b & ca\\ 1 & c & ab \end{vmatrix} =\lambda (a-b)(b-c)(c-a) \qquad [\text{where } \lambda \text{ is a constant}]$

$\displaystyle \begin{vmatrix} 1 & 0 & 2\\ 1 & 1 & 0\\ 1 & 2 & 0 \end{vmatrix} =2\lambda \qquad [\text{Putting } a=0,\ b=1,\ c=2 \text{ to find } \lambda]$

$\displaystyle \Rightarrow 2=2\lambda$

$\displaystyle \Rightarrow \lambda=1$

$\displaystyle \text{Hence,} \qquad \begin{vmatrix} 1 & a & bc\\ 1 & b & ca\\ 1 & c & ab \end{vmatrix} =(a-b)(b-c)(c-a)$

$\\$

$\displaystyle \textbf{Question 5: } \text{Evaluate:}$

$\displaystyle \begin{vmatrix} x+\lambda & x & x\\ x & x+\lambda & x\\ x & x & x+\lambda \end{vmatrix}$

$\displaystyle \text{Answer:}$

$\displaystyle \Delta= \begin{vmatrix} x+\lambda & x & x\\ x & x+\lambda & x\\ x & x & x+\lambda \end{vmatrix}$

$\displaystyle = \begin{vmatrix} \lambda & 0 & x\\ -\lambda & \lambda & x\\ 0 & -\lambda & x+\lambda \end{vmatrix} \quad \big[\text{Applying } C_1\to C_1-C_2,\ C_2\to C_2-C_3\big]$

$\displaystyle = \begin{vmatrix} \lambda & 0 & x\\ -\lambda & 0 & 2x+\lambda\\ 0 & -\lambda & x+\lambda \end{vmatrix} \quad \big[\text{Applying } R_1 \text{ to } R_2+R_3\big]$

$\displaystyle = \lambda \begin{vmatrix} 0 & 2x+\lambda\\ -\lambda & x+\lambda \end{vmatrix} + x \begin{vmatrix} -\lambda & 0\\ 0 & -\lambda \end{vmatrix}$

$\displaystyle = \lambda\big[\lambda(2x+\lambda)\big]+x\lambda^2$

$\displaystyle = \lambda^2(2x+\lambda+\lambda x)$

$\displaystyle = 3\lambda^2x+\lambda^3$

$\displaystyle = \lambda^2(3x+\lambda)$

$\\$

$\displaystyle \textbf{Question 6: } \text{Evaluate:}$

$\displaystyle \begin{vmatrix} a & b & c\\ c & a & b\\ b & c & a \end{vmatrix} \hspace{2.0cm} \text{[CBSE 2004]}$

$\displaystyle \text{Answer:}$

$\displaystyle \Delta= \begin{vmatrix} a & b & c\\ c & a & b\\ b & c & a \end{vmatrix}$

$\displaystyle = a(a^2-bc)-b(ca-b^2)+c(c^2-ba)$

$\displaystyle = a^3-abc-bca+b^3+c^3-abc$

$\displaystyle = a^3+b^3+c^3-3abc$

$\displaystyle = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

$\\$

$\displaystyle \textbf{Question 7: } \text{Evaluate:}$

$\displaystyle \begin{vmatrix} x & 1 & 1\\ 1 & x & 1\\ 1 & 1 & x \end{vmatrix}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let}$

$\displaystyle \Delta= \begin{vmatrix} x & 1 & 1\\ 1 & x & 1\\ 1 & 1 & x \end{vmatrix}$

$\displaystyle = \begin{vmatrix} x-1 & 1-x & 0\\ 1 & x & 1\\ 0 & 1-x & x-1 \end{vmatrix} \quad \big[\text{Applying } R_1\to R_1-R_2 \text{ and } R_3\to R_3-R_2\big]$

$\displaystyle = (x-1)^2 \begin{vmatrix} 1 & -1 & 0\\ 1 & x & 1\\ 0 & -1 & 1 \end{vmatrix}$

$\displaystyle = (x-1)^2 \begin{vmatrix} 1 & -1 & 0\\ 1 & x+1 & 1\\ 0 & 0 & 1 \end{vmatrix} \quad \big[\text{Applying } C_2\to C_2+C_3\big]$

$\displaystyle = (x-1)^2(x+1+1) \quad \big[\text{Expanding along last row}\big]$

$\displaystyle = (x-1)^2(x+2)$

$\displaystyle \therefore\ \Delta=(x-1)^2(x+2)$

$\\$

$\displaystyle \textbf{Question 8: } \text{Evaluate:}$

$\displaystyle \begin{vmatrix} 0 & xy^{2} & xz^{2}\\ x^{2}y & 0 & yz^{2}\\ x^{2}z & zy^{2} & 0 \end{vmatrix}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let}$

$\displaystyle \Delta= \begin{vmatrix} 0 & xy^2 & xz^2\\ x^2y & 0 & yz^2\\ x^2z & zy^2 & 0 \end{vmatrix}$

$\displaystyle = x^2y^2z^2 \begin{vmatrix} 0 & x & x\\ y & 0 & y\\ z & z & 0 \end{vmatrix} \quad \big[\text{Taking } x^2 \text{ common from } C_1,\ y^2 \text{ common from } C_2 \text{ and } z^2 \text{ common from } C_3\big]$

$\displaystyle = x^3y^3z^3 \begin{vmatrix} 0 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 0 \end{vmatrix} \quad \big[\text{Taking } x \text{ common from } R_1,\ y \text{ common from } R_2 \text{ and } z \text{ common from } R_3\big]$

$\displaystyle = x^3y^3z^3 \begin{vmatrix} 0 & 0 & 1\\ 1 & -1 & 1\\ 1 & 1 & 0 \end{vmatrix} \quad \big[\text{Applying } C_2\to C_2-C_3\big]$

$\displaystyle = x^3y^3z^3(1+1) \quad \big[\text{Expanding along first row}\big]$

$\displaystyle = 2x^3y^3z^3$

$\displaystyle \therefore\ \Delta=2x^3y^3z^3$

$\\$

$\displaystyle \textbf{Question 9: } \text{Evaluate:}$

$\displaystyle \begin{vmatrix} a+x & y & z\\ x & a+y & z\\ x & y & a+z \end{vmatrix}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let}$

$\displaystyle \Delta= \begin{vmatrix} a+x & y & z\\ x & a+y & z\\ x & y & a+z \end{vmatrix}$

$\displaystyle = \begin{vmatrix} a+x+y+z & y & z\\ a+x+y+z & a+y & z\\ a+x+y+z & y & a+z \end{vmatrix} \quad \big[\text{Applying } C_1\to C_1+C_2+C_3\big]$

$\displaystyle = (a+x+y+z) \begin{vmatrix} 1 & y & z\\ 1 & a+y & z\\ 1 & y & a+z \end{vmatrix} \quad \big[\text{Taking } (a+x+y+z) \text{ common from } C_1\big]$

$\displaystyle = (a+x+y+z) \begin{vmatrix} 1 & y & z\\ 0 & a & 0\\ 0 & 0 & a \end{vmatrix} \quad \big[\text{Applying } R_2\to R_2-R_1 \text{ and } R_3\to R_3-R_1\big]$

$\displaystyle = (a+x+y+z)a^2 \quad \big[\text{Expanding along first column}\big]$

$\displaystyle \therefore\ \Delta=(a+x+y+z)a^2$

$\\$

$\displaystyle \textbf{Question 10: }$

$\displaystyle \Delta=\begin{vmatrix} 1 & x & x^{2}\\ 1 & y & y^{2}\\ 1 & z & z^{2} \end{vmatrix}, \quad \Delta_{1}=\begin{vmatrix} 1 & 1 & 1\\ yz & zx & xy\\ x & y & z \end{vmatrix}$ ,

then prove that

$\displaystyle \Delta+\Delta_{1}=0$ .

$\displaystyle \text{Answer:}$

$\displaystyle \Delta+\Delta_1= \begin{vmatrix} 1 & x & x^2\\ 1 & y & y^2\\ 1 & z & z^2 \end{vmatrix} + \begin{vmatrix} 1 & 1 & 1\\ yz & zx & xy\\ x & y & z \end{vmatrix}$

$\displaystyle = \begin{vmatrix} 1 & x & x^2\\ 1 & y & y^2\\ 1 & z & z^2 \end{vmatrix} + \begin{vmatrix} 1 & yz & x\\ 1 & zx & y\\ 1 & xy & z \end{vmatrix} \quad \big[\text{Interchanging rows and columns in } \Delta_1\big]$

$\displaystyle = \begin{vmatrix} 1 & x & x^2\\ 1 & y & y^2\\ 1 & z & z^2 \end{vmatrix} - \begin{vmatrix} 1 & x & yz\\ 1 & y & zx\\ 1 & z & xy \end{vmatrix} \quad \big[\text{Applying } C_2 \leftrightarrow C_3 \text{ in } \Delta_1\big]$

$\displaystyle = \begin{vmatrix} 1 & x & x^2\\ 0 & y-x & y^2-x^2\\ 0 & z-x & z^2-x^2 \end{vmatrix} - \begin{vmatrix} 1 & x & yz\\ 0 & y-x & zx-yz\\ 0 & z-x & xy-yz \end{vmatrix} \quad \big[\text{Applying } R_2\to R_2-R_1 \text{ and } R_3\to R_3-R_1\big]$

$\displaystyle = (y-x)(z-x) \begin{vmatrix} 1 & x & x^2\\ 0 & 1 & y+x\\ 0 & 1 & z+x \end{vmatrix} - (y-x)(z-x) \begin{vmatrix} 1 & x & yz\\ 0 & 1 & -z\\ 0 & 1 & -y \end{vmatrix} \quad \big[\text{Taking } (y-x) \text{ common from } R_2 \text{ and } (z-x) \text{ common from } R_3\big]$

$\displaystyle = (y-x)(z-x)\big[(z+x-y-x)\big] - (y-x)(z-x)\big[(-y+z)\big] \quad \big[\text{Expanding along first column}\big]$

$\displaystyle = (y-x)(z-x)(z-y)(1-1)$

$\displaystyle = 0$

$\displaystyle \therefore\ \Delta+\Delta_1=0.$

$\\$

$\displaystyle \textbf{Question 11: } \text{Prove the following identity:}$

$\displaystyle \begin{vmatrix} a & b & c\\ a-b & b-c & c-a\\ b+c & c+a & a+b \end{vmatrix} = a^3+b^3+c^3-3abc \hspace{2.0cm} \text{[CBSE 2009]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{L.H.S.}= \begin{vmatrix} a & b & c\\ a-b & b-c & c-a\\ b+c & c+a & a+b \end{vmatrix}$

$\displaystyle \text{Apply } C_1\to C_1+C_2+C_3$

$\displaystyle = \begin{vmatrix} a+b+c & b & c\\ 0 & b-c & c-a\\ 2(a+b+c) & c+a & a+b \end{vmatrix}$

$\displaystyle \text{Taking } (a+b+c) \text{ common from } C_1 \text{ we get,}$

$\displaystyle =(a+b+c) \begin{vmatrix} 1 & b & c\\ 0 & b-c & c-a\\ 2 & c+a & a+b \end{vmatrix}$

$\displaystyle \text{Applying, } R_3\to R_3-2R_1$

$\displaystyle =(a+b+c) \begin{vmatrix} 1 & b & c\\ 0 & b-c & c-a\\ 0 & c+a-2b & a+b-2c \end{vmatrix}$

$\displaystyle =(a+b+c)\big[(b-c)(a+b-2c)-(c-a)(c+a-2b)\big]$

$\displaystyle = a^3+b^3+c^3-3abc$

$\displaystyle \text{As, L.H.S.=R.H.S.}$

$\displaystyle \text{Hence, proved.}$

$\\$

$\displaystyle \textbf{Question 12: } \text{Prove the following identity:}$

$\displaystyle \begin{vmatrix} b+c & a-b & a\\ c+a & b-c & b\\ a+b & c-a & c \end{vmatrix} =3abc-a^3-b^3-c^3$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let LHS }=\Delta= \begin{vmatrix} b+c & a-b & a\\ c+a & b-c & b\\ a+b & c-a & c \end{vmatrix}$

$\displaystyle \Delta =(b+c) \begin{vmatrix} b-c & b\\ c-a & c \end{vmatrix} -(a-b) \begin{vmatrix} c+a & b\\ a+b & c \end{vmatrix} +a \begin{vmatrix} c+a & b-c\\ a+b & c-a \end{vmatrix} \quad [\text{Expanding}]$

$\displaystyle =(b+c)\{bc-c^2-bc+ab\} -(a-b)\{c^2+ac-ab-b^2\} +a\{c^2-a^2-ab+ac-b^2+bc\}$

$\displaystyle =bc^2-c^3+abc-ac^2-a^2c+a^2b+ab^2+bc^2+abc-ab^2-b^3+ac^2-a^3-a^2c-ab^2+abc$

$\displaystyle \Rightarrow \Delta=3abc-a^3-b^3-c^3 \quad [\text{Simplifying}]$

$\displaystyle = \text{RHS}$

$\\$

$\displaystyle \textbf{Question 13: } \text{Prove the following identity:}$

$\displaystyle \begin{vmatrix} a+b & b+c & c+a\\ b+c & c+a & a+b\\ c+a & a+b & b+c \end{vmatrix} = 2\begin{vmatrix} a & b & c\\ b & c & a\\ c & a & b \end{vmatrix} \hspace{2.0cm} \text{[CBSE 2004, 2006, 2007]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let }\Delta= \begin{vmatrix} a+b & b+c & c+a\\ b+c & c+a & a+b\\ c+a & a+b & b+c \end{vmatrix}$

$\displaystyle \text{Using the property of determinants that if each element of a row or column is expressed as the sum of two or more quantities,}$

$\displaystyle \text{the determinant is expressed as the sum of two or more determinants, we get}$

$\displaystyle \Delta= \begin{vmatrix} a & b & c\\ b & c & a\\ c & a & b \end{vmatrix} + \begin{vmatrix} b & c & a\\ c & a & b\\ a & b & c \end{vmatrix}$

$\displaystyle = \begin{vmatrix} a & b & c\\ b & c & a\\ c & a & b \end{vmatrix} +(-1) \begin{vmatrix} a & c & b\\ b & a & c\\ c & b & a \end{vmatrix} \quad \big[\text{Applying } C_1\leftrightarrow C_3 \text{ in second determinant to get negative value of the determinant}\big]$

$\displaystyle = \begin{vmatrix} a & b & c\\ b & c & a\\ c & a & b \end{vmatrix} +(-1)(-1) \begin{vmatrix} a & b & c\\ b & c & a\\ c & a & b \end{vmatrix} \quad \big[\text{Applying } C_2\leftrightarrow C_3\big]$

$\displaystyle =2 \begin{vmatrix} a & b & c\\ b & c & a\\ c & a & b \end{vmatrix} =\text{RHS}$

$\\$

$\displaystyle \textbf{Question 14: } \text{Prove the following identity:}$

$\displaystyle \begin{vmatrix} a+b+2c & a & b\\ c & b+c+2a & b\\ c & a & c+a+2b \end{vmatrix} =2(a+b+c)^3 \hspace{2.0cm} \text{[CBSE 2006, 2008, 2014]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let LHS }=\Delta= \begin{vmatrix} a+b+2c & a & b\\ c & b+c+2a & b\\ c & a & c+a+2b \end{vmatrix}$

$\displaystyle \Rightarrow \Delta= \begin{vmatrix} 2a+2b+2c & a & b\\ 2a+2b+2c & b+c+2a & b\\ 2a+2b+2c & a & c+a+2b \end{vmatrix} \quad \big[\text{Applying } C_1\to C_1+C_2+C_3\big]$

$\displaystyle =2(a+b+c) \begin{vmatrix} 1 & a & b\\ 1 & b+c+2a & b\\ 1 & a & c+a+2b \end{vmatrix} \quad \big[\text{Taking out } 2(a+b+c) \text{ common from } C_1\big]$

$\displaystyle =2(a+b+c) \begin{vmatrix} 1 & a & b\\ 0 & b+c+a & 0\\ 0 & -b-c-a & a+b+c \end{vmatrix} \quad \big[\text{Applying } R_2\to R_2-R_1 \text{ and } R_3\to R_3-R_1\big]$

$\displaystyle =2(a+b+c) \begin{vmatrix} 1 & a & b\\ 0 & a+b+c & 0\\ 0 & -(a+b+c) & a+b+c \end{vmatrix}$

$\displaystyle =2(a+b+c)(a+b+c)(a+b+c) \begin{vmatrix} 1 & a & b\\ 0 & 1 & 0\\ 0 & -1 & 1 \end{vmatrix} \quad \big[\text{Taking out } (a+b+c) \text{ common from } R_2 \text{ and } R_3\big]$

$\displaystyle =2(a+b+c)^3\{1(1-0)\} \quad \big[\text{Expanding along } C_1\big]$

$\displaystyle =2(a+b+c)^3$

$\displaystyle =\text{RHS}$

$\\$

$\displaystyle \textbf{Question 15: } \text{Prove the following identity:}$

$\displaystyle \begin{vmatrix} a-b-c & 2a & 2a\\ 2b & b-c-a & 2b\\ 2c & 2c & c-a-b \end{vmatrix} =(a+b+c)^3$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let LHS }=\Delta= \begin{vmatrix} a-b-c & 2a & 2a\\ 2b & b-c-a & 2b\\ 2c & 2c & c-a-b \end{vmatrix}$

$\displaystyle \Rightarrow \Delta= \begin{vmatrix} a+b+c & a+b+c & a+b+c\\ 2b & b-c-a & 2b\\ 2c & 2c & c-a-b \end{vmatrix} \quad \big[\text{Applying } R_1\to R_1+R_2+R_3\big]$

$\displaystyle =(a+b+c) \begin{vmatrix} 1 & 1 & 1\\ 2b & b-c-a & 2b\\ 2c & 2c & c-a-b \end{vmatrix}$

$\displaystyle =(a+b+c) \begin{vmatrix} 0 & 1 & 1\\ b+c+a & b-c-a & 2b\\ 0 & 2c & c-a-b \end{vmatrix} \quad \big[\text{Applying } C_1\to C_1-C_2\big]$

$\displaystyle =(a+b+c) \left\{(a+b+c) \begin{vmatrix} 1 & 1\\ 2c & c-a-b \end{vmatrix} \right\} \quad \big[\text{Expanding along } C_1\big]$

$\displaystyle =(a+b+c)^3$

$\displaystyle =\text{RHS}$

$\\$

$\displaystyle \textbf{Question 16: } \text{Prove the following identity:}$

$\displaystyle \begin{vmatrix} 1 & b+c & b^2+c^2\\ 1 & c+a & c^2+a^2\\ 1 & a+b & a^2+b^2 \end{vmatrix} =(a-b)(b-c)(c-a) \hspace{2.0cm} \text{[CBSE 2002]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let LHS }=\Delta= \begin{vmatrix} 1 & b+c & b^2+c^2\\ 1 & c+a & c^2+a^2\\ 1 & a+b & a^2+b^2 \end{vmatrix}$

$\displaystyle \Rightarrow \Delta= \begin{vmatrix} 0 & (b+c)-(c+a) & (b^2+c^2)-(c^2+a^2)\\ 0 & (c+a)-(a+b) & (c^2+a^2)-(a^2+b^2)\\ 1 & a+b & a^2+b^2 \end{vmatrix} \quad \big[\text{Applying } R_1\to R_1-R_2 \text{ and } R_2\to R_2-R_3\big]$

$\displaystyle = \begin{vmatrix} 0 & b-a & b^2-a^2\\ 0 & c-b & c^2-b^2\\ 1 & a+b & a^2+b^2 \end{vmatrix}$

$\displaystyle =(-1)^2 \begin{vmatrix} 0 & a-b & a^2-b^2\\ 0 & b-c & b^2-c^2\\ 1 & a+b & a^2+b^2 \end{vmatrix} \quad \big[\text{Taking out } (-1) \text{ common from } R_1 \text{ and } R_2\big]$

$\displaystyle =(a-b)(b-c) \begin{vmatrix} 0 & 1 & a+b\\ 0 & 1 & b+c\\ 1 & a+b & a^2+b^2 \end{vmatrix}$

$\displaystyle =(a-b)(b-c) \left\{ 1\times \begin{vmatrix} 1 & a+b\\ 1 & b+c \end{vmatrix} \right\} \quad \big[\text{Expanding along } C_1\big]$

$\displaystyle =(a-b)(b-c)(c-a)$

$\displaystyle =\text{RHS}$

$\\$

$\displaystyle \textbf{Question 17: } \text{Prove the following identity:}$

$\displaystyle \begin{vmatrix} a & a+b & a+2b\\ a+2b & a & a+b\\ a+b & a+2b & a \end{vmatrix} =9(a+b)b^2 \hspace{2.0cm} \text{[CBSE 2002, 2013, 2017]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let LHS }=\Delta= \begin{vmatrix} a & a+b & a+2b\\ a+2b & a & a+b\\ a+b & a+2b & a \end{vmatrix}$

$\displaystyle \Delta= \begin{vmatrix} 3a+3b & 3a+3b & 3a+3b\\ a+2b & a & a+b\\ a+b & a+2b & a \end{vmatrix} \quad \big[\text{Applying } R_1\to R_1+R_2+R_3\big]$

$\displaystyle =3(a+b) \begin{vmatrix} 1 & 1 & 1\\ a+2b & a & a+b\\ a+b & a+2b & a \end{vmatrix} \quad \big[\text{Taking out } 3(a+b) \text{ common from } R_1\big]$

$\displaystyle =3(a+b) \begin{vmatrix} 0 & 0 & 1\\ 2b & -b & a+b\\ -b & 2b & a \end{vmatrix} \quad \big[\text{Applying } C_1\to C_1-C_2 \text{ and } C_2\to C_2-C_3\big]$

$\displaystyle =3(a+b)b^2 \begin{vmatrix} 0 & 0 & 1\\ 2 & -1 & a+b\\ -1 & 2 & a \end{vmatrix} \quad \big[\text{Taking out } b \text{ common from } C_1 \text{ and } C_2\big]$

$\displaystyle =3(a+b)b^2\times 3$

$\displaystyle =9(a+b)b^2$

$\displaystyle =\text{RHS}$

$\\$

$\displaystyle \textbf{Question 18: } \text{Prove the following identity:}$

$\displaystyle \begin{vmatrix} 1 & a & bc\\ 1 & b & ca\\ 1 & c & ab \end{vmatrix} = \begin{vmatrix} 1 & a & a^2\\ 1 & b & b^2\\ 1 & c & c^2 \end{vmatrix}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let LHS }=\Delta= \begin{vmatrix} 1 & a & bc\\ 1 & b & ca\\ 1 & c & ab \end{vmatrix}$

$\displaystyle =\frac{1}{abc} \begin{vmatrix} a & a^2 & abc\\ b & b^2 & bca\\ c & c^2 & abc \end{vmatrix} \quad \big[\text{Applying } R_1\to aR_1,\ R_2\to bR_2,\ R_3\to cR_3 \text{ and then dividing it by } abc\big]$

$\displaystyle =\frac{abc}{abc} \begin{vmatrix} a & a^2 & 1\\ b & b^2 & 1\\ c & c^2 & 1 \end{vmatrix} \quad \big[\text{Taking out } abc \text{ common from } C_3\big]$

$\displaystyle =(-1) \begin{vmatrix} 1 & a^2 & a\\ 1 & b^2 & b\\ 1 & c^2 & c \end{vmatrix} \quad \big[\text{Interchanging } C_3 \text{ and } C_1 \text{ to get -ve value of original determinant}\big]$

$\displaystyle =(-1)(-1) \begin{vmatrix} 1 & a & a^2\\ 1 & b & b^2\\ 1 & c & c^2 \end{vmatrix} \quad \big[\text{Applying } C_2\leftrightarrow C_3\big]$

$\displaystyle = \begin{vmatrix} 1 & a & a^2\\ 1 & b & b^2\\ 1 & c & c^2 \end{vmatrix}$

$\displaystyle =\text{RHS}$

$\\$

$\displaystyle \textbf{Question 19: } \text{Prove the following identity:}$

$\displaystyle \begin{vmatrix} z & x & y\\ z^2 & x^2 & y^2\\ z^4 & x^4 & y^4 \end{vmatrix} = \begin{vmatrix} x & y & z\\ x^2 & y^2 & z^2\\ x^4 & y^4 & z^4 \end{vmatrix} = \begin{vmatrix} x^2 & y^2 & z^2\\ x^4 & y^4 & z^4\\ x & y & z \end{vmatrix} =xyz(x-y)(y-z)(z-x)(x+y+z).$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let }\Delta_1= \begin{vmatrix} z & x & y\\ z^2 & x^2 & y^2\\ z^4 & x^4 & y^4 \end{vmatrix},\quad \Delta_2= \begin{vmatrix} x & y & z\\ x^2 & y^2 & z^2\\ x^4 & y^4 & z^4 \end{vmatrix},\quad \Delta_3= \begin{vmatrix} x^2 & y^2 & z^2\\ x^4 & y^4 & z^4\\ x & y & z \end{vmatrix}$

$\displaystyle \text{and }\Delta_4=xyz(x-y)(y-z)(z-x)(x+y+z).$

$\displaystyle \text{Now, }\Delta_1= \begin{vmatrix} z & x & y\\ z^2 & x^2 & y^2\\ z^4 & x^4 & y^4 \end{vmatrix}$

$\displaystyle \text{Using the property that if two rows (or columns) of a determinant are interchanged,}$

$\displaystyle \text{the value of the determinant becomes negative, we get}$

$\displaystyle \Rightarrow \Delta_1=(-1) \begin{vmatrix} x & z & y\\ x^2 & z^2 & y^2\\ x^4 & z^4 & y^4 \end{vmatrix} \quad[\because\ C_1\leftrightarrow C_2]$

$\displaystyle =(-1)(-1) \begin{vmatrix} x & y & z\\ x^2 & y^2 & z^2\\ x^4 & y^4 & z^4 \end{vmatrix} \quad[\because\ C_2\leftrightarrow C_3]$

$\displaystyle = \begin{vmatrix} x & y & z\\ x^2 & y^2 & z^2\\ x^4 & y^4 & z^4 \end{vmatrix} =\Delta_2\qquad (1)$

$\displaystyle =(-1) \begin{vmatrix} x^2 & y^2 & z^2\\ x & y & z\\ x^4 & y^4 & z^4 \end{vmatrix} \quad \big[\text{Applying } R_1 \leftrightarrow R_2\big]$

$\displaystyle =(-1)(-1) \begin{vmatrix} x^2 & y^2 & z^2\\ x^4 & y^4 & z^4\\ x & y & z \end{vmatrix} \quad \big[\text{Applying } R_2 \leftrightarrow R_3\big]$

$\displaystyle = \begin{vmatrix} x^2 & y^2 & z^2\\ x^4 & y^4 & z^4\\ x & y & z \end{vmatrix} =\Delta_3 \qquad (2)$

$\displaystyle \text{Thus,}$

$\displaystyle \Delta_1=\Delta_2=\Delta_3 \quad [\text{From eqs. }(1)\text{ and }(2)]$

$\displaystyle \Delta_2= \begin{vmatrix} x & y & z\\ x^2 & y^2 & z^2\\ x^4 & y^4 & z^4 \end{vmatrix}$

$\displaystyle =xyz \begin{vmatrix} 1 & 1 & 1\\ x & y & z\\ x^3 & y^3 & z^3 \end{vmatrix} \quad \big[\text{Taking out common factor } x \text{ from } C_1,\ y \text{ from } C_2 \text{ and } z \text{ from } C_3\big]$

$\displaystyle =xyz \begin{vmatrix} 0 & 0 & 1\\ x-y & y-z & z\\ x^3-y^3 & y^3-z^3 & z^3 \end{vmatrix} \quad \big[\text{Applying } C_1\to C_1-C_2 \text{ and } C_2\to C_2-C_3\big]$

$\displaystyle =xyz(x-y)(y-z) \begin{vmatrix} 0 & 0 & 1\\ 1 & 1 & z\\ x^2+2xy+y^2 & y^2+2yz+z^2 & z^3 \end{vmatrix} \quad \big[\because\ (a^3-b^3)=(a-b)(a^2+ab+b^2)\big] \ \big[\text{Taking out common factor } (x-y) \text{ from } C_1 \text{ and } (y-z) \text{ from } C_2\big]$

$\displaystyle =xyz(x-y)(y-z) \left\{ 1\times \begin{vmatrix} 1 & 1\\ x^2+xy+y^2 & y^2+yz+z^2 \end{vmatrix} \right\} \quad \big[\text{Expanding along } R_1\big]$

$\displaystyle =xyz(x-y)(y-z)\{\,y^2+yz+z^2-x^2-xy-y^2\,\}$

$\displaystyle =xyz(x-y)(y-z)\{\,yz-xy+z^2-x^2\,\}$

$\displaystyle =xyz(x-y)(y-z)\{\,y(z-x)+(z-x)(z+x)\,\}$

$\displaystyle =xyz(x-y)(y-z)(z-x)(y+x+z)$

$\displaystyle =xyz(x-y)(y-z)(z-x)(x+y+z)$

$\displaystyle =\Delta_4$

$\displaystyle \text{Thus,}$

$\displaystyle \Delta_1=\Delta_2=\Delta_3=\Delta_4$

$\\$

$\displaystyle \textbf{Question 20: } \text{Prove the following identity:}$

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

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let LHS }=\Delta= \begin{vmatrix} (b+c)^2 & a^2 & bc\\ (c+a)^2 & b^2 & ca\\ (a+b)^2 & c^2 & ab \end{vmatrix}$

$\displaystyle = \begin{vmatrix} (b+c)^2-(c+a)^2 & a^2-b^2 & bc-ca\\ (c+a)^2-(a+b)^2 & b^2-c^2 & ca-ab\\ (a+b)^2 & c^2 & ab \end{vmatrix} \quad \big[\text{Applying } R_1\to R_1-R_2 \text{ and } R_2\to R_2-R_3\big]$

$\displaystyle = \begin{vmatrix} (b-a)(b+2c+a) & (a+b)(a-b) & c(b-a)\\ (c-b)(b+2a+c) & (b-c)(b+c) & a(c-b)\\ (a+b)^2 & c^2 & ab \end{vmatrix}$

$\displaystyle =(a-b)(b-c) \begin{vmatrix} -(b+2c+a) & a+b & -c\\ -(b+2a+c) & b+c & -a\\ (a+b)^2 & c^2 & ab \end{vmatrix} \quad \big[\text{Applying } x^2-y^2=(x+y)(x-y)\ \text{and taking out }(a-b)\text{ common from }R_1\text{ and }(b-c)\text{ from }R_2\big]$

$\displaystyle =(a-b)(b-c) \begin{vmatrix} -2(b+c+a) & a+b & -c\\ -2(b+a+c) & b+c & -a\\ (a+b)^2-c^2 & c^2 & ab \end{vmatrix} \quad \big[\text{Applying } C_1\to C_1-C_2\big]$

$\displaystyle =(a-b)(b-c) \begin{vmatrix} -2(b+c+a) & a+b & -c\\ -2(b+a+c) & b+c & -a\\ (a+b+c)(a+b-c) & c^2 & ab \end{vmatrix} \quad \big[\text{Applying } x^2-y^2=(x+y)(x-y)\ \text{in } C_1\big]$

$\displaystyle =(a-b)(b-c)(a+b+c) \begin{vmatrix} -2 & a+b & -c\\ -2 & b+c & -a\\ a+b-c & c^2 & ab \end{vmatrix} \quad \big[\text{Taking out } (a+b+c) \text{ common from } C_1\big]$

$\displaystyle =(a-b)(b-c)(a+b+c) \begin{vmatrix} -2 & a+b & -c\\ 0 & c-a & c-a\\ a+b-c & c^2 & ab \end{vmatrix} \quad \big[\text{Applying } R_2\to R_2-R_1\big]$

$\displaystyle =(a-b)(b-c)(a+b+c)(c-a) \begin{vmatrix} -2 & a+b & -c\\ 0 & 1 & 1\\ a+b-c & c^2 & ab \end{vmatrix} \quad \big[\text{Taking out } (c-a) \text{ common from } R_2\big]$

$\displaystyle =(a-b)(b-c)(a+b+c)(c-a) \begin{vmatrix} -2 & a+b & -c\\ 0 & 1 & 1\\ a+b-c & c^2-ab & ab \end{vmatrix} \quad \big[\text{Applying } C_2\to C_2-C_3\big]$

$\displaystyle =(a-b)(b-c)(a+b+c)(c-a) \left\{ (-1) \begin{vmatrix} -2 & a+b+c\\ a+b-c & c^2-ab \end{vmatrix} \right\} \quad \big[\text{Expanding along } R_2\big]$

$\displaystyle =(a-b)(b-c)(a+b+c)(c-a) \begin{vmatrix} -2 & a+b+c & -c\\ 0 & 0 & 1\\ a+b-c & c^2-ab & ab \end{vmatrix} \quad \big[\text{Applying } C_2\to C_2-C_3\big]$

$\displaystyle =(a-b)(b-c)(a+b+c)(c-a) \left\{(-1) \begin{vmatrix} -2 & a+b+c\\ a+b-c & c^2-ab \end{vmatrix} \right\} \quad \big[\text{Expanding along } R_2\big]$

$\displaystyle =-(a-b)(b-c)(a+b+c)(c-a) \{-2c^2+2ab-a^2-b^2-2ab+c^2\}$

$\displaystyle =-(a-b)(b-c)(a+b+c)(c-a)(-a^2-b^2-c^2)$

$\displaystyle =(a-b)(b-c)(a+b+c)(c-a)(a^2+b^2+c^2)$

$\displaystyle =\text{RHS}$

$\\$

$\displaystyle \textbf{Question 21: } \text{Prove the following identity:}$

$\displaystyle \begin{vmatrix} (a+1)(a+2) & a+2 & 1\\ (a+2)(a+3) & a+3 & 1\\ (a+3)(a+4) & a+4 & 1 \end{vmatrix} =-2$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let LHS }=\Delta= \begin{vmatrix} (a+1)(a+2) & a+2 & 1\\ (a+2)(a+3) & a+3 & 1\\ (a+3)(a+4) & a+4 & 1 \end{vmatrix}$

$\displaystyle = \begin{vmatrix} (a+1)(a+2)-(a+2)(a+3) & (a+2)-(a+3) & 0\\ (a+2)(a+3)-(a+3)(a+4) & (a+3)-(a+4) & 0\\ (a+3)(a+4) & a+4 & 1 \end{vmatrix} \quad \big[\text{Applying } C_1\to C_1-C_2 \text{ and } C_2\to C_2-C_3\big]$

$\displaystyle = \begin{vmatrix} -2(a+2) & -1 & 0\\ -2(a+3) & -1 & 0\\ (a+3)(a+4) & a+4 & 1 \end{vmatrix}$

$\displaystyle = \left\{ 1\times \begin{vmatrix} -2(a+2) & -1\\ -2(a+3) & -1 \end{vmatrix} \right\} \quad \big[\text{Expanding along } C_3\big]$

$\displaystyle = \big[\,2(a+2)-2(a+3)\,\big]$

$\displaystyle = 4+2a-2a-6$

$\displaystyle = -2$

$\displaystyle =\text{RHS}$

$\displaystyle \text{Hence proved.}$

$\\$

$\displaystyle \textbf{Question 22: } \text{Prove the following identity:}$

$\displaystyle \begin{vmatrix} a^2 & a^2-(b-c)^2 & bc\\ b^2 & b^2-(c-a)^2 & ca\\ c^2 & c^2-(a-b)^2 & ab \end{vmatrix} =(a-b)(b-c)(c-a)(a+b+c)(a^2+b^2+c^2) \hspace{2.0cm} \text{[CBSE 2012]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let LHS }=\Delta= \begin{vmatrix} a^2 & a^2-(b-c)^2 & bc\\ b^2 & b^2-(c-a)^2 & ca\\ c^2 & c^2-(a-b)^2 & ab \end{vmatrix}$

$\displaystyle \Rightarrow \Delta= \begin{vmatrix} a^2 & -(b-c)^2 & bc\\ b^2 & -(c-a)^2 & ca\\ c^2 & -(a-b)^2 & ab \end{vmatrix} \quad \big[\text{Applying } C_2\to C_2-C_1\big]$

$\displaystyle =(-1) \begin{vmatrix} a^2 & (b-c)^2 & bc\\ b^2 & (c-a)^2 & ca\\ c^2 & (a-b)^2 & ab \end{vmatrix}$

$\displaystyle = \begin{vmatrix} a^2 & b^2+c^2 & bc\\ b^2 & c^2+a^2 & ca\\ c^2 & a^2+b^2 & ab \end{vmatrix} \quad \big[\text{Applying } C_2\to C_2-2C_3\big]$

$\displaystyle = \begin{vmatrix} a^2+b^2+c^2 & b^2+c^2 & bc\\ b^2+c^2+a^2 & c^2+a^2 & ca\\ c^2+a^2+b^2 & a^2+b^2 & ab \end{vmatrix} \quad \big[\text{Applying } C_1\to C_1+C_2\big]$

$\displaystyle =-(a^2+b^2+c^2) \begin{vmatrix} 1 & b^2+c^2 & bc\\ 1 & c^2+a^2 & ca\\ 1 & a^2+b^2 & ab \end{vmatrix}$

$\displaystyle =-(a^2+b^2+c^2) \begin{vmatrix} 1 & b^2+c^2 & bc\\ 0 & (c^2+a^2)-(b^2+c^2) & ca-bc\\ 0 & (a^2+b^2)-(b^2+c^2) & ab-bc \end{vmatrix} \quad \big[\text{Applying } R_2\to R_2-R_1 \text{ and } R_3\to R_3-R_1\big]$

$\displaystyle =-(a^2+b^2+c^2) \begin{vmatrix} 1 & b^2+c^2 & bc\\ 0 & a^2-b^2 & c(a-b)\\ 0 & a^2-c^2 & b(a-c) \end{vmatrix}$

$\displaystyle =-(a^2+b^2+c^2)(a-b)(a-c) \begin{vmatrix} 1 & b^2+c^2 & bc\\ 0 & a+b & c\\ 0 & a+c & b \end{vmatrix} \quad \big[\text{Taking } (a-b) \text{ common from } R_2 \text{ and } (a-c) \text{ common from } R_3\big]$

$\displaystyle =(a^2+b^2+c^2)(a-b)(c-a) \left\{ 1\times \begin{vmatrix} a+b & c\\ a+c & b \end{vmatrix} \right\} \quad [\because\ (c-a)=-(a-c)] \quad [\text{Expanding along } C_1]$

$\displaystyle =(a^2+b^2+c^2)(a-b)(c-a)(ab+b^2-ac-c^2)$

$\displaystyle =(a^2+b^2+c^2)(a-b)(c-a)\{a(b-c)+(b+c)(b-c)\}$

$\displaystyle =(a-b)(c-a)(b-c)(a+b+c)(a^2+b^2+c^2)$

$\displaystyle =\text{RHS}$

$\displaystyle \text{Hence proved.}$

$\\$

$\displaystyle \textbf{Question 23: } \text{Prove the following identity:}$

$\displaystyle \begin{vmatrix} 1 & a^2+bc & a^3\\ 1 & b^2+ca & b^3\\ 1 & c^2+ab & c^3 \end{vmatrix} =-(a-b)(b-c)(c-a)(a^2+b^2+c^2) \hspace{2.0cm} \text{[CBSE 2008]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let LHS }=\Delta= \begin{vmatrix} 1 & a^2+bc & a^3\\ 1 & b^2+ca & b^3\\ 1 & c^2+ab & c^3 \end{vmatrix}$

$\displaystyle \Rightarrow \Delta= \begin{vmatrix} 0 & (a^2+bc)-(b^2+ca) & a^3-b^3\\ 0 & (b^2+ca)-(c^2+ab) & b^3-c^3\\ 1 & c^2+ab & c^3 \end{vmatrix} \quad \big[\text{Applying } R_1\to R_1-R_2 \text{ and } R_2\to R_2-R_3\big]$

$\displaystyle = \begin{vmatrix} 0 & a^2-b^2-ca+bc & a^3-b^3\\ 0 & b^2-c^2-ab+ca & b^3-c^3\\ 1 & c^2+ab & c^3 \end{vmatrix}$

$\displaystyle = \begin{vmatrix} 0 & (a-b)(a+b-c) & (a-b)(a^2+ab+b^2)\\ 0 & (b-c)(b+c-a) & (b-c)(b^2+bc+c^2)\\ 1 & c^2+ab & c^3 \end{vmatrix}$

$\displaystyle =(a-b)(b-c) \begin{vmatrix} 0 & a+b-c & a^2+ab+b^2\\ 0 & b+c-a & b^2+bc+c^2\\ 1 & c^2+ab & c^3 \end{vmatrix} \quad \big[\text{Taking out }(a-b)\text{ common from }R_1\text{ and }(b-c)\text{ from }R_2\big]$

$\displaystyle =(a-b)(b-c) \begin{vmatrix} 0 & a+b-c & a^2+ab+b^2\\ 0 & (b+c-a)-(a+b-c) & (b^2+bc+c^2)-(a^2+ab+b^2)\\ 1 & c^2+ab & c^3 \end{vmatrix} \quad \big[\text{Applying } R_2\to R_2-R_1\big]$

$\displaystyle =(a-b)(b-c) \begin{vmatrix} 0 & a+b-c & a^2+ab+b^2\\ 0 & c-a & c^2-a^2\\ 1 & c^2+ab & c^3 \end{vmatrix}$

$\displaystyle =(a-b)(b-c)(c-a) \begin{vmatrix} 0 & a+b-c & a^2+ab+b^2\\ 0 & 1 & a+c\\ 1 & c^2+ab & c^3 \end{vmatrix} \quad \big[\text{Taking out }(c-a)\text{ common from }R_2\big]$

$\displaystyle =(a-b)(b-c)(c-a) \left\{ (-1) \begin{vmatrix} a+b-c & a^2+ab+b^2\\ 1 & a+c \end{vmatrix} \right\} \quad \big[\text{Expanding along }C_1\big]$

$\displaystyle =-(a-b)(b-c)(c-a)\{(a+b-c)(a+c)-(a^2+ab+b^2)\}$

$\displaystyle =-(a-b)(b-c)(c-a)(a^2+b^2+c^2)$

$\displaystyle =\text{RHS}$

$\\$

$\displaystyle \textbf{Question 24: } \text{Prove the following identity:}$

$\displaystyle \begin{vmatrix} a^2 & bc & ac+c^2\\ a^2+ab & b^2 & ac\\ ab & b^2+bc & c^2 \end{vmatrix} =4a^2b^2c^2 \hspace{2.0cm} \text{[CBSE 2014, 2015]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let LHS }=\Delta= \begin{vmatrix} a^2 & bc & ac+c^2\\ a^2+ab & b^2 & ac\\ ab & b^2+bc & c^2 \end{vmatrix}$

$\displaystyle \Delta =abc \begin{vmatrix} a & c & a+c\\ a+b & b & a\\ b & b+c & c \end{vmatrix} \quad \big[\text{Taking out } a,b,c \text{ common from } C_1,C_2,C_3\big]$

$\displaystyle =abc \begin{vmatrix} a & c & 0\\ a+b & b & -2b\\ b & b+c & -2b \end{vmatrix} \quad \big[\text{Applying } C_3\to C_3-C_2-C_1\big]$

$\displaystyle =(abc)(-2b) \begin{vmatrix} a & c & 0\\ a+b & b & 1\\ b & b+c & 1 \end{vmatrix} \quad \big[\text{Taking } (-2b) \text{ common from } C_3\big]$

$\displaystyle =(abc)(-2b) \begin{vmatrix} a & c & 0\\ a & -c & 0\\ b & b+c & 1 \end{vmatrix} \quad \big[\text{Applying } R_2\to R_2-R_3\big]$

$\displaystyle =(abc)(-2b) \left\{ 1\times \begin{vmatrix} a & c\\ a & -c \end{vmatrix} \right\} \quad \big[\text{Expanding along } C_3\big]$

$\displaystyle =(abc)(-2b)(-2ac)$

$\displaystyle =4a^2b^2c^2$

$\displaystyle =\text{RHS}$

$\\$

$\displaystyle \textbf{Question 25: } \text{Prove the following identity:}$

$\displaystyle \left| \begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array} \right| = 16(3x+4)$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let LHS }=\Delta= \left| \begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array} \right|$

$\displaystyle = \left| \begin{array}{ccc} 3x+4 & 3x+4 & 3x+4 \\ x & x+4 & x \\ x & x & x+4 \end{array} \right| \quad [\text{Applying } R_1 \rightarrow R_1+R_2+R_3]$

$\displaystyle = (3x+4) \left| \begin{array}{ccc} 1 & 1 & 1 \\ x & x+4 & x \\ x & x & x+4 \end{array} \right| \quad [\text{Taking out } (3x+4) \text{ common from } R_1]$

$\displaystyle = (3x+4) \left| \begin{array}{ccc} 1 & 0 & 0 \\ x & 4 & 0 \\ x & 0 & 4 \end{array} \right| \quad [\text{Applying } C_2 \rightarrow C_2-C_1 \text{ and } C_3 \rightarrow C_3-C_1]$

$\displaystyle = (3x+4)(4^2) \quad [\text{Expanding along } R_1]$

$\displaystyle = 16(3x+4)$

$\displaystyle = \text{RHS}$

$\\$

$\displaystyle \textbf{Question 26: } \text{Prove the following identity:}$

$\displaystyle \left| \begin{array}{ccc} 1 & 1+p & 1+p+q \\ 2 & 3+2p & 4+3p+2q \\ 3 & 6+3p & 10+6p+3q \end{array} \right| = 1$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let LHS }=\Delta= \left| \begin{array}{ccc} 1 & 1+p & 1+p+q \\ 2 & 3+2p & 4+3p+2q \\ 3 & 6+3p & 10+6p+3q \end{array} \right|$

$\displaystyle = \left| \begin{array}{ccc} 1 & 1 & 1+p \\ 2 & 3 & 4+3p \\ 3 & 6 & 10+6p \end{array} \right| + \left| \begin{array}{ccc} 1 & p & q \\ 2 & 2p & 2q \\ 3 & 3p & 3q \end{array} \right|$

$\displaystyle = \left| \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 3 & 6 & 10 \end{array} \right| + \left| \begin{array}{ccc} 1 & 1 & p \\ 2 & 3 & 3p \\ 3 & 6 & 6p \end{array} \right| + (pq) \left| \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 3 & 3 & 3 \end{array} \right| \quad [\text{Taking out } pq \text{ common from last determinant}]$

$\displaystyle = \left| \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 3 & 6 & 10 \end{array} \right| + (p) \left| \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 3 & 3 \\ 3 & 6 & 6 \end{array} \right| + 0 \quad [\text{Taking out } p \text{ common from second determinant}]$

$\displaystyle = \left| \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 3 & 6 & 10 \end{array} \right| + 0 \quad [\because \text{Determinant with two identical columns is zero}]$

$\displaystyle = \left| \begin{array}{ccc} 1 & 0 & 0 \\ 2 & 1 & 2 \\ 3 & 3 & 7 \end{array} \right| \quad [\text{Applying } C_2 \rightarrow C_2 - C_1 \text{ and } C_3 \rightarrow C_3 - C_1]$

$\displaystyle = \left\{ 1 \times \left| \begin{array}{cc} 1 & 2 \\ 3 & 7 \end{array} \right| \right\} \quad [\text{Expanding along } R_1]$

$\displaystyle = 7 - 6$

$\displaystyle = 1$

$\displaystyle = \text{RHS}$

$\\$

$\displaystyle \textbf{Question 27: } \text{Prove the following identity:}$

$\displaystyle \left| \begin{array}{ccc} a & b-c & c-b \\ a-c & b & c-a \\ a-b & b-a & c \end{array} \right| = (a+b-c)(b+c-a)(c+a-b)$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let LHS }=\Delta= \left| \begin{array}{ccc} a & b-c & c-b \\ a-c & b & c-a \\ a-b & b-a & c \end{array} \right|$

$\displaystyle \Delta= \left| \begin{array}{ccc} a & 0 & c-b+a \\ a-c & b+c-a & 0 \\ a-b & b+c-a & c+a-b \end{array} \right| \quad [\text{Applying } C_2 \rightarrow C_2+C_3 \text{ and } C_3 \rightarrow C_1+C_3]$

$\displaystyle = (b+c-a)(c+a-b) \left| \begin{array}{ccc} a & 0 & 1 \\ a-c & 1 & 0 \\ a-b & 1 & 1 \end{array} \right| \quad [\text{Taking out common factor from } C_2 \text{ and } C_3]$

$\displaystyle = (b+c-a)(c+a-b) \left\{ a \left| \begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array} \right| + 1 \left| \begin{array}{cc} a-c & 1 \\ a-b & 1 \end{array} \right| \right\} \quad [\text{Expanding along } R_1]$

$\displaystyle = (a+b-c)(b+c-a)(c+a-b)$

$\displaystyle = \text{RHS}$

$\\$

$\displaystyle \textbf{Question 28: } \text{Prove the following identity:}$

$\displaystyle \left| \begin{array}{ccc} a^{2} & 2ab & b^{2} \\ b^{2} & a^{2} & 2ab \\ 2ab & b^{2} & a^{2} \end{array} \right| = (a^{3}+b^{3})^{2}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let LHS }=\Delta= \left| \begin{array}{ccc} a^{2} & 2ab & b^{2} \\ b^{2} & a^{2} & 2ab \\ 2ab & b^{2} & a^{2} \end{array} \right|$

$\displaystyle = a^{2} \left| \begin{array}{cc} a^{2} & 2ab \\ b^{2} & a^{2} \end{array} \right| - (2ab) \left| \begin{array}{cc} b^{2} & 2ab \\ 2ab & a^{2} \end{array} \right| + b^{2} \left| \begin{array}{cc} b^{2} & a^{2} \\ 2ab & b^{2} \end{array} \right| \quad \text{Expanding}$

$\displaystyle = a^{2}(a^{4}-2ab^{3}) - (2ab)(b^{2}a^{2}-4a^{2}b^{2}) + b^{2}(b^{4}-2a^{3}b)$

$\displaystyle = a^{6}-2a^{3}b^{3}-2a^{3}b^{3}+8a^{3}b^{3}+b^{6}-2a^{3}b^{3}$

$\displaystyle = a^{6}+2a^{3}b^{3}+b^{6}$

$\displaystyle = (a^{3})^{2}+2a^{3}b^{3}+(b^{3})^{2}$

$\displaystyle = (a^{3}+b^{3})^{2}$

$\displaystyle = \text{RHS}$

$\displaystyle \text{Hence proved.}$

$\\$

$\displaystyle \textbf{Question 29: } \text{Prove the following identity:}$

$\displaystyle \left| \begin{array}{ccc} a^{2}+1 & ab & ac \\ ab & b^{2}+1 & bc \\ ca & cb & c^{2}+1 \end{array} \right| = 1+a^{2}+b^{2}+c^{2} \hspace{2.0cm} \text{[CBSE 2014]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let LHS }=\Delta= \left| \begin{array}{ccc} a^{2}+1 & ab & ac \\ ab & b^{2}+1 & bc \\ ca & cb & c^{2}+1 \end{array} \right|$

$\displaystyle = (abc) \left| \begin{array}{ccc} a+\frac{1}{a} & b & c \\ a & b+\frac{1}{b} & c \\ a & b & c+\frac{1}{c} \end{array} \right| \quad \text{Taking out } a,b,c \text{ common from } R_1,R_2,R_3$

$\displaystyle = (abc) \left| \begin{array}{ccc} a+\frac{1}{a} & b & c \\ -\frac{1}{a} & \frac{1}{b} & 0 \\ -\frac{1}{a} & 0 & \frac{1}{c} \end{array} \right| \quad \text{Applying } R_2 \rightarrow R_2-R_1 \text{ and } R_3 \rightarrow R_3-R_1$

$\displaystyle = (abc)\left(\frac{1}{abc}\right) \left| \begin{array}{ccc} a^{2}+1 & b^{2} & c^{2} \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array} \right| \quad \text{Applying } C_1 \rightarrow aC_1,\; C_2 \rightarrow bC_2,\; C_3 \rightarrow cC_3$

$\displaystyle = \left| \begin{array}{ccc} a^{2}+1 & b^{2} & c^{2} \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array} \right|$

$\displaystyle = (-1) \left| \begin{array}{cc} b^{2} & c^{2} \\ 1 & 0 \end{array} \right| + (1) \left| \begin{array}{cc} a^{2}+1 & b^{2} \\ -1 & 1 \end{array} \right| \quad \text{Expanding along } R_3$

$\displaystyle = (-1)(-c^{2})+(a^{2}+1+b^{2})$

$\displaystyle = a^{2}+b^{2}+c^{2}+1$

$\displaystyle = \text{RHS}$

$\\$

$\displaystyle \textbf{Question 30: } \text{Prove the following identity:}$

$\displaystyle \left| \begin{array}{ccc} 1 & a & a^{2} \\ a^{2} & 1 & a \\ a & a^{2} & 1 \end{array} \right| = (a^{3}-1)^{2} \hspace{2.0cm} \text{[CBSE 2013, 2014, 2015]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let LHS }=\Delta= \left| \begin{array}{ccc} 1 & a & a^{2} \\ a^{2} & 1 & a \\ a & a^{2} & 1 \end{array} \right|$

$\displaystyle \Delta= \left| \begin{array}{ccc} 1+a^{2}+a & 1+a^{2}+a & 1+a^{2}+a \\ a^{2} & 1 & a \\ a & a^{2} & 1 \end{array} \right| \quad \text{Applying } R_1 \rightarrow R_1+R_2+R_3$

$\displaystyle = (1+a^{2}+a) \left| \begin{array}{ccc} 1 & 1 & 1 \\ a^{2} & 1 & a \\ a & a^{2} & 1 \end{array} \right| \quad \text{Applying } C_2 \rightarrow C_2-C_1 \text{ and } C_3 \rightarrow C_3-C_1$

$\displaystyle = (1+a^{2}+a) \left| \begin{array}{ccc} 1 & 0 & 0 \\ a^{2} & 1-a^{2} & a-a^{2} \\ a & a^{2}-a & 1-a \end{array} \right|$

$\displaystyle = (1+a^{2}+a) \left| \begin{array}{ccc} 1 & 0 & 0 \\ a^{2} & (1-a)(1+a) & a(1-a) \\ a & a(a-1) & 1-a \end{array} \right|$

$\displaystyle = (1+a^{2}+a)(a-1)(a-1) \left| \begin{array}{ccc} 1 & 0 & 0 \\ a^{2} & -(1+a) & -a \\ a & a & -1 \end{array} \right| \quad \text{Taking out } (a-1) \text{ common from } C_2 \text{ and } C_3$

$\displaystyle = (a^{3}-1) \left\{ (a-1) \left| \begin{array}{ccc} 1 & 0 & 0 \\ a^{2} & -(1+a) & -a \\ a & a & -1 \end{array} \right| \right\} \quad \because (1+a^{2}+a)(a-1)=(a^{3}-1)$

$\displaystyle = (a^{3}-1)\{(a-1)(1+a+a^{2})\}$

$\displaystyle = (a^{3}-1)(a^{3}-1)$

$\displaystyle = (a^{3}-1)^{2}$

$\displaystyle = \text{RHS}$

$\displaystyle \text{Hence proved.}$

$\\$

$\displaystyle \textbf{Question 31: } \text{Prove the following identity:}$

$\displaystyle \left| \begin{array}{ccc} a+b+c & -c & -b \\ -c & a+b+c & -a \\ -b & -a & a+b+c \end{array} \right| = 2(a+b)(b+c)(c+a)$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let LHS }=\Delta= \left| \begin{array}{ccc} a+b+c & -c & -b \\ -c & a+b+c & -a \\ -b & -a & a+b+c \end{array} \right|$

$\displaystyle = \left| \begin{array}{ccc} a & -c & -b \\ b & a+b+c & -a \\ c & -a & a+b+c \end{array} \right| \quad \text{Applying } C_1 \rightarrow C_1+C_2+C_3$

$\displaystyle = \left| \begin{array}{ccc} a+b & a+b & -(a+b) \\ b+c & b+c & b+c \\ c & -a & a+b+c \end{array} \right| \quad \text{Applying } R_1 \rightarrow R_1+R_2 \text{ and } R_2 \rightarrow R_2+R_3$

$\displaystyle = (a+b)(b+c) \left| \begin{array}{ccc} 1 & 1 & -1 \\ 1 & 1 & 1 \\ c & -a & a+b+c \end{array} \right| \quad \text{Taking out common factor from } R_1 \text{ and } R_2$

$\displaystyle = (a+b)(b+c) \left| \begin{array}{ccc} 0 & 0 & -2 \\ 1 & 1 & 1 \\ c & -a & a+b+c \end{array} \right| \quad \text{Applying } R_1 \rightarrow R_1-R_2$

$\displaystyle = (a+b)(b+c)\{(-2)(-a-c)\} \quad \text{Expanding along } R_1$

$\displaystyle = 2(a+b)(b+c)(a+c)$

$\displaystyle = \text{RHS}$

$\displaystyle \text{Hence proved.}$

$\\$

$\displaystyle \textbf{Question 32: } \text{Prove the following identity:}$

$\displaystyle \left| \begin{array}{ccc} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{array} \right| = 4abc \hspace{2.0cm} \text{[CBSE 2006]}$

$\displaystyle \text{Answer:}$

$\displaystyle \Delta= \left| \begin{array}{ccc} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{array} \right|$

$\displaystyle = \left| \begin{array}{ccc} 0 & -2c & -2b \\ b & c+a & b \\ c & c & a+b \end{array} \right| \quad \text{Applying } R_1 \rightarrow R_1-(R_2+R_3)$

$\displaystyle = \left| \begin{array}{ccc} 0 & -2c & -2b \\ b & c+a-b & 0 \\ c & 0 & a+b-c \end{array} \right| \quad \text{Applying } C_2 \rightarrow C_2-C_1 \text{ and } C_3 \rightarrow C_3-C_1$

$\displaystyle = 0 \left| \begin{array}{cc} c+a-b & 0 \\ 0 & a+b-c \end{array} \right| - (-2c) \left| \begin{array}{cc} b & 0 \\ c & a+b-c \end{array} \right| + (-2b) \left| \begin{array}{cc} b & c+a-b \\ c & 0 \end{array} \right| \quad \text{Expanding along } R_1$

$\displaystyle = 2c\{b(a+b-c)-0\}-2b\{0-c(c+a-b)\}$

$\displaystyle = 2bc(a+b-c)-2bc(b-c-a)$

$\displaystyle = 2bc\{(a+b-c)-(b-c-a)\}$

$\displaystyle = 4abc$

$\displaystyle \text{Hence proved.}$

$\\$

$\displaystyle \textbf{Question 33: } \text{Prove the following identity:}$

$\displaystyle \left| \begin{array}{ccc} b^{2}+c^{2} & ab & ac \\ ba & c^{2}+a^{2} & bc \\ ca & cb & a^{2}+b^{2} \end{array} \right| = 4a^{2}b^{2}c^{2}$

$\displaystyle \text{Answer:}$

$\displaystyle \Delta= \left| \begin{array}{ccc} b^{2}+c^{2} & ab & ac \\ ba & c^{2}+a^{2} & bc \\ ca & cb & a^{2}+b^{2} \end{array} \right|$

$\displaystyle = \left| \begin{array}{ccc} a(b^{2}+c^{2}) & a^{2}b & a^{2}c \\ b^{2}a & b(c^{2}+a^{2}) & b^{2}c \\ c^{2}a & c^{2}b & c(a^{2}+b^{2}) \end{array} \right| \quad \text{Multiplying the three rows by } a,b \text{ and } c$

$\displaystyle = \frac{abc}{abc} \left| \begin{array}{ccc} b^{2}+c^{2} & a^{2} & a^{2} \\ b^{2} & c^{2}+a^{2} & b^{2} \\ c^{2} & c^{2} & a^{2}+b^{2} \end{array} \right| \quad \text{Taking out } a,b,c \text{ common from the three columns}$

$\displaystyle = \left| \begin{array}{ccc} 2(b^{2}+c^{2}) & 2(a^{2}+c^{2}) & 2(a^{2}+b^{2}) \\ b^{2} & c^{2}+a^{2} & b^{2} \\ c^{2} & c^{2} & a^{2}+b^{2} \end{array} \right| \quad \text{Applying } R_1 \rightarrow R_1+R_2+R_3$

$\displaystyle = 2 \left| \begin{array}{ccc} b^{2}+c^{2} & a^{2}+c^{2} & a^{2}+b^{2} \\ -\,c^{2} & 0 & -\,a^{2} \\ -\,b^{2} & -\,a^{2} & 0 \end{array} \right| \quad \text{Taking out 2 common and applying } R_2 \rightarrow R_2-R_1,\; R_3 \rightarrow R_3-R_1$

$\displaystyle = 2 \left| \begin{array}{ccc} 0 & c^{2} & b^{2} \\ -\,c^{2} & 0 & -\,a^{2} \\ -\,b^{2} & -\,a^{2} & 0 \end{array} \right| \quad \text{Applying } R_1 \rightarrow R_1+R_2+R_3$

$\displaystyle = 2\big[-c^{2}(-a^{2}b^{2})+b^{2}(c^{2}a^{2})\big] \quad \text{Expanding along } R_1$

$\displaystyle = 4a^{2}b^{2}c^{2}$

$\\$

$\displaystyle \textbf{Question 34: } \text{Prove the following identity:}$

$\displaystyle \left| \begin{array}{ccc} 0 & b^{2}a & c^{2}a \\ a^{2}b & 0 & c^{2}b \\ a^{2}c & b^{2}c & 0 \end{array} \right| = 2a^{3}b^{3}c^{3} \hspace{2.0cm} \text{[CBSE 2003]}$

$\displaystyle \text{Answer:}$

$\displaystyle \Delta= \left| \begin{array}{ccc} 0 & b^{2}a & c^{2}a \\ a^{2}b & 0 & c^{2}b \\ a^{2}c & b^{2}c & 0 \end{array} \right|$

$\displaystyle = \frac{1}{abc} \left| \begin{array}{ccc} 0 & b^{3}a & c^{3}a \\ a^{3}b & 0 & c^{3}b \\ a^{3}c & b^{3}c & 0 \end{array} \right| \quad \text{Multiplying the three columns by } a,b \text{ and } c$

$\displaystyle = \frac{abc}{abc} \left| \begin{array}{ccc} 0 & b^{3} & c^{3} \\ a^{3} & 0 & c^{3} \\ a^{3} & b^{3} & 0 \end{array} \right| \quad \text{Taking out } a,b,c \text{ common from the three rows}$

$\displaystyle = b^{3} \left| \begin{array}{cc} a^{3} & c^{3} \\ a^{3} & 0 \end{array} \right| + c^{3} \left| \begin{array}{cc} a^{3} & 0 \\ a^{3} & b^{3} \end{array} \right| \quad \text{Expanding along } R_1$

$\displaystyle = b^{3}(-a^{3}c^{3})+c^{3}(a^{3}b^{3})$

$\displaystyle = 2a^{3}b^{3}c^{3}$

$\\$

$\displaystyle \textbf{Question 35: } \text{Prove the following identity:}$

$\displaystyle \left| \begin{array}{ccc} \dfrac{a^{2}+b^{2}}{c} & c & c \\ a & \dfrac{b^{2}+c^{2}}{a} & a \\ b & b & \dfrac{c^{2}+a^{2}}{b} \end{array} \right| = 4abc$

$\displaystyle \text{Answer:}$

$\displaystyle \Delta= \left| \begin{array}{ccc} \dfrac{a^{2}+b^{2}}{c} & c & c \\ a & \dfrac{b^{2}+c^{2}}{a} & a \\ b & b & \dfrac{c^{2}+a^{2}}{b} \end{array} \right|$

$\displaystyle = \frac{1}{abc} \left| \begin{array}{ccc} a^{2}+b^{2} & c^{2} & c^{2} \\ a^{2} & b^{2}+c^{2} & a^{2} \\ b^{2} & b^{2} & c^{2}+a^{2} \end{array} \right| \quad \text{Multiplying } R_1,R_2,R_3 \text{ by } c,a,b \text{ and dividing by } abc$

$\displaystyle = \frac{1}{abc} \left| \begin{array}{ccc} a^{2}+b^{2} & c^{2}-a^{2}-b^{2} & c^{2}-a^{2}-b^{2} \\ a^{2} & b^{2}+c^{2}-a^{2} & 0 \\ b^{2} & 0 & c^{2}+a^{2}-b^{2} \end{array} \right| \quad \text{Applying } C_2 \rightarrow C_2-C_1,\; C_3 \rightarrow C_3-C_1$

$\displaystyle = \frac{1}{abc} \left| \begin{array}{ccc} 0 & -2b^{2} & -2a^{2} \\ a^{2} & b^{2}+c^{2}-a^{2} & 0 \\ b^{2} & 0 & c^{2}+a^{2}-b^{2} \end{array} \right| \quad \text{Applying } R_1 \rightarrow R_1-R_2-R_3$

$\displaystyle = \frac{1}{abc} \left[ -\,a^{2} \left| \begin{array}{cc} -2b^{2} & -2a^{2} \\ 0 & c^{2}+a^{2}-b^{2} \end{array} \right| + b^{2} \left| \begin{array}{cc} -2b^{2} & -2a^{2} \\ b^{2}+c^{2}-a^{2} & 0 \end{array} \right| \right] \quad \text{Expanding along } C_1$

$\displaystyle = \frac{1}{abc} \left[ -\,a^{2}\{-2b^{2}(c^{2}+a^{2}-b^{2})\} + b^{2}\{2a^{2}(b^{2}+c^{2}-a^{2})\} \right]$

$\displaystyle = \frac{1}{abc} \left[ 2a^{2}b^{2}c^{2} +2a^{4}b^{2} -2a^{2}b^{4} +2a^{2}b^{4} +2a^{2}b^{2}c^{2} -2a^{4}b^{2} \right]$

$\displaystyle = \frac{1}{abc}\,4a^{2}b^{2}c^{2}$

$\displaystyle = 4abc$

$\displaystyle \text{Hence proved.}$

$\\$

$\displaystyle \textbf{Question 36: } \text{Prove the following identity:}$

$\displaystyle \left| \begin{array}{ccc} -bc & b^{2}+bc & c^{2}+bc \\ a^{2}+ac & -ac & c^{2}+ac \\ a^{2}+ab & b^{2}+ab & -ab \end{array} \right| = (ab+bc+ca)^{3}$

$\displaystyle \text{Answer:}$

$\displaystyle \Delta= \left| \begin{array}{ccc} -bc & b^{2}+bc & c^{2}+bc \\ a^{2}+ac & -ac & c^{2}+ac \\ a^{2}+ab & b^{2}+ab & -ab \end{array} \right|$

$\displaystyle = \frac{1}{abc} \left| \begin{array}{ccc} -abc & ab^{2}+abc & ac^{2}+abc \\ a^{2}b+abc & -abc & c^{2}b+abc \\ a^{2}c+abc & b^{2}c+abc & -abc \end{array} \right| \quad \text{Applying } R_1\!\to\! aR_1,\; R_2\!\to\! bR_2,\; R_3\!\to\! cR_3$

$\displaystyle = \frac{abc}{abc} \left| \begin{array}{ccc} -bc & ab+ac & ac+ab \\ ab+bc & -ac & cb+ab \\ ac+bc & bc+ac & -ab \end{array} \right| \quad \text{Taking out } a,b,c \text{ common from the three columns}$

$\displaystyle = \left| \begin{array}{ccc} ab+bc+ca & ab+bc+ca & ab+bc+ca \\ ab+bc & -ac & cb+ab \\ ac+bc & bc+ac & -ab \end{array} \right| \quad \text{Applying } R_1 \rightarrow R_1+R_2+R_3$

$\displaystyle = (ab+bc+ca) \left| \begin{array}{ccc} 1 & 1 & 1 \\ ab+bc & -ac & cb+ab \\ ac+bc & bc+ac & -ab \end{array} \right|$

$\displaystyle = (ab+bc+ca) \left| \begin{array}{ccc} 0 & 0 & 1 \\ 0 & -(ab+bc+ac) & cb+ab \\ ac+bc+ab & bc+ac+ab & -ab \end{array} \right| \quad \text{Applying } C_1\!\to\! C_1-C_3,\; C_2\!\to\! C_2-C_3$

$\displaystyle = (ab+bc+ca) \left| \begin{array}{cc} -(ab+bc+ac) & cb+ab \\ bc+ac+ab & -ab \end{array} \right|$

$\displaystyle = (ab+bc+ca)(ab+bc+ac)^{2}$

$\displaystyle = (ab+bc+ca)^{3}$

$\displaystyle \text{Hence proved.}$

$\\$

$\displaystyle \textbf{Question 37: } \text{Prove the following identity:}$

$\displaystyle \left| \begin{array}{ccc} x+\lambda & 2x & 2x \\ 2x & x+\lambda & 2x \\ 2x & 2x & x+\lambda \end{array} \right| = (5x+\lambda)(\lambda-x)^{2} \hspace{2.0cm} \text{[CBSE 2014]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{LHS: } \left| \begin{array}{ccc} x+\lambda & 2x & 2x \\ 2x & x+\lambda & 2x \\ 2x & 2x & x+\lambda \end{array} \right|$

$\displaystyle = \left| \begin{array}{ccc} x+\lambda & 2x & 2x \\ 2x-x-\lambda & x+\lambda-2x & 0 \\ 2x-x-\lambda & 0 & x+\lambda-2x \end{array} \right| \quad \text{Applying } R_2 \rightarrow R_2-R_1 \text{ and } R_3 \rightarrow R_3-R_1$

$\displaystyle = \left| \begin{array}{ccc} x+\lambda & 2x & 2x \\ -(\lambda-x) & \lambda-x & 0 \\ -(\lambda-x) & 0 & \lambda-x \end{array} \right|$

$\displaystyle = (\lambda-x)^{2} \left| \begin{array}{ccc} x+\lambda & 2x & 2x \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array} \right| \quad \text{Taking } (\lambda-x) \text{ common from } R_2 \text{ and } R_3$

$\displaystyle = (\lambda-x)^{2} \left[ -1(-2x)+1(x+\lambda+2x) \right] \quad \text{Expanding along the last row}$

$\displaystyle = (\lambda-x)^{2}(\lambda+5x)$

$\displaystyle = \text{RHS}$

$\displaystyle \therefore \left| \begin{array}{ccc} x+\lambda & 2x & 2x \\ 2x & x+\lambda & 2x \\ 2x & 2x & x+\lambda \end{array} \right| = (\lambda-x)^{2}(\lambda+5x)$

$\\$

$\displaystyle \textbf{Question 38: } \text{Prove the following identity:}$

$\displaystyle \left| \begin{array}{ccc} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{array} \right| = (5x+4)(4-x)^2 \hspace{2.0cm} \text{[CBSE 2007, 2011]}$

$\displaystyle \text{Answer:}$

$\displaystyle \Delta = \left| \begin{array}{ccc} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{array} \right|$

$\displaystyle = \left| \begin{array}{ccc} 5x+4 & 5x+4 & 5x+4 \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{array} \right| \quad [\text{Applying } R_1 \to R_1 + R_2 + R_3]$

$\displaystyle = (5x+4) \left| \begin{array}{ccc} 1 & 1 & 1 \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{array} \right| \quad [\text{Take out } 5x+4 \text{ common from } R_1]$

$\displaystyle = (5x+4) \left| \begin{array}{ccc} 1 & 0 & 0 \\ 2x & 4-x & 0 \\ 2x & 0 & 4-x \end{array} \right| \quad [\text{Applying } C_2 \to C_2 - C_1 \text{ and } C_3 \to C_3 - C_1]$

$\displaystyle = (5x+4)(4-x)^2 \quad [\text{Expanding along } R_1]$

$\displaystyle \text{Hence proved.}$

$\\$

$\displaystyle \textbf{Question 39: } \text{Prove the following identity:}$

$\displaystyle \left| \begin{array}{ccc} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{array} \right| = 4xyz$

$\displaystyle \text{Answer:}$

$\displaystyle \text{LHS: } \\ \\ \left| \begin{array}{ccc} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{array} \right| = \left| \begin{array}{ccc} y+z-z-y & z-(z+x)-x & y-x-(x+y) \\ z & z+x & x \\ y & x & x+y \end{array} \right| \\ \\ \;[\text{Applying } R_1 \to R_1 - R_2 - R_3] \\ \\ = \left| \begin{array}{ccc} 0 & -2x & -2x \\ z & z+x & x \\ y & x & x+y \end{array} \right| = -2x \left| \begin{array}{ccc} 0 & 1 & 1 \\ z & z+x & x \\ y & x & x+y \end{array} \right|\\ \\ \;[\text{Taking } -2x \text{ common from } R_1] \\ \\ = -2x \left| \begin{array}{ccc} 0 & 0 & 1 \\ z & z & x \\ y & -y & x+y \end{array} \right| \\ \\ \;[\text{Applying } C_2 \to C_2 - C_3] \\ \\ = -2x(-zy-zy) = 4xyz = \text{RHS} \\ \\ \;\therefore\; \left| \begin{array}{ccc} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{array} \right| = 4xyz$

$\\$

$\displaystyle \textbf{Question 40: } \text{Prove the following identity:}$

$\displaystyle \left| \begin{array}{ccc} -a\left(b^{2}+c^{2}-a^{2}\right) & 2b^{3} & 2c^{3} \\ 2a^{3} & -b\left(c^{2}+a^{2}-b^{2}\right) & 2c^{3} \\ 2a^{3} & 2b^{3} & -c\left(a^{2}+b^{2}-c^{2}\right) \end{array} \right| = abc\left(a^{2}+b^{2}+c^{2}\right)^{3}$

$\displaystyle \text{Answer:}$

$\displaystyle \Delta = \left| \begin{array}{ccc} -a(b^{2}+c^{2}-a^{2}) & 2b^{3} & 2c^{3} \\ 2a^{3} & -b(c^{2}+a^{2}-b^{2}) & 2c^{3} \\ 2a^{3} & 2b^{3} & -c(a^{2}+b^{2}-c^{2}) \end{array} \right| \\ \\ = abc \left| \begin{array}{ccc} -a^{2}+b^{2}+c^{2} & 2b^{2} & 2c^{2} \\ 2a^{2} & -b^{2}+c^{2}+a^{2} & 2c^{2} \\ 2a^{2} & 2b^{2} & -c^{2}+a^{2}+b^{2} \end{array} \right| \\ \\ \;[\text{Taking out } a,b,c \text{ common from } C_1,C_2,C_3] \\ \\ = abc \left| \begin{array}{ccc} a^{2}+b^{2}+c^{2} & 2b^{2} & 2c^{2} \\ a^{2}+b^{2}+c^{2} & -c^{2}-a^{2}+b^{2} & 2c^{2} \\ a^{2}+b^{2}+c^{2} & 2b^{2} & -a^{2}-b^{2}+c^{2} \end{array} \right| \\ \\ \;[\text{Applying } C_1 \to C_1+C_2+C_3] \\ \\ = abc(a^{2}+b^{2}+c^{2}) \left| \begin{array}{ccc} 1 & 2b^{2} & 2c^{2} \\ 1 & -c^{2}-a^{2}+b^{2} & 2c^{2} \\ 1 & 2b^{2} & -a^{2}-b^{2}+c^{2} \end{array} \right| \\ \\ \;[\text{Taking out } a^{2}+b^{2}+c^{2} \text{ from } C_1] \\ \\ = abc(a^{2}+b^{2}+c^{2}) \left| \begin{array}{ccc} 1 & 2b^{2} & 2c^{2} \\ 0 & -a^{2}-b^{2}-c^{2} & 0 \\ 0 & 0 & -a^{2}-b^{2}-c^{2} \end{array} \right| \\ \\ \;[\text{Applying } R_2 \to R_2-R_1,\; R_3 \to R_3-R_1] \\ \\ = abc(a^{2}+b^{2}+c^{2})^{3} \;[\text{ }] \quad \text{Hence proved.}$

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$\displaystyle \textbf{Question 41: } \text{Prove the following identity:}$

$\displaystyle \left| \begin{array}{ccc} 1+a & 1 & 1 \\ 1 & 1+a & a \\ 1 & 1 & 1+a \end{array} \right| = a^{3}+3a^{2}$

$\displaystyle \text{Answer:}$

$\displaystyle \Delta = \left| \begin{array}{ccc} 1+a & 1 & 1 \\ 1 & 1+a & 1 \\ 1 & 1 & 1+a \end{array} \right| \\ \\ = (1+a) \left| \begin{array}{cc} 1+a & 1 \\ 1 & 1+a \end{array} \right| - 1 \left| \begin{array}{cc} 1 & 1 \\ 1 & 1+a \end{array} \right| + 1 \left| \begin{array}{cc} 1 & 1+a \\ 1 & 1 \end{array} \right| \\ \\ \;[\text{Expanding}] \\ \\ = (1+a)\big[(1+a)^2-1\big] -1(1+a-1) +(1-1-a) \\ \\ = (1+a)(1+a^2+2a-1)-a-a \\ \\ = 1+a+a^2+a^3+2a+2a^2-2a \\ \\ = a^3+3a^2$

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$\displaystyle \textbf{Question 42: } \text{Prove the following identity:}$

$\displaystyle \left| \begin{array}{ccc} 2y & y-z-x & 2y \\ 2z & 2z & z-x-y \\ x-y-z & 2x & 2x \end{array} \right| = (x+y+z)^{3} \hspace{2.0cm} \text{[CBSE 20014}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{LHS} \\[6pt] = \left| \begin{array}{ccc} 2y & y-z-x & 2y \\ 2z & 2z & z-x-y \\ x-y-z & 2x & 2x \end{array} \right| \\[10pt] = \left| \begin{array}{ccc} 2y+2z+x-y-z & y-z-x+2z+2x & 2y+z-x-y+2x \\ 2z & 2z & z-x-y \\ x-y-z & 2x & 2x \end{array} \right| \;[\text{Applying } R_1 \to R_1+R_2+R_3] \\[10pt] = \left| \begin{array}{ccc} x+y+z & x+y+z & x+y+z \\ 2z & 2z & z-x-y \\ x-y-z & 2x & 2x \end{array} \right| \\[10pt] = (x+y+z) \left| \begin{array}{ccc} 1 & 1 & 1 \\ 2z & 2z & z-x-y \\ x-y-z & 2x & 2x \end{array} \right| \;[\text{Taking } (x+y+z) \text{ common from } R_1] \\[10pt] = (x+y+z) \left| \begin{array}{ccc} 0 & 1 & 1 \\ 0 & 2z & z-x-y \\ -x-y-z & 2x & 2x \end{array} \right| \;[\text{Applying } C_1 \to C_1-C_2] \\[10pt] = (x+y+z)^2 \left| \begin{array}{ccc} 0 & 1 & 1 \\ 0 & 2z & z-x-y \\ -1 & 2x & 2x \end{array} \right| \;[\text{Taking } (x+y+z) \text{ common from } C_1] \\[10pt] = (x+y+z)^2[-1(z-x-y-2z)] \;[\text{Expanding along first column}] \\[10pt] = (x+y+z)^3 \\[8pt] = \text{RHS} \\[10pt] \therefore \left| \begin{array}{ccc} 2y & y-z-x & 2y \\ 2z & 2z & z-x-y \\ x-y-z & 2x & 2x \end{array} \right| = (x+y+z)^3$

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$\displaystyle \textbf{Question 43: } \text{Prove the following identity:}$

$\displaystyle \left| \begin{array}{ccc} y+z & x & y \\ z+x & z & x \\ x+y & y & z \end{array} \right| = (x+y+z)(x-z)^2 \hspace{2.0cm} \text{[CBSE 2007]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } \Delta = \left| \begin{array}{ccc} y+z & x & y \\ z+x & z & x \\ x+y & y & z \end{array} \right| \\[12pt] = \left| \begin{array}{ccc} y+z-x-y & x & y \\ z+x-z-x & z & x \\ x+y-y-z & y & z \end{array} \right| \;[\text{Applying } C_1 \to C_1-C_2-C_3] \\[12pt] = \left| \begin{array}{ccc} z-x & x & y \\ 0 & z & x \\ x-z & y & z \end{array} \right| \\[12pt] = (x-z) \left| \begin{array}{ccc} -1 & x & y \\ 0 & z & x \\ 1 & y & z \end{array} \right| \\[12pt] = (x-z)\Bigl\{-1(z^2-xy)-x(0\cdot z-x\cdot1)+y(0\cdot y-z\cdot1)\Bigr\} \\[10pt] = (x-z)\{xy-z^2+x^2-yz\} \\[10pt] = (x-z)\{(x^2-z^2)+y(x-z)\} \\[10pt] = (x-z)\{(x-z)(x+z)+y(x-z)\} \\[10pt] = (x-z)^2(x+y+z) \\[12pt] \therefore\; \left| \begin{array}{ccc} y+z & x & y \\ z+x & z & x \\ x+y & y & z \end{array} \right| = (x+y+z)(x-z)^2.$

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$\displaystyle \textbf{Question 44: } \text{Prove the following identity:}$

$\displaystyle \left| \begin{array}{ccc} a+x & y & z \\ x & a+y & z \\ x & y & a+z \end{array} \right| = a^{2}(a+x+y+z) \hspace{2.0cm} \text{[CBSE 2014]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{LHS} \\[6pt] = \left| \begin{array}{ccc} a+x & y & z \\ x & a+y & z \\ x & y & a+z \end{array} \right| \\[10pt] = \left| \begin{array}{ccc} a+x+y+z & y & z \\ a+x+y+z & a+y & z \\ a+x+y+z & y & a+z \end{array} \right| \;[\text{Applying } C_1 \text{ to } C_1+C_2+C_3] \\[10pt] = (a+x+y+z) \left| \begin{array}{ccc} 1 & y & z \\ 1 & a+y & z \\ 1 & y & a+z \end{array} \right| \;[\text{Taking } (a+x+y+z) \text{ common from } C_1] \\[10pt] = (a+x+y+z) \left| \begin{array}{ccc} 1 & y & z \\ 0 & a & 0 \\ 0 & 0 & a \end{array} \right| \;[\text{Applying } R_2 \text{ to } R_2-R_1 \text{ and } R_3 \text{ to } R_3-R_1] \\[10pt] = (a+x+y+z)a^{2} \;[\text{Expanding along first column}] \\[10pt] = a^{2}(a+x+y+z) \\[10pt] = \text{RHS} \\[10pt] \therefore \left| \begin{array}{ccc} a+x & y & z \\ x & a+y & z \\ x & y & a+z \end{array} \right| = a^{2}(a+x+y+z)$

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$\displaystyle \textbf{Question 45: } \text{Prove the following identity:}$

$\displaystyle \left| \begin{array}{ccc} a^{3} & 2 & a \\ b^{3} & 2 & b \\ c^{3} & 2 & c \end{array} \right| = 2(a-b)(b-c)(c-a)(a+b+c) \hspace{2.0cm} \text{[CBSE 2015]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{LHS} \\[6pt] = \left| \begin{array}{ccc} a^{3} & 2 & a \\ b^{3} & 2 & b \\ c^{3} & 2 & c \end{array} \right| \\[10pt] = \left| \begin{array}{ccc} a^{3} & 2 & a \\ b^{3}-a^{3} & 0 & b-a \\ c^{3}-a^{3} & 0 & c-a \end{array} \right| \;[\text{Applying } R_2 \to R_2-R_1 \text{ and } R_3 \to R_3-R_1] \\[12pt] = -(a-b)(c-a) \left| \begin{array}{ccc} a^{3} & 2 & a \\ b^{2}+a^{2}+ab & 0 & 1 \\ c^{2}+a^{2}+ac & 0 & 1 \end{array} \right| \;[\text{Taking } (b-a) \text{ common from } R_2 \text{ and } (c-a) \text{ common from } R_3] \\[12pt] = -(a-b)(c-a) \left| \begin{array}{ccc} a^{3} & 2 & a \\ b^{2}-c^{2}+ab-ac & 0 & 0 \\ c^{2}+a^{2}+ac & 0 & 1 \end{array} \right| \;[\text{Applying } R_2 \to R_2-R_3] \\[12pt] = -(a-b)(c-a) \left| \begin{array}{ccc} a^{3} & 2 & a \\ (b-c)(a+b+c) & 0 & 0 \\ c^{2}+a^{2}+ac & 0 & 1 \end{array} \right| \\[10pt] = -(a-b)(c-a)(b-c)(a+b+c) \left| \begin{array}{ccc} a^{3} & 2 & a \\ 1 & 0 & 0 \\ c^{2}+a^{2}+ac & 0 & 1 \end{array} \right| \;[\text{Taking } (b-c)(a+b+c) \text{ common from } R_2] \\[12pt] = -(a-b)(c-a)(b-c)(a+b+c)(-2) \;[\text{Expanding along second column}] \\[10pt] = 2(a-b)(c-a)(b-c)(a+b+c) \\[10pt] = \text{RHS} \\[10pt] \therefore \left| \begin{array}{ccc} a^{3} & 2 & a \\ b^{3} & 2 & b \\ c^{3} & 2 & c \end{array} \right| = 2(a-b)(b-c)(c-a)(a+b+c)$

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$\displaystyle \textbf{Question 46: }$

$\displaystyle \text{Without expanding, prove that } \\[6pt] \left| \begin{array}{ccc} a & b & c \\ x & y & z \\ p & q & r \end{array} \right| = \left| \begin{array}{ccc} x & y & z \\ p & q & r \\ a & b & c \end{array} \right| = \left| \begin{array}{ccc} y & b & q \\ x & a & p \\ z & c & r \end{array} \right|$

$\displaystyle \text{Answer:}$

$\displaystyle \left| \begin{array}{ccc} x & y & z \\ p & q & r \\ a & b & c \end{array} \right| \;x \leftrightarrow{R_2 \leftrightarrow R_3}\; - \left| \begin{array}{ccc} x & y & z \\ a & b & c \\ p & q & r \end{array} \right| \; x \leftrightarrow{R_1 \leftrightarrow R_2}\; \left| \begin{array}{ccc} a & b & c \\ x & y & z \\ p & q & r \end{array} \right|$

$\displaystyle \left| \begin{array}{ccc} y & b & q \\ x & a & p \\ z & c & r \end{array} \right| = \left| \begin{array}{ccc} y & x & z \\ b & a & c \\ q & p & r \end{array} \right| \;x \leftrightarrow{C_1 \leftrightarrow C_2}\; - \left| \begin{array}{ccc} x & y & z \\ a & b & c \\ p & q & r \end{array} \right| \; x \leftrightarrow{R_1 \leftrightarrow R_2}\; \left| \begin{array}{ccc} a & b & c \\ x & y & z \\ p & q & r \end{array} \right|$

$\displaystyle \text{Hence proved.}$

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$\displaystyle \textbf{Question 47: }$

$\displaystyle \text{Show that } \\[6pt] \left| \begin{array}{ccc} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{array} \right| =0 \quad \text{where } a,b,c \text{ are in A.P.} \hspace{2.0cm} \text{[CBSE 2005]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given: } a,b,c \text{ are in A.P.} \\[6pt] 2b=a+c \\[10pt] \Delta = \left| \begin{array}{ccc} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{array} \right| \;[\text{Applying } R_2 = 2R_2] \\[10pt] \Delta = \frac{1}{2} \left| \begin{array}{ccc} x+1 & x+2 & x+a \\ 2x+4 & 2x+6 & 2x+2b \\ x+3 & x+4 & x+c \end{array} \right| \\[10pt] \Delta = \frac{1}{2} \left| \begin{array}{ccc} x+1 & x+2 & x+a \\ 0 & 0 & 0 \\ x+3 & x+4 & x+c \end{array} \right| \;[:\,2b=a+c]\; [\text{Applying } R_2 \to R_2-(R_1+R_3)] \\[10pt] \Delta = 0$

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$\displaystyle \textbf{Question 48: }$

$\displaystyle \text{Show that } \\[6pt] \left| \begin{array}{ccc} x-3 & x-4 & x-\alpha \\ x-2 & x-3 & x-\beta \\ x-1 & x-2 & x-\gamma \end{array} \right| =0, \quad \text{where } \alpha,\beta,\gamma \text{ are in A.P.} \hspace{2.0cm} \text{[CBSE 2007]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given: } \alpha,\beta,\gamma \text{ are in A.P.} \\[6pt] \text{Now,} \\[6pt] 2\beta=\alpha+\gamma \\[12pt] \Delta = \left| \begin{array}{ccc} x-3 & x-4 & x-\alpha \\ x-2 & x-3 & x-\beta \\ x-1 & x-2 & x-\gamma \end{array} \right| \\[12pt] \Delta = \frac{1}{2} \left| \begin{array}{ccc} x-3 & x-4 & x-\alpha \\ 2x-4 & 2x-6 & 2x-2\beta \\ x-1 & x-2 & x-\gamma \end{array} \right| \;[\text{Applying } R_2 \to 2R_2] \\[12pt] \Delta = \frac{1}{2} \left| \begin{array}{ccc} x-3 & x-4 & x-\alpha \\ 0 & 0 & -2\beta+\alpha+\gamma \\ x-1 & x-2 & x-\gamma \end{array} \right| \;[:\,2\beta=\alpha+\gamma]\; [\text{Applying } R_2 \to R_2-(R_1+R_3)] \\[12pt] \Delta = \frac{1}{2} \left| \begin{array}{ccc} x-3 & x-4 & x-\alpha \\ 0 & 0 & 0 \\ x-1 & x-2 & x-\gamma \end{array} \right| \\[12pt] \Delta = 0$

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$\displaystyle \textbf{Question 49: }$

$\displaystyle \text{If } a,b,c \text{ are real numbers such that} \\[6pt] \left| \begin{array}{ccc} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array} \right| =0, \ \text{then show that either} \\[8pt] a+b+c=0 \ \text{or},\ a=b=c.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } \Delta = \left| \begin{array}{ccc} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array} \right| \\[10pt] = \left| \begin{array}{ccc} 2(a+b+c) & 2(a+b+c) & 2(a+b+c) \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array} \right| \;[\text{Applying } R_1 \to R_1+R_2+R_3] \\[10pt] = 2(a+b+c) \left| \begin{array}{ccc} 1 & 1 & 1 \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array} \right| \\[10pt] = 2(a+b+c) \left| \begin{array}{ccc} 1 & 0 & 0 \\ c+a & b-a & b-c \\ a+b & c-a & c-b \end{array} \right| \;[\text{Applying } C_2 \to C_2-C_1 \text{ and } C_3 \to C_3-C_1] \\[10pt] = 2(a+b+c) \left\{ \begin{array}{cc} b-c & b-a \\ c-a & c-b \end{array} \right\} \\[10pt] = 2(a+b+c)\{(b-c)(c-b)-(b-a)(c-a)\} \\[10pt] = -2(a+b+c)\{a^{2}+b^{2}+c^{2}-ab-bc-ca\} \\[10pt] = -(a+b+c)\{2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca\} \\[10pt] = -(a+b+c)\{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\} \\[12pt] \text{But } \Delta=0 \;[\text{Given}] \\[8pt] \Rightarrow -(a+b+c)\{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\}=0 \\[8pt] \Rightarrow \text{Either } (a+b+c)=0 \text{ or } (a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0 \\[8pt] \Rightarrow (a+b+c)=0 \text{ or } a=b=c \\[10pt] \text{Hence proved.}$

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$\displaystyle \textbf{Question 50: }$

$\displaystyle \text{If } \left| \begin{array}{ccc} p & b & c \\ a & q & c \\ a & b & r \end{array} \right| =0, \ \text{find the value of } \frac{p}{p-a} + \frac{q}{q-b} + \frac{r}{r-c}, \quad p\ne a,\ q\ne b,\ r\ne c. \hspace{2.0cm} \text{[CBSE 2014]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } \\[6pt] \Delta = \left| \begin{array}{ccc} p & b & c \\ a & q & c \\ a & b & r \end{array} \right| \\[12pt] \text{Now,} \\[6pt] \Delta = \left| \begin{array}{ccc} p & b & c \\ a & q & c \\ a & b & r \end{array} \right| \\[12pt] = \left| \begin{array}{ccc} p & b & c \\ 0 & q-b & c-r \\ a & b & r \end{array} \right| \;[\text{Applying } R_2 \to R_2-R_3] \\[12pt] = p\{r(q-b)-b(c-r)\} + a\{b(c-r)-c(q-b)\} \;[\text{Expanding along first column}] \\[12pt] = pr(q-b)+pb(r-c)-ab(r-c)-ac(q-b) \\[12pt] = (pr-ac)(q-b)+b(p-a)(r-c)$

$\displaystyle \text{Since, } \Delta = 0. \\[10pt] \therefore\; (pr-ac)(q-b)+b(p-a)(r-c)=0 \\[12pt] \Rightarrow \frac{pr-ac}{(p-a)(r-c)}+\frac{b}{q-b}=0 \\[12pt] \Rightarrow \frac{pr-ar+ar-ac}{(p-a)(r-c)}+\frac{b}{q-b}=0 \\[12pt] \Rightarrow \frac{r(p-a)+a(r-c)}{(p-a)(r-c)}+\frac{b}{q-b}=0 \\[12pt] \Rightarrow \frac{r}{r-c}+\frac{a}{p-a}+\frac{b}{q-b}=0 \\[12pt] \Rightarrow \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c} = \frac{p}{p-a}+\frac{q}{q-b} -\frac{a}{p-a}-\frac{b}{q-b}$

$\displaystyle \Rightarrow \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c} = \frac{p-a}{p-a}+\frac{q-b}{q-b} \\[12pt] \Rightarrow \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=2 \\[16pt] \text{Hence, the value of } \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c} \text{ is } 2.$

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$\displaystyle \textbf{Question 51: }$

$\displaystyle \text{Show that } x=2 \text{ is a root of the equation} \\[6pt] \left| \begin{array}{ccc} x & -6 & -1 \\ 2-3x & x & x-3 \\ -3 & 2x & x+2 \end{array} \right| =0 \ \text{and solve it completely.}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } \Delta = \left| \begin{array}{ccc} x & -6 & -1 \\ 2 & -3x & x-3 \\ -3 & 2x & x+2 \end{array} \right| \\[12pt] = \left| \begin{array}{ccc} x & -6 & -1 \\ 2 & -3x & x-3 \\ -3-x & 2x+6 & x+3 \end{array} \right| \;[\text{Applying } R_3 \to R_3-R_1] \\[12pt] = (x+3) \left| \begin{array}{ccc} x & -6 & -1 \\ 2 & -3x & x-3 \\ -1 & 2 & 1 \end{array} \right| \\[12pt] = (x+3) \left| \begin{array}{ccc} x-2 & 3x-6 & -x+2 \\ 2 & -3x & x-3 \\ -1 & 2 & 1 \end{array} \right| \;[\text{Applying } R_1 \to R_1-R_2] \\[12pt] = (x+3)(x-2) \left| \begin{array}{ccc} 1 & 3 & -1 \\ 2 & -3x & x-3 \\ -1 & 2 & 1 \end{array} \right| \\[12pt] = (x+3)(x-2) \left| \begin{array}{ccc} 1 & 3 & 0 \\ 2 & -3x & x-1 \\ -1 & 2 & 0 \end{array} \right| \;[\text{Applying } C_3 \to C_3+C_1] \\[12pt] = (x+3)(x-2)(x-1) \left| \begin{array}{ccc} 1 & 3 & 0 \\ 2 & -3x & 1 \\ -1 & 2 & 0 \end{array} \right| \\[12pt] = (x+3)(x-2)(x-1) \left\{ -1 \left| \begin{array}{cc} 1 & 3 \\ -1 & 2 \end{array} \right| \right\} \;[\text{Expanding along } C_3] \\[12pt] = -5(x+3)(x-2)(x-1) \\[14pt] \Rightarrow x=2,\,-3,\,1$

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$\displaystyle \textbf{Question 52: } \text{Solve the following Determinnt equations: }$

$\displaystyle \textbf{(i) } \left| \begin{array}{ccc} x+a & b & c \\ a & x+b & c \\ a & b & x+c \end{array} \right| =0 \hspace{2.0cm} \text{[CBSE 2006]}$ $\displaystyle \textbf{(ii) } \left| \begin{array}{ccc} x+a & x & x \\ x & x+a & x \\ x & x & x+a \end{array} \right| =0,\ a\ne 0 \hspace{2.0cm} \text{[CBSE 2011]}$

$\displaystyle \textbf{(iii) } \left| \begin{array}{ccc} 3x-8 & 3 & 3 \\ 3 & 3x-8 & 3 \\ 3 & 3 & 3x-8 \end{array} \right| =0 \hspace{2.0cm} \text{[CBSE 2008]}$ $\displaystyle \textbf{(iv) } \left| \begin{array}{ccc} 1 & x & x^{2} \\ 1 & a & a^{2} \\ 1 & b & b^{2} \end{array} \right| =0, \quad a \ne b$

$\displaystyle \textbf{(v) } \left| \begin{array}{ccc} x+1 & 3 & 5 \\ 2 & x+2 & 5 \\ 2 & 3 & x+4 \end{array} \right| =0$ $\displaystyle \textbf{(vi) } \left| \begin{array}{ccc} 1 & x & x^{3} \\ 1 & b & b^{3} \\ 1 & c & c^{3} \end{array} \right| =0, \quad b \ne c$

$\displaystyle \textbf{(vii) } \left| \begin{array}{ccc} 15-2x & 11-3x & 7-x \\ 11 & 17 & 14 \\ 10 & 16 & 13 \end{array} \right| =0$

$\displaystyle \textbf{(viii) } \left| \begin{array}{ccc} 1 & 1 & x \\ p+1 & p+1 & p+x \\ 3 & x+1 & x+2 \end{array} \right| =0$ $\displaystyle \textbf{(ix) } \left| \begin{array}{ccc} 3 & -2 & \sin(3\theta) \\ -7 & 8 & \cos(2\theta) \\ -11 & 14 & 2 \end{array} \right| =0$

$\displaystyle \text{Answer:}$

$\displaystyle \textbf{(i) }$

$\displaystyle \text{Let } \Delta = \left| \begin{array}{ccc} x+a & b & c \\ a & x+b & c \\ a & b & x+c \end{array} \right| \\[12pt] = \left| \begin{array}{ccc} x+a+b+c & b & c \\ x+a+b+c & x+b & c \\ x+a+b+c & b & x+c \end{array} \right| \;[\text{Applying } C_1 \to C_1+C_2+C_3] \\[12pt] = (x+a+b+c) \left| \begin{array}{ccc} 1 & b & c \\ 1 & x+b & c \\ 1 & b & x+c \end{array} \right| \\[12pt] = (x+a+b+c) \left| \begin{array}{ccc} 1 & b & c \\ 0 & x & 0 \\ 1 & b & x+c \end{array} \right| \;[\text{Applying } R_2 \to R_2-R_1] \\[12pt] = (x+a+b+c) \left| \begin{array}{ccc} 1 & b & c \\ 0 & x & 0 \\ 0 & 0 & x \end{array} \right| \;[\text{Applying } R_3 \to R_3-R_1] \\[12pt] \Delta = (x+a+b+c)(x^{2}-0)=0 \;[\text{Given}] \\[12pt] \Rightarrow x^{2}=0 \text{ or } x+a+b+c=0 \\[12pt] \Rightarrow x=0 \text{ or } x=-(a+b+c)$

$\displaystyle \textbf{(ii) }$

$\displaystyle \text{Let } \Delta = \left| \begin{array}{ccc} x+a & x & x \\ x & x+a & x \\ x & x & x+a \end{array} \right| \\[12pt] = \left| \begin{array}{ccc} 3x+a & x & x \\ 3x+a & x+a & x \\ 3x+a & x & x+a \end{array} \right| \;[\text{Applying } C_1 \to C_1+C_2+C_3] \\[12pt] = (3x+a) \left| \begin{array}{ccc} 1 & x & x \\ 1 & x+a & x \\ 1 & x & x+a \end{array} \right| \\[12pt] = (3x+a) \left| \begin{array}{ccc} 1 & x & x \\ 0 & a & 0 \\ 1 & x & x+a \end{array} \right| \;[\text{Applying } R_2 \to R_2-R_1] \\[12pt] = (3x+a) \left| \begin{array}{ccc} 1 & x & x \\ 0 & a & 0 \\ 0 & 0 & a \end{array} \right| \;[\text{Applying } R_3 \to R_3-R_1] \\[12pt] \Delta = (3x+a)(a^{2}-0)=0 \\[12pt] \Rightarrow 3x+a=0 \\[8pt] \Rightarrow x=-\frac{a}{3}$

$\displaystyle \textbf{(iii) }$

$\displaystyle \text{Let } \Delta = \left| \begin{array}{ccc} 3x-8 & 3 & 3 \\ 3 & 3x-8 & 3 \\ 3 & 3 & 3x-8 \end{array} \right| \\[12pt] = \left| \begin{array}{ccc} 3x-2 & 3 & 3 \\ 3x-2 & 3x-8 & 3 \\ 3x-2 & 3 & 3x-8 \end{array} \right| \;[\text{Applying } C_1=C_1+C_2+C_3] \\[12pt] = (3x-2) \left| \begin{array}{ccc} 1 & 3 & 3 \\ 1 & 3x-8 & 3 \\ 1 & 3 & 3x-8 \end{array} \right| \\[12pt] = (3x-2) \left| \begin{array}{ccc} 1 & 3 & 3 \\ 0 & 3x-11 & 0 \\ 1 & 3 & 3x-8 \end{array} \right| \;[\text{Applying } R_2 \to R_2-R_1] \\[12pt] = (3x-2) \left| \begin{array}{ccc} 1 & 3 & 3 \\ 0 & 3x-11 & 0 \\ 0 & 0 & 3x-11 \end{array} \right| \;[\text{Applying } R_3 \to R_3-R_1] \\[12pt] \Delta = (3x-2)(3x-11)^2=0 \\[12pt] \Rightarrow x=\frac{2}{3},\;\frac{11}{3},\;\frac{11}{3}$

$\displaystyle \textbf{(iv) }$

$\displaystyle \text{Let } \Delta = \left| \begin{array}{ccc} 1 & x & x^{2} \\ 1 & a & a^{2} \\ 1 & b & b^{2} \end{array} \right| \\[12pt] = \left| \begin{array}{ccc} 1 & x & x^{2} \\ 0 & x-a & x^{2}-a^{2} \\ 1 & b & b^{2} \end{array} \right| \;[\text{Applying } R_2 \to R_1-R_2] \\[12pt] = \left| \begin{array}{ccc} 1 & x & x^{2} \\ 0 & x-a & x^{2}-a^{2} \\ 0 & x-b & x^{2}-b^{2} \end{array} \right| \;[\text{Applying } R_3 \to R_1-R_3] \\[12pt] = (x-a)(x-b) \left| \begin{array}{ccc} 1 & x & x^{2} \\ 0 & 1 & x+a \\ 0 & 1 & x+b \end{array} \right| \\[12pt] \Delta = (x-a)(x-b)(x+b-x-a)=0 \\[12pt] \Rightarrow x=a,\;b$

$\displaystyle \textbf{(v) }$

$\displaystyle \text{Let } \Delta = \left| \begin{array}{ccc} x+1 & 3 & 5 \\ 2 & x+2 & 5 \\ 2 & 3 & x+4 \end{array} \right| \\[12pt] = \left| \begin{array}{ccc} x+9 & 3 & 5 \\ x+9 & x+2 & 5 \\ x+9 & 3 & x+4 \end{array} \right| \;[\text{Applying } C_1 = C_1+C_2+C_3] \\[12pt] = (x+9) \left| \begin{array}{ccc} 1 & 3 & 5 \\ 1 & x+2 & 5 \\ 1 & 3 & x+4 \end{array} \right| \\[12pt] = (x+9) \left| \begin{array}{ccc} 1 & 3 & 5 \\ 0 & x-1 & 0 \\ 1 & 3 & x+4 \end{array} \right| \;[\text{Applying } R_2 \to R_2-R_1] \\[12pt] = (x+9) \left| \begin{array}{ccc} 1 & 3 & 5 \\ 0 & x-1 & 0 \\ 0 & 0 & x-1 \end{array} \right| \;[\text{Applying } R_3 \to R_3-R_1] \\[12pt] \Delta = (x+9)(x-1)^2=0 \\[12pt] \Rightarrow x=-9,\;1,\;1$

$\displaystyle \textbf{(vi) }$

$\displaystyle \text{Let } \Delta = \left| \begin{array}{ccc} 1 & x & x^{3} \\ 1 & b & b^{3} \\ 1 & c & c^{3} \end{array} \right| \\[12pt] = \left| \begin{array}{ccc} 1 & x & x^{3} \\ 0 & b-x & b^{3}-x^{3} \\ 1 & c & c^{3} \end{array} \right| \;[\text{Applying } R_2 \to R_2-R_1] \\[12pt] = \left| \begin{array}{ccc} 1 & x & x^{3} \\ 0 & b-x & b^{3}-x^{3} \\ 0 & c-x & c^{3}-x^{3} \end{array} \right| \;[\text{Applying } R_3 \to R_3-R_1] \\[12pt] = \left| \begin{array}{ccc} 1 & x & x^{3} \\ 0 & x-b & x^{3}-b^{3} \\ 0 & x-c & x^{3}-c^{3} \end{array} \right| \\[12pt] = (x-b)(x-c) \left| \begin{array}{ccc} 1 & x & x^{2} \\ 0 & 1 & x^{2}+xb+b^{2} \\ 0 & 1 & x^{2}+xc+c^{2} \end{array} \right| \\[12pt] \Delta = (x-b)(x-c)\bigl[x(c-b)-b^{2}+c^{2}\bigr]=0 \\[12pt] \Rightarrow x=b,\;c,\;-(b+c)$

$\displaystyle \textbf{(vii) }$

$\displaystyle \text{Let } \Delta = \left| \begin{array}{ccc} 15-2x & 11-3x & 7-x \\ 11 & 17 & 14 \\ 10 & 16 & 13 \end{array} \right| =0 \\[12pt] \Rightarrow \left| \begin{array}{ccc} 15-2x-14+2x & 11-3x & 7-x \\ 11-28 & 17 & 14 \\ 10-26 & 16 & 13 \end{array} \right| =0 \;[\text{Applying } C_1 \to C_1-2C_3] \\[12pt] \Rightarrow \left| \begin{array}{ccc} 1 & 11-3x & 7-x \\ -17 & 17 & 14 \\ -16 & 16 & 13 \end{array} \right| =0 \\[12pt] \Rightarrow \left| \begin{array}{ccc} 12-3x & 4-2x & 7-x \\ 0 & 3 & 14 \\ 0 & 3 & 13 \end{array} \right| =0 \;[\text{Applying } C_1 \to C_1+C_2 \text{ and } C_2 \to C_2-C_3] \\[12pt] \Rightarrow (12-3x)\bigl(3\times13-3\times14\bigr)=0 \\[12pt] \Rightarrow (12-3x)(-3)=0 \\[12pt] \Rightarrow 12-3x=0 \\[10pt] \Rightarrow 3x=12 \\[10pt] \Rightarrow x=4$

$\\$

$\displaystyle \textbf{(viii) }$

$\displaystyle \text{Let } \Delta = \left| \begin{array}{ccc} 3 & -2 & \sin(3\theta) \\ -7 & 8 & \cos(2\theta) \\ -11 & 14 & 2 \end{array} \right| =0 \\[12pt] \Rightarrow \left| \begin{array}{ccc} 1 & -2 & \sin(3\theta) \\ 1 & 8 & \cos(2\theta) \\ 3 & 14 & 2 \end{array} \right| =0 \;[\text{Applying } C_1 \to C_1+C_2] \\[12pt] \Rightarrow \left| \begin{array}{ccc} 1 & -2 & \sin(3\theta) \\ 0 & 10 & \cos(2\theta)-\sin(3\theta) \\ 0 & 20 & 2-3\sin(3\theta) \end{array} \right| =0 \;[\text{Applying } R_2 \to R_2-R_1 \text{ and } R_3 \to R_3-3R_1] \\[12pt] \Rightarrow 10(2-3\sin(3\theta))-20\cos(2\theta)-\sin(3\theta)=0 \\[12pt] \Rightarrow 20-10\sin(3\theta)-20\cos(2\theta)=0 \\[12pt] \Rightarrow \sin(3\theta)+2\cos(2\theta)-2=0 \\[12pt] \Rightarrow 3\sin\theta-4\sin^{3}\theta+2-4\sin^{2}\theta-2=0 \\[12pt] \Rightarrow -\sin\theta(4\sin^{2}\theta+4\sin\theta-3)=0 \\[12pt] \Rightarrow \sin\theta=0 \text{ or } 4\sin^{2}\theta+4\sin\theta-3=0 \\[12pt] \Rightarrow \theta=n\pi \text{ or } (2\sin\theta+3)(2\sin\theta-1)=0 \\[12pt] \Rightarrow \theta=n\pi \text{ or } \sin\theta=-\frac{3}{2} \text{ or } \sin\theta=\frac{1}{2} \\[12pt] \Rightarrow \theta=n\pi \text{ or } \theta=n\pi+(-1)^{n}\frac{\pi}{6}, \quad n\in\mathbb{Z}$

$\displaystyle \textbf{(ix) }$

$\displaystyle \text{Let } \Delta = \left| \begin{array}{ccc} 1 & 1 & x \\ p+1 & p+1 & p+x \\ 3 & x+1 & x+2 \end{array} \right| \\[12pt] = \left| \begin{array}{ccc} 1 & 1 & x \\ p & p & p \\ 3 & x+1 & x+2 \end{array} \right| \;[\text{Applying } R_2 \to R_2-R_1] \\[12pt] = p \left| \begin{array}{ccc} 1 & 1 & x \\ 1 & 1 & 1 \\ 3 & x+1 & x+2 \end{array} \right| \\[12pt] = p \left| \begin{array}{ccc} 1 & 1 & x \\ 1 & 1 & 1 \\ 2 & x & 2 \end{array} \right| \;[\text{Applying } R_3 \to R_3-R_1] \\[12pt] = p \left| \begin{array}{ccc} 0 & 1 & x \\ 0 & 1 & 1 \\ 2-x & x & 2 \end{array} \right| \;[\text{Applying } C_1 \to C_1-C_2] \\[12pt] = p \left\{ (2-x) \left| \begin{array}{cc} 1 & x \\ 1 & 1 \end{array} \right| \right\} \;[\text{Expanding along } C_1] \\[12pt] = p(2-x)(1-x)=0 \\[12pt] \Rightarrow x=1,\;2$

$\displaystyle \textbf{Question 53: }$

$\displaystyle \text{If } a,b \text{ and } c \text{ are all non-zero and} \left| \begin{array}{ccc} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{array} \right| =0, \ \text{then prove that} \\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1=0. \hspace{2.0cm} \text{[CBSE 2016]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{We have,} \\[6pt] \left| \begin{array}{ccc} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{array} \right| =0 \\[12pt] C_1 \to C_1-C_2 \\[8pt] \left| \begin{array}{ccc} a & 1 & 1 \\ -b & 1+b & 1 \\ 0 & 1 & 1+c \end{array} \right| =0 \\[12pt] C_2 \to C_2-C_3 \\[8pt] \left| \begin{array}{ccc} a & 0 & 1 \\ -b & b & 1 \\ 0 & -c & 1+c \end{array} \right| =0 \\[14pt] \text{Expanding along } R_1, \text{ we get} \\[8pt] a(b+bc+c)+1(bc)=0 \\[12pt] \Rightarrow ab+abc+ac+bc=0 \\[12pt] \text{Dividing by } abc, \text{ we get} \\[10pt] \frac{1}{c}+1+\frac{1}{b}+\frac{1}{a}=0 \\[12pt] \therefore\; \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1=0$

$\\$

$\displaystyle \textbf{Question 54: }$

$\displaystyle \text{If } \left| \begin{array}{ccc} a & b-y & c-z \\ a-x & b & c-z \\ a-x & b-y & c \end{array} \right| =0, \ \text{then using properties of determinants, find the value of } \\ \frac{a}{x}+\frac{b}{y}+\frac{c}{z}, \ \text{where } x,y,z\ne 0. \hspace{2.0cm} \text{[CBSE 2017]}$

$\displaystyle \text{Answer:}$

$\displaystyle \left| \begin{array}{ccc} a & b-y & c-z \\ a-x & b & c-z \\ a-x & b-y & c \end{array} \right| =0 \\[12pt] R_1 \to R_1-R_2 \\[8pt] \left| \begin{array}{ccc} x & -y & 0 \\ a-x & b & c-z \\ a-x & b-y & c \end{array} \right| =0 \\[12pt] R_2 \to R_2-R_3 \\[8pt] \Rightarrow \left| \begin{array}{ccc} x & -y & 0 \\ 0 & y & -z \\ a-x & b-y & c \end{array} \right| =0 \\[14pt] \text{Expanding along first row, we get} \\[8pt] x(yc+zb-zy)+y(0-za+zx)=0 \\[12pt] \Rightarrow xyc+xzb-xyz+zya-xyz=0 \\[12pt] \Rightarrow xyc+xzb-2xyz+zya=0 \\[12pt] \text{Dividing by } xyz, \text{ we get} \\[10pt] \frac{c}{z}+\frac{b}{y}-2+\frac{a}{x}=0 \\[12pt] \therefore\; \frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2$

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Class 12: Determinants – Exercise 6.2

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