81. Determinants Solutions Part 3

101. R.H.S. \(= \frac{1}{abc}\begin{vmatrix} a & abc & a(b + c) \\ b & abc & b(c + a) \\ c & abc & c(a + b) \end{vmatrix}[R_1\rightarrow aR_1; R_2\rightarrow bR_2; R_3\rightarrow cR_3]\)

     \(= -\frac{abc}{abc}\begin{vmatrix}1 & a & ab + ac \\ 1 & b & bc + ba \\ 1 & c & ca + cb\end{vmatrix}\) Taking \(abc\) out and then applying \(C_1\leftrightarrow C_2\)

     \(= -\begin{vmatrix} 1 & a & -bc \\ 1 & b & - ca \\ 1 & c & -ab \end{vmatrix}[C_3\rightarrow C_3 - (ab + bc + ca)C_1]\)

     \(= \begin{vmatrix}1 & a & bc \\ 1 & b & ca \\ 1 & c & ab\end{vmatrix} = \frac{1}{abc}\begin{vmatrix}a & a^2 & abc \\ b & b^2 & abc \\ c & c^2 & abc\end{vmatrix}[R_1\rightarrow aR_1; R_2\rightarrow bR_2; R_3\rightarrow cR_3]\)

     \(= \frac{abc}{abc}\begin{vmatrix}a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1\end{vmatrix}\)

     \(= \begin{vmatrix}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{vmatrix}[C_2\leftrightarrow C_3; C_1\leftrightarrow C_2]\)

102. \(\Delta = \begin{vmatrix}x^2 & x + 1 & x - 2 \\ 2x^2 & 3x & 3x - 3 \\ x^2 & 2x - 1 & 2x - 1\end{vmatrix} + \begin{vmatrix} x & x + 1 & x - 2 \\ 3x - 1 & 3x & 3x - 3 \\ 2x + 3 & 2x - 1 & 2x - 1\end{vmatrix}\)

     \(= \begin{vmatrix}2x^2 & 3x & 3x - 3 \\ 2x^2 & 3x & 3x - 3 \\ x^2 & 2x - 1 & 2x - 1\end{vmatrix}[R_1\rightarrow R_1 + R_3] + \\\begin{vmatrix}2 & 3 & x - 2 \\ 2 & 3 & 3x - 3 \\ 4 & 0 & 2x -1\end{vmatrix}[C_1\rightarrow C_1-C_3;C_2\rightarrow C_2 - C_3]\)

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

     \(= xA + B\) where \(A = \begin{vmatrix}2 & 3 & 1 \\ 2 & 3 & 3 \\ 4 & 0 & 2\end{vmatrix}\) and \(B = \begin{vmatrix}2 & 3 & -2 \\ 2 & 3 & -3 \\ 4 & 0 & -1\end{vmatrix}\) which are determinants of 3rd order independent of \(x\)

103. \(\sum_{r = 1}^n D_r = D_1 + D_2 + \ldots + D_n\)

     \(= \begin{vmatrix}\sum_{r=1}^nr & x & \frac{n(n + 1)}{2} \\ \sum_{r=1}^n(2r - 1) & y & n^2 \\ \sum_{r=1}^n(3r - 2) & z & \frac{n(3n - 1)}{2}\end{vmatrix}\)

     \(= \begin{vmatrix}\frac{(n(n + 1))}{2} & x & \frac{(n(n + 1))}{2} \\ n^2 & y & n^2 \\ \frac{n(3n - 1)}{2} & z & \frac{n(3n - 1)}{2}\end{vmatrix}\)

     \(= 0\) because first and third columns are identical.

104. \(\Delta = \begin{vmatrix}-5 & 3 + 5i & \frac{3}{2} - 4i \\ 3 - 5i & 8 & 4 + 5i \\ \frac{3}{2} + 4i & 4 - 5i & 9\end{vmatrix}\)

     \(\overline{\Delta} = \begin{vmatrix}-5 & 3 - 5i & \frac{3}{2} + 4i \\ 3 + 5i & 8 & 4 - 5i \\ \frac{3}{2} - 4i & 4 + 5i & 9\end{vmatrix}\)

     Exchanging rows and columns

     \(\overline{\Delta} = \Delta \therefore \Delta\) is purely real.

105. Putting \(b = -c,\) we have

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

     \(= \begin{vmatrix}c - a & a - c & a - c \\ a - c & 2c & 0 \\ c + a & 0 & -2c\end{vmatrix}[R_1 \rightarrow R_1 + R_3]\)

     \(= \begin{vmatrix}c - a & 0 & 0 \\ a - c & a + c & a - c \\ c + a & a + c & a - c\end{vmatrix}[C_2\rightarrow C_2 + C_1; C_3\rightarrow C_3 + C_1]\)

     \(= (c - a)[(a ^2 - c^2) - (a^2 - c^2)] = 0\)

     Hence, \(b + c\) is a factor of \(\Delta.\) Similarly it can be proven that \(a + b\) and \(c + a\) are factors of \(\Delta.\)

     We see that, upon expansion of determinant, each term of the L.H.S. and R.H.S. is a homogeneous expression in \(a,b,c\) of 3rd degree.

     Let \(\begin{vmatrix}-2a & a + b & b + c \\ b + a & -2b & b + c \\ c + a & c + b & -2c\end{vmatrix} = k(b + c)(c + a)(a + b)\) where \(k\) is independent of \(a,b,c\)

     Putting \(a = 0, b = 1, c = 1\) we get

     \(\begin{vmatrix}0 & 1 & 1 \\ 1 & -2 & 2 \\ 1 & 2 & -2 \end{vmatrix} = 2k\)

     \(k = 4\)

     Thus, we have proven the required condition.

106. \(F'(a) = \begin{vmatrix}f_1'(a) & f_2'(a) & f_3'(x) \\ g_1(a) & g_2(x) & g_3(a) \\ h_1(a) & h_2(a) & h_3(a)\end{vmatrix} + \begin{vmatrix}f_1(a) & f_2(a) & f_3(x) \\ g_1'(a) & g_2'(x) & g_3'(a) \\ h_1(a) & h_2(a) & h_3(a)\end{vmatrix} + \begin{vmatrix}f_1(a) & f_2(a) & f_3(x) \\ g_1(a) & g_2(x) & g_3(a) \\ h_1'(a) & h_2'(a) & h_3'(a)\end{vmatrix}\)

     \(= 0 + 0 + 0\)

     \(\because f_r(a)=g_r(a)=h_r(a), r=1,2,3\) in the first determinant last two, in the second determinant first and third, in the third determinant first two, rows are identical. Therefore, all determinants are zero.

107. Since \(f(x) = 0\) is a quadratic equation with repeated root \(\alpha, \therefore f(x) = a_r(x - \alpha)^2\) where \(a\) is a constant.

     Clearly \(\Delta(x)\) is a polynomial of degree having a maximum value of \(5\)

     \(\Delta(\alpha) = \begin{vmatrix}A(\alpha) & B(\alpha) & C(\alpha) \\ A(\alpha) & B(\alpha) & C(\alpha) \\ A'(\alpha) & B'(\alpha) & C'(\alpha)\end{vmatrix}\)

     \(\Delta(\alpha) = 0\) [ \(\because R_1\) and \(R_2\) are identical]

     \(\Delta'(\alpha) = \begin{vmatrix}A'(\alpha) & B'(\alpha) & C'(\alpha) \\ A(\alpha) & B(\alpha) & C(\alpha) \\ A'(\alpha) & B'(\alpha) & C'(\alpha)\end{vmatrix} = 0\) [ \(\because R_1\) and \(R_3\) are identical]

     Thus, we can say that \(\Delta(x) = 0\) has two roots equal to \(\alpha\)

     \(\Rightarrow \Delta(x) = (x - \alpha)^2g(x),\) where \(g(x)\) is a polynomial of degree \(3\) at most.

     Now it can be easily proven that \(\Delta(x)\) is divisible by \(f(x).\)

108. Let \(\Delta\) be the determinant. Then,

     \(\frac{d\Delta}{d\theta} = \begin{vmatrix}-\sin(\theta + \alpha) & -\sin(\theta + \beta) & -\sin(\theta + \gamma) \\ \sin(\theta + \alpha) & \sin(\theta + \beta) & \sin(\theta + \gamma) \\ \sin(\beta + \gamma) & \sin(\gamma - \alpha) & \sin(\alpha - \beta)\end{vmatrix} + \begin{vmatrix}\cos(\theta + \alpha) & cos(\theta + \beta) & \cos(\theta + \gamma) \\ \cos(\theta + \alpha) & cos(\theta + \beta) & \cos(\theta + \gamma) \\ \sin(\beta + \gamma) & \sin(\gamma - \alpha) & \sin(\alpha - \beta)\end{vmatrix} + \\ \begin{vmatrix}\cos(\theta + \alpha) & cos(\theta + \beta) & \cos(\theta + \gamma) \\ \cos(\theta + \alpha) & cos(\theta + \beta) & \cos(\theta + \gamma) \\ 0 & 0 & 0\end{vmatrix}\)

     \(= 0 + 0 + 0\)

     Thus, \(\Delta\) is a constant, which will be independent of \(\theta\)

109. \(\Delta = \begin{vmatrix}f & g & h \\ xf' + f & xg' + g & xh' + h \\ x^2f'' + 4xf' + 2f & x^2g'' + 4xg' + 2g & x^2h'' + 4xh' + 2h\end{vmatrix}\)

     \(= \begin{vmatrix}f & g & h \\ xf' & xg' & xh' \\ x^2f'' + 4xf' & x^2g'' + 2xg' & x^2h'' + 2xh'\end{vmatrix}[R_2\rightarrow R_2 - R_1; R_3\rightarrow R_3 - 2R_1]\)

     \(= \begin{vmatrix}f & g & h \\ xf' & xg' & xh' \\ x^2f'' & x^2g'' & x^2h''\end{vmatrix}[R_3 \rightarrow R_3 - 4R_2]\)

     \(= x^3\begin{vmatrix}f & g & h\\ f' & g' & h' \\ f'' & g'' & h''\end{vmatrix}\)

     \(\Delta' = \begin{vmatrix}f' & g ' & h' \\ f' & g ' & h' \\ x^3f'' & x^3g'' & x^3h''\end{vmatrix} + \begin{vmatrix}f & g & h \\ f'' & g'' & h'' \\ x^3f'' & x^3g'' & x^3h''\end{vmatrix} + \begin{vmatrix}f & g & h \\ f' & g' & h' \\ (x^2f'')' & (x^2g'')' & (x^2h'')'\end{vmatrix}\)

     \(= 0 + 0 + \begin{vmatrix}f & g & h \\ f' & g' & h' \\ (x^2f'')' & (x^2g'')' & (x^2h'')'\end{vmatrix}\) because two rows of first two determinants are equal.

110. \(\frac{d^n\{f(x)\}}{dx^n} = \begin{vmatrix}\frac{d^nx^n}{dx^n} & \frac{d^n\sin x}{dx^n} & \frac{d^n\cos x}{dx^n} \\ n! & \sin \frac{n\pi}{2} & \cos \frac{n\pi}{2} \\ a & a ^2 & a^2\end{vmatrix}\)

     \(y = x^n, y_1 = \frac{dy}{dx} = nx^{n - 1}, y_2 = \frac{d^2y}{dy^2} = n(n - 1)x^{n - 1}, \ldots y_n = n(n - 1)\ldots 3.2.1 = n!\)

     \(y = \sin x, y_1 = \cos x = \sin\left(\frac{\pi}{2} + x\right), y_2 = \cos\left(\frac{\pi}{2} + x\right) = \sin\left(\frac{\pi}{2} + \frac{\pi}{2} + x\right)\)

     Proceeding in the same way \(y_n = \sin\left(\frac{n\pi}{2} + x\right)\)

     \(y = \cos x, y_1 = -\sin x = \cos \left(\frac{\pi}{2} + x\right), y_2 = -sin\left(\frac{\pi}{2} + x\right) = \cos\left(2\frac{\pi}{2} + x\right)\)

     Proceeding in the same way \(y_n = \cos\left(n\frac{\pi}{2} + x\right)\)

     \(\frac{d^n\{f(x)\}}{dx^n} = \begin{vmatrix}n! &\sin\left(\frac{n\pi}{2} + x\right) & \cos\left(\frac{n\pi}{2} + x\right) \\ n! & \sin\frac{n\pi}{2} & \cos\frac{n\pi}{2} \\ a & a^2 & a^3\end{vmatrix}\)

     \(= \begin{vmatrix}n! &\sin\frac{n\pi}{2} & \cos\frac{n\pi}{2} \\ n! & \sin\frac{n\pi}{2} & \cos\frac{n\pi}{2} \\ a & a^2 & a^3\end{vmatrix} = 0\) because first two rows are identical.

111. \(\Delta = \begin{vmatrix}\cos A\cos P + \sin A\sin P & \cos A\cos Q + \sin A\sin Q & \cos A\cos R + \sin A\sin R \\ \cos B\cos P + \sin B\sin P & \cos B\cos Q + \sin B\sin Q & \cos B\cos R + \sin B\sin R \\ \cos C\cos P + \sin C\sin P & \cos C\cos Q + \sin C\sin Q & \cos C\cos R + \sin C\sin R \end{vmatrix}\)

     \(= \begin{vmatrix}\cos A & \sin A & 0 \\ \cos B & \sin B & 0 \\ \cos C & \sin C & 0\end{vmatrix} + \begin{vmatrix}\cos P & \sin P & 0 \\ \cos Q & \sin Q & 0 \\ \cos R & \sin R & 0\end{vmatrix}\)

     \(= 0 + 0 = 0\)

112. We know that \(\begin{vmatrix}a & b & c \\ b & c & a \\ c & a & b\end{vmatrix} = 3abc - a^3 - b^3 - c^3\)

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

     \(= \begin{vmatrix}a & b & c \\ b & c & a \\ c & a & b\end{vmatrix}\begin{vmatrix}-a & c & b \\ -b & a & c \\ -c & b & a\end{vmatrix}\)

     \(= \begin{vmatrix}2bc - a^2 & c^2 & b^2 \\ c^2 & 2bc - b^2 & a^2 \\ b^2 & a^2 & 2bc - c^2\end{vmatrix}\)

113. L.H.S. \(= \begin{vmatrix}\sin\alpha & \cos\alpha & 0 \\ \sin\beta & \cos\beta & 0 \\ \sin\gamma & \cos\gamma & 0 \end{vmatrix} \begin{vmatrix} \sin\alpha & \cos\alpha & 0 \\ \sin\beta & \cos\beta & 0 \\ \sin\gamma & \cos\gamma & 0 \end{vmatrix}\)

     \(=0.0 = 0\)

114. \(\Delta = \begin{vmatrix}3 & m \\ 2 & -5\end{vmatrix} = -(15 + 2m)\)

     Case I: When \(\Delta = 0, m = \frac{-15}{2}\)

     \(\Delta_1 = \begin{vmatrix}m & m \\ 20 & -5\end{vmatrix} = -25m\neq 0\)

     Hence, given system of equation has no solution when \(m = \frac{-15}{2}\)

     Case II When \(m \neq \frac{-15}{2}\)

     \(\Delta_2 = \begin{vmatrix}2 & m \\ 2 & 20\end{vmatrix} = 2(30 - m)\)

     \(x = \frac{\Delta_1}{\Delta} = \frac{25m}{15 + 2m} > 0[\because x >0]\)

     \(\Rightarrow -\infty < m < \frac{-15}{2}\) or \(0 < m < \infty\)

     \(y = \frac{\Delta_2}{\Delta} = \frac{2(m - 30)}{15 + 2m} > 0[\because y > 0]\)

     \(\Rightarrow -\infty < m < \frac{-15}{2}\) or \(30 < m < \infty\)

     Combining both we get, \(-\infty < m < \frac{-15}{2}\) or \(30 < m < \infty\)

115. \(\Delta = \begin{vmatrix}3 & -1 & 4 \\ 1 & 2 & -3 \\ 6 & 5 & \lambda\end{vmatrix}\)

     \(= 7(\lambda + 5)\)

     Case I When \(\lambda \neq 5 \Rightarrow \Delta \neq 0\) which means the system of equations has unique solution.

     Case II When \(\lambda = -5 \Rightarrow \Delta = 0\)

     Also, \(\Delta_1 = \begin{vmatrix}3 & -1 & 4 \\ -2 & 2 & -3 \\ 3 & 5 & -5\end{vmatrix} = 0, \Delta_2 = \begin{vmatrix}3 & 3 & 4 \\ 1 & -2 & -3 \\ 6 & -3 & -5\end{vmatrix} = 0\)

     \(\Delta_3 = \begin{vmatrix}3 & -1 & 3 \\ 1 & 2 & -2 \\ 6 & 5 & -3\end{vmatrix} = 0\)

     Since all the determinants are zero, in this case we have infinite solutions for given system of equations.

     Putting the value of \(\lambda\) the set of equation becomes

     \(3x - y + 4z = 3; x + 2y - 3z = -2; 6x + 5y - 5z = -3\)

     From first two equations we get, \(z = \frac{4 - 7x}{5}\)

     Substituting this in first we get \(y = \frac{1 - 13x}{5}\)

     Thus the set of solutions is \(x = t, y = \frac{1 - 13t}{5}, z = \frac{4 - 7t}{5}\)

116. \(\Delta = \begin{vmatrix}2 & p & 6 \\ 1 & 2 & q \\ 1 & 1 & 3\end{vmatrix} = (p - 2)(q - 3)\)

     \(\Delta_1 = \begin{vmatrix}8 & p & 8 \\ 5 & 2 & q \\ 4 & 1 & 3\end{vmatrix} = (p - 2)(4q - 15)\)

     \(\Delta_2 = \begin{vmatrix} 2 & 8 & 6 \\ 1 & 5 & 1 \\ 1 & 4 & 3\end{vmatrix} = 0\)

     \(\Delta_3 = \begin{vmatrix}2 & p & 8 \\ 1 & 2 & 5 \\ 1 & 1 & 4\end{vmatrix} = p -2\)

     Case I When \(\Delta \neq 0\) i.e. \(p \neq 2, q\neq 3,\) given system of equations has unique solution.

     Case II When \(\Delta = 0, p =2, q = 3\)

     When \(p = 2 \Rightarrow \Delta_1 = 0, \Delta_2 = 0, \Delta_3 = 0\)

     Thus, given system of equations has inifinite solutions.

     When \(q = 3\Rightarrow \Delta_1\neq 0\)

     Thus, given system of equations has no solutions.

117. For non-trivial solution

     \(\Delta = 0\) or \(\begin{vmatrix}\lambda & \sin\alpha & \cos\alpha \\ 1 & \cos\alpha & \sin\alpha \\ -1 & \sin\alpha & \cos\alpha \end{vmatrix} = 0\)

     \(\Rightarrow \lambda = \sin 2\alpha + \cos 2\alpha\)

     If \(\lambda = 1, \sin 2\alpha + \cos 2\alpha = 1\)

     \(\sin 2\alpha = 1 - \cos 2\alpha = 2\sin^2\alpha\)

     \(2\sin\alpha(\cos\alpha - \sin\alpha) = 0\)

     \(\therefore \sin\alpha = 0\) or \(\tan\alpha = 1\)

     \(\therefore \alpha = n\pi\) or \(\alpha = n\pi + \frac{\pi}{4}, n\in I\)

118. \(= (a + b + c)\begin{vmatrix}1 & b + c & a^2 \\ 1 & c + a & b^2 \\ 1 & a + b & c^2\end{vmatrix}[C_1\rightarrow C_1 + C_2]\)

     \(=(a + b + c)\begin{vmatrix}1 & b + a & a^2 \\ 0 & a - b & b^2 - a^2 \\ 0 & a - c & c^2 - a^2\end{vmatrix}[R_3\rightarrow R_3 - R_1; R_2\rightarrow R_2 - R_1]\)

     \(= (a + b + c)[(a - b)(c^2 - a^2) - (a - c)(b^2 - a^2)]\)

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

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

119. \(\Delta = \begin{vmatrix}\sqrt{13} + \sqrt{3} & 2\sqrt{5} & \sqrt{5} \\ \sqrt{15} - \sqrt{6} & 5 - 2\sqrt{10} & 0 \\ 3 - \sqrt{15} & \sqrt{15} - 10 & 0\end{vmatrix}[R_2\rightarrow R_2 - \sqrt{2}R_1;R_3\rightarrow R_3 - \sqrt{5}R_1]\)

     \(= 15\sqrt{2} - 25\sqrt{3}\)

120. \(\Delta = \begin{vmatrix}x & x(x^2 + 1) & x \\ y & y(y^2 + 1) & y \\ z & z(z^2 + 1) & z\end{vmatrix} + \begin{vmatrix}x & x(x^2 + 1) & 1 \\ y & y(y^2 + 1) & 1 \\ z & z(z^2 + 1) & 1\end{vmatrix}\)

     Since first and third columns of first determinant are identical.

     \(\Rightarrow \Delta = \begin{vmatrix}x & x(x^2 + 1) & 1 \\ y & y(y^2 + 1) & 1 \\ z & z(z^2 + 1) & 1\end{vmatrix}\)

     \(= \begin{vmatrix}x & x^3 & 1 \\ y & y^3 & 1 \\ z & z^3 & 1\end{vmatrix} + \begin{vmatrix}x & x^3 & x \\ y & y^3 & y \\ z & z^3 & z\end{vmatrix}\)

     Again second and third columns are identical in second determinant.

     \(\Delta = \begin{vmatrix}x & x^3 & 1 \\ y & y^3 & 1 \\ z & z^3 & 1\end{vmatrix}\)

     \(= (x - y)(y - z)(z - x)(x + y + z)\)

121. Let \(a\) and \(d\) be the first term and common difference of corresponding A.P.

     \(\frac{1}{x} = a + (l - 1)d, \frac{1}{y} = a + (2m - 1)d, \frac{1}{z} = a + (3n - 1)d\)

     \(\Delta = \frac{1}{xyz}\begin{vmatrix}x & y & z \\ l & 2m & 3n \\ 1 & 1 & 1\end{vmatrix}\)

     \(= \frac{1}{xyz}\begin{vmatrix}a + (l - 1)d & a + (2m - 1)d & a + (3n - 1)d\\ l & 2m & 3n \\ 1 & 1 & 1\end{vmatrix}\)

     \(= \frac{1}{xyz}\begin{vmatrix}ld - l & 2md - 2m & 2nd - 3n \\ l & 2m & 3n \\ 1 & 1 & 1\end{vmatrix}[R_1\rightarrow aR_3]\)

     \(= \frac{1}{xyz}\begin{vmatrix}0 & 0 & 0 \\ l& 2m & 3n \\ 1 & 1 & 1\end{vmatrix}[R_1\rightarrow (d -1)R_2]\)

     \(= 0\)

122. \(\Delta = \begin{vmatrix}1 & a^2 & a^3 \\ 0 & b^2 - a^2 & b^3 - a^3 \\ 0 & c^2 -a^2& c^3 - a^3\end{vmatrix}[R_2 \rightarrow R_2 - R_1; R_3 \rightarrow R_3 - R_1]\)

     \(= (b^2 - a^2)(c^3 - a^3) - (c^2 - a^2)(b^3 - a^3)\)

     \(= (b - a)(c - a)[(b + a)(c^2 + ac + a^2) - (c + a)(b^2 + ab + a^2)]\)

     \(= (b - a)(c - a)(bc^2 + abc + a^2b + ac^2 + a^2c + a^3 - b^2c - abc - a^2c - ab^2 - a^2b - a^3)\)

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

     \(= (b - a)(c - a)[bc(c - b) + a(c^2 - b^2)]\)

     \(= (b - a)(c - a)(c - b)(ab + bc + ca)\)

     We know that \(\begin{vmatrix}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{vmatrix} = (b - a)(c - a)(c - b)\)

     Hence, L.H.S. = R.H.S.

123. \(\Delta = \begin{vmatrix}b^2 + c^2 & a^2 & bc \\ c^2 + a^2 & b^2 & ca \\ a^2 + b^2 & c^2 & ab\end{vmatrix}[C_1\rightarrow -2C_3]\)

     \(= (a^2 + b^2 + c^2)\begin{vmatrix}1 & a^2 & bc \\ 1 & b^2 & ca \\ 1 & c^2 & ab\end{vmatrix}[C_1\rightarrow C_1 + C_2 + C_3]\)

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

     \(\Delta = (a^2 + b^2 + c^2)(a + b + c)(a - b)(b - c)(c - a)\)

124. This problem can be solved like \(123\) and has the same answer.

125. Let \(a_1b_1c_1 = 100\times a_1 + 10\times b_1 + c_1 = pk\) where \(p\in R\)

     \(a_2b_2c_2 = 100\times a_2 + 10\times b_2 + c_2 = qk\) where \(q\in R\)

     \(a_3b_3c_3 = 100\times a_3 + 10\times b_3 + c_3 = rk\) where \(r\in R\)

     \(\Delta = \begin{vmatrix}a_1 & b_1 & pk \\ a_2 & b_2 & qk \\ a_3 & b_3 & rk\end{vmatrix}[C_3\rightarrow 100C_1 + 10C_2 + C_3]\)

     Thus, given determinant is divisible by \(k\)

126. \(\Delta = \begin{vmatrix}a_1 & a_1x + b_1 & c_1 \\ a_2 & a_2x + b_2 & c_2 \\ a_3 & a_3x + b_3 & c_3\end{vmatrix} + \begin{vmatrix}b_1x & a_1x + b_1 & c_1 \\ b_2x & a_2x + b_2 & c_2 \\ b_3x & a_3x + b_3 & c_3\end{vmatrix}\)

     \(= \begin{vmatrix}a_1 & a_1x + b_1 & c_1 \\ a_2 & a_2x + b_2 & c_2 \\ a_3 & a_3x + b_3 & c_3\end{vmatrix} + x\begin{vmatrix}b_1 & a_1x + b_1 & c_1 \\ b_2 & a_2x + b_2 & c_2 \\ b_3 & a_3x + b_3 & c_3\end{vmatrix}\)

     \(= x\begin{vmatrix}a_1 & a_1 & c_1 \\ a_2 & a_2 & c_2 \\ a_3 & a_3 & c_3\end{vmatrix} + \begin{vmatrix}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_3 \\ a_3 & b_3 & c_3\end{vmatrix} + \\ x\begin{vmatrix}b_1 & a_1x & c_1 \\ b_2 & a_2x & c_2 \\ b_3 & a_3x & c_3 \end{vmatrix} + \begin{vmatrix}b_1 & b_1 & c_1 \\ b_2 & b_2 & c_2 \\ b_3 & b_3 & c_3\end{vmatrix}\)

     Clearly, first and last determinants are zero as they have identical columns.

     \(= \begin{vmatrix}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_3 \\ a_3 & b_3 & c_3\end{vmatrix} + x^2\begin{vmatrix}b_1 & a_1 & c_1 \\ b_2 & a_2 & c_2 \\ b_3 & a_3 & c_3\end{vmatrix}\)

     Exchanging first two columns of second determinants

     \(= (1 - x^2)\begin{vmatrix}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{vmatrix}\)

127. \(\Delta = abc\begin{vmatrix}\frac{1}{a} + 1 & \frac{1}{a} & \frac{1}{a} \\ \frac{1}{b} & \frac{1}{b} + 1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c} + 1\end{vmatrix}\)

     \(= abc\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + 1\right) \begin{vmatrix} 1 & 1 & 1 \\ \frac{1}{b} & \frac{1}{b} + 1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c} + 1\end{vmatrix}\)

     \(= abc\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + 1\right) \begin{vmatrix}1 & 0 & 0 \\ \frac{1}{b} & 1 & 0 \\ \frac{1}{c} & 0 & 1\end{vmatrix}[C_2 \rightarrow C_2 - C_1; C_3\rightarrow C_3 - C_1]\)

     \(= abc\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + 1\right)\)

     Now since \(a, b, c\) are roots of \(px^3 + qx^2 + rx + s = 0\)

     \(\therefore px^3 + qx^2 + rx + s = (x - a)(x - b)(x - c)\)

     Comparing coefficients, \(a + b + c = \frac{-q}{p}\)

     \(ab + bc + ca = \frac{r}{p}; abc = \frac{-s}{p}\)

     Thus, \(abc\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + 1\right) = \frac{r - s}{p}\)

128. \(\Delta = \begin{vmatrix}1 & a & a^4 \\ 0 & b - a & b^4 - a^4 \\ 0 & c - a & c^4 - a^4\end{vmatrix}[R_2\rightarrow R_2 - R_1; R_3 \rightarrow R_3 - R-1]\)

     \(= (b - a)(c^4 - a^4) - (c - a)(b^4 - a^4)\)

     \(= (b - a)(c - a)[(c + a)(c^2 + a^2) - (b + a)(b^2 + c^2)] > 0~\forall~a<b<c\)

129. \(\Delta = \begin{vmatrix}a & a^3 & a^4 \\ b & b^3 & b^4 \\ c & c^3 & c^4\end{vmatrix} - \begin{vmatrix}a & a^3 & 1 \\ b & b^3 & 1 \\ c & c^3 & 1\end{vmatrix}\)

     \(= abc \begin{vmatrix}1 & a^2 & a^3 \\ 1 & b^2 & b^3 \\ 1 & c^2 & c^3\end{vmatrix} - \begin{vmatrix}a & a^3 & 1 \\ b & b^3 & 1 \\ c & c^3 & 1\end{vmatrix}\)

     \(= abc \begin{vmatrix}0 & a^2 - b^2 & a^3 - b^3\\ 0 & b^2 - c^2 & b^3 - c^3 \\ 1 & c^2 & c^3\end{vmatrix} - \begin{vmatrix}0 & a - b & a^3 - b^3 \\ 0 & b - c & b^3 - c^3 \\ 1 & c & c^3\end{vmatrix}[R_1 \rightarrow R_1 - R_2; R_2 \rightarrow R_2 -R_3]\)

     \(\Rightarrow abc[(a^2 - b^2)(b^3 - c^3) - (b^2 - c^2)(a^3 - b^3)] - [(a - b)(b^3 - c^3) - (b - c)(a^3 - b^3)] = 0\)

     \(abc(a - b)(b - c)[(a + b)(b^2 + bc + c^2) - (b + c)(a^2 + ab + b^2)] =\\ (a - b)(b - c)(b^2 + bc + c^2 - a^2 - ab - b^2)\)

     Becasue \(a,b,c\) are distinct we \(a - b \neq 0; b - c \neq 0; c - a\neq 0\)

     \(abc(a + b)(b^2 + bc + c^2) - (b + c)(a^2 + ab + b^2) = (b^2 + bc + c^2 - a^2 - ab - b^2)\)

     \(abc(ab^2 + abc + ac^2 + b^3 + b^2c + bc^2 - a^2b - ab^2 - b^3 - a^2c - abc - b^2c) = bc + c^2 - a^2 -ab\)

     \(abc(ac^2 + bc^2 - a^2b - a^2c) = b(c - a) + (c^2 - a^2)\)

     \(abc[ac(c - a) + b(c^2 - a^2)] = (c - a)(a + b + c)\)

     \(abc(ab + bc + ca) = a + b + c\)

130. Taking \(b_1, b_2, b_3\) common from columns and multiplying rows with them, we get

     \(\Delta = \begin{vmatrix}x_1 + a_1b_1 & a_1b_1 & a_1b_1 \\ a_2b_2 & x_2 + a_2b_2 & a_2b_2 \\ a_3b_3 & a_3b_3 & x + a_3b_3\end{vmatrix}\)

     Taking \(x_1,x_2,x_3\) common from rows

     \(= x_1x_2x_3\begin{vmatrix}1 + \frac{a_1b_1}{x_1} & \frac{a_1b_1}{x_1} & \frac{a_1b_1}{x_1} \\ \frac{a_2b_2}{x_2} & 1 + \frac{a_2b_2}{x_2} & \frac{a_2b_2}{x_2} \\ \frac{a_3b_3}{x_3} & \frac{a_3b_3}{x_3} & 1 + \frac{a_3b_3}{x_3}\end{vmatrix}\)

     \(= x_1x_2x_3\left(1 + \frac{a_1b_1}{x_1} + \frac{a_2b_2}{x_2} + \frac{a_3b_3}{c_3}\right) \begin{vmatrix}1 & 1 & 1 \\ \frac{a_2b_2}{x_2} & 1 + \frac{a_2b_2}{x_2} & \frac{a_2b_2}{x_2} \\ \frac{a_3b_3}{x_3} & \frac{a_3b_3}{x_3} & 1 + \frac{a_3b_3}{x_3}\end{vmatrix}[R_1 \rightarrow R_1 + R_2 + R_3]\)

     \(= x_1x_2x_3\left(1 + \frac{a_1b_1}{x_1} + \frac{a_2b_2}{x_2} + \frac{a_3b_3}{c_3}\right)\begin{vmatrix}1 & 0 & 0 \\ \frac{a_2b_2}{x_2} & 1 & 0 \\ \frac{a_3b_3}{x_3} & 0 & 1\end{vmatrix}[C_2\rightarrow C_2 - C_1; C_3 \rightarrow C_3 - C_1]\)

     \(= x_1x_2x_3\left(1 + \frac{a_1b_1}{x_1} + \frac{a_2b_2}{x_2} + \frac{a_3b_3}{c_3}\right)\)

131. This problem is similar to \(92\) and has been left as an exercise.

132. \(\Delta = abc\begin{vmatrix}a & c & a + c \\ a + b & b & a \\ b & b + c & c\end{vmatrix}\)

     \(= abc \begin{vmatrix}-2a & -2b & 0 \\ a + b & b & a \\ b & b + c & c\end{vmatrix}[R_1 \rightarrow R_1 - R_2 - R_3]\)

     \(= 4a^2b^2c^2\)

133. \(\Delta = \begin{vmatrix} 1 + a^2 + b^2 & 0 & -2b \\ 0 & 1 + a^2 + b^2 & 2a \\ b(1 + a^2 + b^2) & -a(1 + a^2 + b^2) & 1 - a^2 - b^2\end{vmatrix} [C_1\rightarrow C_1 - bC_3; C_2\rightarrow C_2 + aC_3]\)

     \(= (1 + a^2 + b^2)^2(1 -a^2 - b^2 + 2a^2) + 2b^2(1 + a^2 + b^2)^2\)

     \(= (1 + a^2 + b^2)^3\)

134. We know that \(P = \frac{a + b + c}{a}; A = \sqrt{s(s - a)(s - b)(s - c)}\)

     After that this problem is same as \(93\) and you just need to substitute for the values of \(A\) and \(P\).

135. Taking \(a,b,c\) common from rows and multiplying with columns we get the same determinant as in problem \(90\) and can be solved similarly.

136. \(\Delta = \begin{vmatrix}x^3 & 6x^2a + 2a^3 & (x - a)^3 \\ y^3 & 6y^2a + 2a^3 & (y - a)^3 \\ z^3 & 6z^2a + 2a^3 & (z - a)^3\end{vmatrix} [C_2 \rightarrow C_2 - C_3]\)

     \(= 2\begin{vmatrix}x^3 & 3x^2a + a^3 & (x - a)^3 \\ y^3 & 3y^2a + a^3 & (y - a)^3 \\ z^3 & 3z^2a + a^3 & (z - a)^3\end{vmatrix}\)

     \(= 2 \begin{vmatrix}x^3 & 3x^2a + a^3 & 3xa^2 \\ y^3 & 3y^2a + a^3 & 3ya^2 \\ z^3 & 3z^2a + a^3 & 3za^2\end{vmatrix} [C_3 \rightarrow C_3 - C_1 - C_2]\)

     \(= 2 \begin{vmatrix}x^3 & 3x^2a + a^3 & 3xa^2 \\ y^3 - x^3 & 3(y^2 - x^2) & 3a^2(y - x) \\ z^3 - x^3 & 3(z^2 - x^2) & 3a^2(z - x)\end{vmatrix} [R_2\rightarrow R_2 - R_1; R_3 \rightarrow R_2 - R_1]\)

     Now upon expasion desired condition can be proven easily.

137. \(\Delta = \begin{vmatrix} 1- x & a & 0 \\ a & a^2 - x & x \\ a^2 & a^3 & -x\end{vmatrix}[C_3 \rightarrow C_3 -aC_2]\)

     \(= x\begin{vmatrix}1 - x & a & 0 \\ a + a^2 & a^2 - x + a^3 & 0 \\ a^2 & a^3 & -1 \\ \end{vmatrix}[R_2 \rightarrow R_2 + R_3]\)

     \(= x[a(a + a^2) - (1 - x)(a^2 + a^3 - x)]\)

     \(= x(a^2 + a^3 - a^2 - a^3 + x + xa^2 + xa^3 - x^2)\)

     \(= x^2(1 + a ^2 + a^3)- x^3\)

138. \(y_1 = p\cos px, y_2 = -p^2\sin px, y_3 = -p^3\cos px, y_4 = p^4 \sin px\)

     \(y_5 = p^5 \cos px, y_6 = -p^6\sin px, y_7 = -p^7\cos px, y_8 = p^8\sin px\)

     \(\Delta = -p^6 \begin{vmatrix}\sin px & p\cos px & -p^2 \sin px \\ -p^3\cos px & p^4 \sin px & p^5 \cos px \\ \sin px & p\cos px & -p^2 \sin px\end{vmatrix}\)

     Clearly first and last rows are identical.

     \(\Delta = 0\)

139. \(\Delta = \begin{vmatrix}1 & 0 & -\sin\theta \\ 0 & 1 & \cos\theta \\ \sin\theta & -\cos\theta & 0\end{vmatrix}[C_1 \rightarrow C_1 - \cos\theta C_3; C_2 \rightarrow C_2 + \sin\theta C_3]\)

     \(= \cos^2\theta + \sin^2\theta = 1\)

140. \(\Delta = \begin{vmatrix}\cos\alpha & \sin\alpha\cos\beta & 0 \\ -\sin\alpha & \cos\alpha\cos\beta & 0 \\ 0 & -\sin\beta & \frac{1}{\cos\beta} \end{vmatrix}[C_3 \rightarrow C_3 - \tan\beta C_2]\)

     \(= \frac{1}{\cos\beta}[\cos\beta(\cos^2\alpha + \sin^2\alpha)] = 1\)

141. Multiplying columns with \(a, b, c,\) we get

     \(\Delta = \frac{1}{abc}\begin{vmatrix}a(a^2 + x) & ab^2 & ac^2 \\ a^2b & b(b^2 + x) & bc^2 \\ a^2c & b^2c & c(c^2 + x)\end{vmatrix}\)

     Now taking out \(a,b,c\) from rows, we have

     \(= \begin{vmatrix}a^2 + x & b^2 & c^2 \\ a^2 & b^2 + x & c^2 \\ a^2 & b^2 & c^2 + x\end{vmatrix}\)

     \(= \begin{vmatrix}x & 0 & -x \\ 0 & x & -x \\ a^2 & b^2 & c^2 + x\end{vmatrix}[R_1\rightarrow R_1 - R_3; R_2\rightarrow R_2 - R_3]\)

     \(= \begin{vmatrix}x & 0 & 0 \\ 0 & x & -x \\ a^2 & b^2 & a^2 + c^2 + x\end{vmatrix}[C_3 \rightarrow C_3 + C_1]\)

     \(\Rightarrow x^2(a^2 + b^2 + c^2 + x) = 0\)

     \(\Rightarrow x=0, -(a^2 + b^2 + c^2)\)

142. By observation if we put \(x = n,\) then first two columns are same. Thus, it is one of the solutions.

     Similarly,if we put \(x = n,\) then first and third columns are same, hence it is second solution.

     If we take \(\frac{x!}{r!(x + 2 - r)!}\) common from first column then we get a quadratic equation which will have two roots and we have found both of them.

143. \(\Delta = \frac{1}{a^2}\begin{vmatrix}u + a^2x & aw' - bu & av' - cu \\ w' + abx & av - bw' & au' - cw' \\ v' + acx & au' - bv' & aw - cv'\end{vmatrix}[C_2\rightarrow aC_2 - bC_1; C_3\rightarrow aC_3 - cC_1]\)

     \(\Rightarrow x = - \begin{vmatrix}u & aw' - bu & av' - cu \\ w' & av - bw' & au' - cw' \\ v' & au' - bv' & aw - cv' \end{vmatrix} \div \begin{vmatrix}a^2 & aw' - bu & av' - cu \\ ab & av - bw' & au' - cw' \\ ac & au' - bv' & aw - cv'\end{vmatrix}\)

144. We know that value of the determinant in denominator is \((a - b)(b - c)(c - a) = k\) (say)

     \(f(a, b, c) = \begin{vmatrix}f(a) - f(b)& f(b) - f(c) & f(c) \\ 0 & 0 & 1 \\ a - b & b - c & c\end{vmatrix}[C_1\rightarrow C_1 - C_2; C_2\rightarrow C_2 - C_3] \div k\)

     \(= -(b - c)(f(a) - f(b)) - (a - b)(f(b) - f(c))\div k\)

     \(= -(a - b)(b - c)\left[\frac{f(a) - f(b)}{a - b} - \frac{f(b) - f(c)}{(b - c)}\right]\div k\)

     \(= -(a - b)(b - c)(f(a, b) - f(b, c))\div k\)

     \(= (a - b)(b - c)(c - a)\frac{f(b, c) - f(a, b)}{c - a} = (a - b)(b - c)(c - a)f(a, b, c)\div (a - b)(b - c)(c - a)\)

     \(= f(a, b, c)\)

145. Becasue \(A,B,C\) are angles of a triangle. \(A + B + C = \pi\)

     Also, \(e^{i\pi} = \cos\pi + i\sin\pi = -1\)

     Taking \(e^{iA}, e^{iB}, e^{iC}\) common from \(R_1, R_2, R_3,\) we get

     \(\Delta = e^{i(A + B + C)}\begin{vmatrix}e^{iA} & e^{-i(A +C)} & e^{-i(A + B)} \\ e^{-i(B + C)} & e^{iB} & e^{-i(A + B)} \\ e^{-i(B + C)} & e^{-i(A + C)} & e^{iC}\end{vmatrix}\)

     \(= -\begin{vmatrix}e^{iA} & -e^{iB} & -e^{iC} \\ -e^{iA} & e^{iB} & -e^{iC} \\ -e^{iA} & -e^{iB} & e^{iC}\end{vmatrix}\)

     Taking \(e^{iA}, e^{iB}, e^{iC}\) common from \(C_1, C_2, C_3,\) we get

     \(= \begin{vmatrix}1 & -1 & -1 \\ -1 & 1 & -1 \\ -1 & -1 & 1\end{vmatrix} = -4\)

     which is purely real.

146. \(\Delta = \begin{vmatrix}1 & \sin A\cos A & \cos^2 A \\ 1 & \sin B\cos B & \cos^2 B \\ 1 & \sin C\cos C & \cos^2 C \end{vmatrix}[C_1 \rightarrow C_1 + C_3]\)

     Performing \(R_3\rightarrow R_3 - R_1; R_2\rightarrow R_2 - R_1,\) we get

     \(= \sin(A - B)\sin(B - C)\sin(C - A)\geq 0\)

     Following the expansion the other condition can also be shown and has been left as an exercise.

147. Performing \(C_1 \rightarrow C_1 - aC_2; C_2\rightarrow C_2 -aC_3]\)

     \(\Delta = \begin{vmatrix}0 & 0 & 1 \\ \cos nx - a\cos(n + 1)x & \cos(n + 1)x - a\cos(n + 2)x & \cos(n + 2)x \\ \sin nx - a\sin(n + 1)x & \sin(n + 1)x - a\sin(n + 2)x & \sin(n + 2)x\end{vmatrix}\)

     \(= \sin(n+1)x\cos nx - a\sin(n+1)x\cos(n+1)x - a\sin(n+2)x\cos nx \\ + a^2\sin(n+2)x\cos(n+1)x -\sin nx\cos(n+1)x +a\sin nx\cos(n + 2)x \\ + a\sin(n + 1)x\cos(n + 1)x - a^2sin(n + 1)x\cos(n + 2)x\)

     \(= \sin(n + 1 - n)x - a\sin(n + 2 - n)x + a^2\sin(n + 2 - n - 1)x\)

     \(= \sin x - a\sin 2x + a^2\sin x\)

     \(= (a62 -2a\cos x + 1)\sin x\)

148. \(\Delta = \begin{vmatrix}2 & \cos^2x & 4\sin 2x \\ 2 & 1 + \cos^2x & 4\sin 2x \\ 1 & \cos^2x & 1 + 4\sin 2x\end{vmatrix}[C_1 \rightarrow C_1 + C_2]\)

     \(= \begin{vmatrix}0 & -1 & 0 \\ 2 & 1 + \cos^2x & 4\sin 2x \\ 1 & \cos^2x & 1 + 4\sin 2x\end{vmatrix}[R_1 \rightarrow R_1 - R_2]\)

     \(= 2 - 4\sin 2x\)

     The above expression has maximum value for \(0 < x < \frac{\pi}{2}\) when \(x = \frac{\pi}{4}\)

149. Expanding the determinant we get \(\Delta = -1 + 2\cos A\cos B \cos C + \cos^2 A + \cos^2 B + \cos^2 C\)

     Consider the expression \(2(\cos^2 A + \cos^2 B + \cos^2 C)\)

     \(= 1 + \cos 2A + 1 + \cos 2B + 1 + \cos 2C\)

     \(= 2 + 2\cos(A + B)\cos(A - B) + 2\cos^2 C\)

     \(= 2 + 2\cos(\pi - C)\cos(A - B) + 2\cos^2C\)

     \(= 2 - 2\cos C[\cos(A - B) - \cos C]\)

     \(= 2 - 2\cos C[\cos(A - B) 0 \cos (\pi - (A + B))]\)

     \(= 2 - 4\cos A\cos B\cos C\)

     Thus, \(\Delta = 0\)

150. Since \(A, B, C\) are angles of an isosceles triangle, let \(A = B\)

     Thus, first two columns become equal leading determinant to be zero.