# 81. Determinants Solutions Part 3

1. 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 $$C_1\leftrightarrow C_2$$

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

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

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

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

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

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

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

$$= xA + B$$ 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 $$x$$

3. $$\sum_{r = 1}^n D_r = D_1 + D_2 + \ldots + D_n$$

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

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

$$= 0$$ because first and third columns are identical.

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

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

Exchanging rows and columns

$$\overline{\Delta} = \Delta \therefore \Delta$$ is purely real.

5. 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.

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

$$= 0 + 0 + 0$$

$$\because f_r(a)=g_r(a)=h_r(a), r=1,2,3$$ 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.

7. 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 $$f(x).$$

8. 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 $$\theta$$

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

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

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

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

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

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

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

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

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

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.

11. $$\Delta = \begin{vmatrix}\cos A\cos P + \sin A\sin P & \cos A\cos Q + \sin A\sin Q & \cos A\cos R + \sin A\sin R \\ \cos B\cos P + \sin B\sin P & \cos B\cos Q + \sin B\sin Q & \cos B\cos R + \sin B\sin R \\ \cos C\cos P + \sin C\sin P & \cos C\cos Q + \sin C\sin Q & \cos C\cos R + \sin C\sin R \end{vmatrix}$$

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

$$= 0 + 0 = 0$$

12. We know that $$\begin{vmatrix}a & b & c \\ b & c & a \\ c & a & b\end{vmatrix} = 3abc - a^3 - b^3 - c^3$$

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

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

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

13. L.H.S. $$= \begin{vmatrix}\sin\alpha & \cos\alpha & 0 \\ \sin\beta & \cos\beta & 0 \\ \sin\gamma & \cos\gamma & 0 \end{vmatrix} \begin{vmatrix} \sin\alpha & \cos\alpha & 0 \\ \sin\beta & \cos\beta & 0 \\ \sin\gamma & \cos\gamma & 0 \end{vmatrix}$$

$$=0.0 = 0$$

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

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 $$30 < m < \infty$$

15. $$\Delta = \begin{vmatrix}3 & -1 & 4 \\ 1 & 2 & -3 \\ 6 & 5 & \lambda\end{vmatrix}$$

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

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 $$x = t, y = \frac{1 - 13t}{5}, z = \frac{4 - 7t}{5}$$

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

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

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

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

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.

17. 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 $$\alpha = n\pi + \frac{\pi}{4}, n\in I$$

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

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

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

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

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

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

$$= 15\sqrt{2} - 25\sqrt{3}$$

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

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.

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

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

21. Let $$a$$ and $$d$$ be the first term and common difference of corresponding A.P.

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

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

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

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

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

$$= 0$$

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

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

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

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

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

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

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

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.

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

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

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

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

24. This problem can be solved like $$123$$ and has the same answer.

25. 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 $$k$$

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

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

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

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

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

27. $$\Delta = abc\begin{vmatrix}\frac{1}{a} + 1 & \frac{1}{a} & \frac{1}{a} \\ \frac{1}{b} & \frac{1}{b} + 1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c} + 1\end{vmatrix}$$

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

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

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

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

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

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

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

29. $$\Delta = \begin{vmatrix}a & a^3 & a^4 \\ b & b^3 & b^4 \\ c & c^3 & c^4\end{vmatrix} - \begin{vmatrix}a & a^3 & 1 \\ b & b^3 & 1 \\ c & c^3 & 1\end{vmatrix}$$

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

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

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

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

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

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

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

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

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

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

30. 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

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

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

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

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

31. This problem is similar to $$92$$ and has been left as an exercise.

32. $$\Delta = abc\begin{vmatrix}a & c & a + c \\ a + b & b & a \\ b & b + c & c\end{vmatrix}$$

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

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

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

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

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

34. 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 $$P$$.

35. 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.

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

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

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

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

Now upon expasion desired condition can be proven easily.

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

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

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

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

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

38. $$y_1 = p\cos px, y_2 = -p^2\sin px, y_3 = -p^3\cos px, y_4 = p^4 \sin px$$

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

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

Clearly first and last rows are identical.

$$\Delta = 0$$

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

$$= \cos^2\theta + \sin^2\theta = 1$$

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

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

41. 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

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

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

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

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

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

42. 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.

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

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

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

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

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

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

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

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

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

45. 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.

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

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.

47. Performing $$C_1 \rightarrow C_1 - aC_2; C_2\rightarrow C_2 -aC_3]$$

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

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

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

$$= \sin x - a\sin 2x + a^2\sin x$$

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

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

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

$$= 2 - 4\sin 2x$$

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

49. 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, $$\Delta = 0$$

50. Since $$A, B, C$$ are angles of an isosceles triangle, let $$A = B$$

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