# 77. Determinants Solutions Part 1

1. Let $$\Delta = \begin{vmatrix}4 & 9 & 7\\3 & 5 & 7\\5 & 4 & 5\end{vmatrix}$$

$$\Delta = \begin{vmatrix}1 & 4 & 0\\3 & 5 & 7\\5 & 4 & 5\end{vmatrix}[R_1\rightarrow R_1 - R_2]$$

$$\Delta = \begin{vmatrix}1 & 4 & 0\\0 & -7 & 7\\0 & -16 & 5\end{vmatrix}[R_2\rightarrow R_2 -3R_1~\text{and}~R_3\rightarrow R_3 - 5R_1]$$

$$= 1(-35 + 112) = 77$$ (expanding along first column)

2. Let $$\Delta = \begin{vmatrix}1 & a & a^2\\1 & b & b^2\\1 & c & c^2\end{vmatrix}$$

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

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

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

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

3. Let $$a = 2, b = 3, c = 4$$

So from previous example, $$\Delta = \begin{vmatrix}1 & a & a^2\\1 & b & b^2\\1 & c & c^2\end{vmatrix} = (a - b)(b - c)(c - a)$$

$$= (2 - 3)(3 - 4)(4 - 2) = 2$$

4. $$\Delta = \begin{vmatrix}a + b + c & b & c\\a + b + c & c & a\\a + b + c & a & b\end{vmatrix}[C_1\rightarrow C_1 + C_2 + C_3]$$

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

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

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

$$= -\frac{1}{2}(a + b + c)[(a - b)^2 + (b -c)^2 + (c - a)^2]$$

$$\because a, b, c$$ are positive $$\therefore a + b + c > 0$$

Again, since $$a, b, c$$ are unequal $$\therefore (a - b)^2 + (b -c)^2 + (c - a)^2 > 0$$

Thus, $$\Delta < 0$$

5. $$\Delta = \begin{vmatrix}a + b + c & a + b & a\\ a + b + c & b + c & b \\ a + b + c & c + a\\ c\end{vmatrix}[C_1\rightarrow C_1 + C_3]$$

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

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

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

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

6. $$\Delta = \begin{vmatrix}1 + a_1 + a_2 + a_3 & a_2 & a_3\\1 + a_1 + a_2 + a_3& 1 + a_2 & a_3\\1 + a_1 + a_2 + a_3 & a_2 & 1 + a_3\end{vmatrix}[C_1\rightarrow C_1 + C_2 + C_3]$$

$$= (1 + a_1 + a_2 + a_3)\begin{vmatrix}1 & a_1 & a_3\\ 1 & 1 + a_2 & a_3\\ 1 & a_2 & 1 + a_3\end{vmatrix}$$

$$= (1 + a_1 + a_2 + a_3)\begin{vmatrix}0 & -1 & 0\\ 0 & 1 & -1 \\ 1 & a_2 & 1 + a_3\end{vmatrix}[R_1\rightarrow R_1 - R_2; R_2\rightarrow R_2 - R_3]$$

$$= (1 + a_1 + a_2 + a_3)\begin{vmatrix}-1 & 0 \\ 1 & -1\end{vmatrix}$$

$$= 1 + a_1 + a_2 + a_3$$

7. $$\Delta = \begin{vmatrix}2(a + b + c) & a & b \\ 2(a + b + c) & b + c + 2a & b \\ 2(a + b + c) & a + c + a + 2b\end{vmatrix}[C_1\rightarrow C_1 + C_2 + C_3]$$

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

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

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

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

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

8. $$\Delta = \begin{vmatrix}2a & 2b & a - b - c \\ 2b & 2c & b - c - a \\ 2c & 2a & c - a - b\end{vmatrix}[C_1\rightarrow C_1 + C_2;C_2\rightarrow C_2 -C_3]$$

$$= 4\begin{vmatrix}a & b & a - b - c \\ b & c & b - c - a \\ c & a & c - a - b\end{vmatrix}$$

$$= 4\begin{vmatrix}a & b & -c \\ b & c & b - c - a \\ c & a & -b\end{vmatrix}[C_3\rightarrow C_3 - C_1 + C_2]$$

Proceeding like exercise 4, we get

$$= -4[(-(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)]$$

$$4(a^3 + b^3 + c^3 - 3abc)$$

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

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

$$= (a + b + c)\begin{vmatrix}1 & 0 & 0 \\ 2b & -b - c- a & 0 \\ 2c & 0 & -c -a -b\end{vmatrix}[C_2\rightarrow C_2 - C_1;C_3\rightarrow C_3 - C_1]$$

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

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

10. $$\Delta = \frac{1}{xyz}\begin{vmatrix}x^2 + y^2 + z^2 \\ x^3 & y^3 & z^3 \\ xyz & xyz & xyz \end{vmatrix}[C_1\rightarrow xC1; C_2\rightarrow yC2; C_2\rightarrow zC_3]$$

$$=\frac{xyz}{xyz}\begin{vmatrix}x^2 + y^2 + z^2 \\ x^3 & y^3 & z^3 \\ 1 & 1 & 1\end{vmatrix}$$

$$= \begin{vmatrix}1 & 1 & 1\\ x^2 & y^2 & z^2\\ x^3 & y^3 & z3 \end{vmatrix}$$ (Doing double row exchange)

$$= \begin{vmatrix}1 & 0 & 0\\ x^2 & y^2 - x^2 & z^2 - x^2 \\ z^3 & y^3 - x^3 & z^3 - x^3\end{vmatrix}[C_2\rightarrow C_2 - C_1; C_3\rightarrow C_3 - C_1]$$

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

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

$$= (y - z)(z - x)\begin{vmatrix}y + x & z - y \\ y^2 + xy + x^2 \\ (z^2 - y^2) + zx - zy\end{vmatrix}$$

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

$$= (x - y)(y - z)(z - x)(xy + yz + zx)$$

11. $$\Delta = \frac{1}{abc}\begin{vmatrix}a(a^2 + 1) & ab^2 & ac^2 \\ a^2b & b(b^2 + 1) & bc^2 \\ a^2c & b^2c & c(c^2 + 1)\end{vmatrix}[C_1\rightarrow aC1; C_2\rightarrow bC2; C_2\rightarrow cC_3]$$

$$= \frac{abc}{abc}\begin{vmatrix}a^2 + 1 & b^2 & c^2 \\ a^2 & b^2 + 1 & c^2 \\ a^2 & b^2 & c^2 + 1\end{vmatrix}$$ [taking $$a, c, c$$ common from rows]

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

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

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

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

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

12. $$\Delta = a_1a_2a_3\begin{vmatrix}\frac{1}{a_1} + 1 & \frac{1}{a_2} & \frac{1}{a_3} \\ \frac{1}{a_1} & \frac{1}{a_2} + 1 & \frac{1}{a_3} \\ \frac{1}{a_1} & \frac{1}{a_2} & \frac{1}{a_3} + 1\end{vmatrix}[C_1\rightarrow \frac{1}{a_1}C_1; C_2\rightarrow \frac{1}{a_2}C_2; C_3\rightarrow \frac{1}{a_3}C_3]$$

$$= a_1a_2a_3\begin{vmatrix}1 + \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} & \frac{1}{a_2} & \frac{1}{a_3} \\ 1 + \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} & \frac{1}{a_2} + 1 & \frac{1}{a_3} \\ 1 + \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} & \frac{1}{a_2} & \frac{1}{a_3} + 1\end{vmatrix}[C_1\rightarrow C_1 + C_2 + C_3]$$

$$=a_1a_2a_3\left(1 + \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3}\right)\begin{vmatrix} 1 & \frac{1}{a_2} & \frac{1}{a_3} \\ 1 & \frac{1}{a_2} + 1 & \frac{1}{a_3} \\ 1 & \frac{1}{b_2} & \frac{1}{a+3} + 1\end{vmatrix}$$

$$=a_1a_2a_3\left(1 + \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3}\right)\begin{vmatrix}1 & \frac{1}{a_2} & \frac{1}{a_3} \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{vmatrix}[R_2\rightarrow R_2 - R_1;R_3\rightarrow R_3 - R_1]$$

$$==a_1a_2a_3\left(1 + \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3}\right)$$

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

$$= \begin{vmatrix}x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1\end{vmatrix} + xyz\begin{vmatrix}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{vmatrix}$$

Performing $$C_2\leftrightarrow C_3; C_1\leftrightarrow C_2$$

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

$$\Rightarrow (1 + xyz)(x - y)(y - z)(z - x) = 0$$

$$\because x, y, z\neq 0$$ as they are different, $$(x - y), (y - z), (z - x)\neq 0$$

$$\therefore 1 + xyz = 0 \Rightarrow xyz = -1$$

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

$$= \frac{1}{c}\begin{vmatrix}0 & 02c & -2b \ 0 & c(c + a - b) & b(c - a - b) \\ c & c & a + b\end{vmatrix}[R_2\rightarrow cR_2 - bR_3]$$

$$= \frac{1}{c}[c(-2bc)(c - a - b - (c + a - b)]$$

$$= 4abc$$

15. $$\Delta = \begin{vmatrix}(b + c)^2 - a^2 & 0 & a^2 \\ b^2 - (c + a)^2 & (c + a)^2 - b^2 & b^2 \\ 0 & c^2 - (a + b)^2 & (a + b)^2\end{vmatrix}[C_1\rightarrow C_1 - C_2; C_2\rightarrow C_2 - C_3]$$

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

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

$$= (a + b + c)^2\begin{vmatrix}b + c - a & 0 & a^2 \\ 0 & c + a - b & b^2 \\ -2b & -2a & 2ab\end{vmatrix}[C_1\rightarrow C_1 + C_2]$$

$$= \frac{(a + b + c)^2}{ab}\begin{vmatrix}a(b + c) & a^2 & a^2 \\ b^2 & b(c + a) & b^2 \\ 0 & 0 & 2ab\end{vmatrix}[C_1\rightarrow aC_1 + C3;C_2\rightarrow bC_2 + C_3]$$

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

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

$$=2abc(a + b + c)^3$$

16. $$\begin{vmatrix}15 - x & 1 & 10 \\ -4 -2x & 0 & 6 \\ -8 & 0 & 3\end{vmatrix} = 0[R_2\rightarrow R_2 -R_1; R_3 \rightarrow R_3 - R_1]$$

$$-36 + 6x = 0 \Rightarrow x = 6$$

17. $$\begin{vmatrix}a + b + c - x & c & b \\ a + b + c - x & b - x & a \\ a + b + c - x & a & c - a\end{vmatrix} = 0[C_1\rightarrow C_1 + C_2 + C_3]$$

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

$$[\because a + b + c = 0]$$

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

$$x[(b - c - x)(c - b - x) - (a - c)(a - b)] = 0$$

$$x[x^2 - a^2 - b^2 - c^2 + ab + bc + ca] = 0$$

$$\therefore$$ Either $$x = 0$$ or

$$x^2 = a^2 + b^2 + c^2 - ab - bc - ca$$

$$= a^2 + b^2 + c^2 - \frac{1}{2}[(a + b + c)^2 - a^2 - b^2 - c^2]$$

$$= \frac{3}{2}(a^2 + b^2 + c^2)[\because a + b + c = 0]$$

$$x = \pm\sqrt{\frac{3}{3}(a^2 + b^2 + c^2)}$$

18. $$D_1 = \begin{vmatrix}a & b & c \\ d & e & f \\ g & h & k\end{vmatrix}$$

$$= \begin{vmatrix}a & b & c \\ tx & ty & tz \\ g & h & k\end{vmatrix}$$

$$= t\begin{vmatrix}a & b & c \\ x & y & z \\ g & h & k\end{vmatrix}$$

$$= t\begin{vmatrix}a & x & g \\ b & y & h \\ c & z & k\end{vmatrix}$$ [Changing rows into corresponding columns]

$$= -t\begin{vmatrix}a & g & x \\ b & h & y \\ c & k & z\end{vmatrix}[C_2\leftrightarrow C_3]$$

$$= -tD_2$$

19. L.H.S. $$= \frac{1}{a^2b^2c^2}\begin{vmatrix}a^3 & a^2bc & a^3bc \\ b^3 & ab^2c & ab^3c \\ c^3 & abc^2 & abc^3\end{vmatrix}[R_1\rightarrow a^2R_1; R_2\rightarrow b^2R_2; R_3\rightarrow c^2R_3]$$

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

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

20. Let $$x$$ be the first term and $$y$$ be the common ratio of G.P., then

$$a = xy^{p -1}; b = xy^{q - 1}; c = xy^{r - 1}$$

$$\Rightarrow \log a = \log x + (p - 1)\log y; \log b = \log x + (q - 1)\log y; \log c = \log x + (r - 1)\log y$$

$$\Delta = \begin{vmatrix}\log x + (p - 1)\log y & p & 1 \\ \ \log x + (q - 1)\log y & q & 1 \\ \log x + (r - 1)\log y & r & 1\end{vmatrix}$$

$$= \begin{vmatrix}(p - 1)\log y & p & 1 \\ (q - 1)\log x & q & 1 \\ (r - 1)\log y & r & 1\end{vmatrix}[C_1\rightarrow C_1 - \log x C_3]$$

$$= \log y \begin{vmatrix}p - 1 & p & 1 \\ q - 1 & q & 1 \\ r - 1 & r & 1\end{vmatrix}$$

$$= 0[C_1\rightarrow C_1 - C_2 + C_3]$$

21. $$\Delta = \begin{vmatrix}2 & 9 & 2 \\ -4 & 5 & 7 \\ 2 & 1 & 6\end{vmatrix}[C_1\rightarrow C_1 - C_3;]$$

$$= \begin{vmatrix}2 & 9 & 2 \\ 0 & 23 & 11 \\ 0 & -8 & 4\end{vmatrix}[R_2\rightarrow R_2 + 2_R1;R_3\rightarrow R_3 - R_2]$$

$$= 2(23 * 4 -(11 * -8)) = 2(92 + 88) = 360$$

22. $$\Delta = \begin{vmatrix}1 & 1 & 17 \\ 3 & 3 & 19 \\ 5 & 5 & 21\end{vmatrix}[C_1\rightarrow C_1 - C_3]$$

$$= 0$$ (because first two columns are identical)

23. $$\Delta = \begin{vmatrix}1 & 4 & 0 \\ 3 & 5 & 7 \\ 5 & 4 & 5\end{vmatrix}[R_1\rightarrow R_1 - R_2]$$

$$= \begin{vmatrix}1 & 0 & 0 \\ 2 & -7 & 7 \\ 5 & -16 & 5\end{vmatrix}[C_2 \rightarrow C_2 - 4C_1]$$

$$= (-7 * 5 - (-16 * 7)) = (112 - 35) = 77$$

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

$$= \begin{vmatrix}1 & 3 & 5 \\ 4 & 5 & 7 \\ 9 & 7 & 9\end{vmatrix}[C_3\rightarrow C_3 - C_2; C_2\rightarrow C_2 - C_1]$$

$$= \begin{vmatrix}1 & 3 & 2 \\ 4 & 5 & 2 \\ 9 & 7 & 2\end{vmatrix}[C_3\rightarrow C_3 - C-2]$$

$$= \begin{vmatrix}1 & 3 & 2 \\ 3 & 2 & 0 \\ 5 & 2 & 0\end{vmatrix}[R_3\rightarrow R_3 - R_2; R_2\rightarrow R_2 - R_1]$$

$$= 2(3 * 2 - 5 * 2) = -8$$

25. $$\Delta = \begin{vmatrix}1 & 0 & 0 \\ 1 & x & 0 \\ 1 & 0 & y\end{vmatrix}[C_3\rightarrow C_3 - C_1; C_2\rightarrow C_2 - C_1]$$

$$= xy$$

26. $$\Delta = \begin{vmatrix}1 & 0 & 0 \\ a & b - a & c - a \\ a^3 & b^3 - a^3 & c^3 - a^3\end{vmatrix}[C_3\rightarrow C_3 - C_1; C_2\rightarrow C_2 - C_1]$$

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

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

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

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

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

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

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

28. $$\Delta = \begin{vmatrix}0 & a - b & a^2 - b^2 + ca - bc \\ 0 & b - c & b^2 - c^2 + ab - ca \\ 1 & c & c^2 - ab\end{vmatrix}[R_2\rightarrow R_2 - R_3; R_1\rightarrow R_1- R_2]$$

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

$$= 0$$ because first two rows are identical.

29. $$\Delta = \begin{vmatrix}1 & bc & bc(b + c) \\ 0 & c(a - b) & c[a^2 + ca - b^2 - bc] \\ 0 & a(b - c) & a[ab + b^2 - c^2 - ac]\end{vmatrix}[R_3\rightarrow R_3 - R_2;R_2\rightarrow R_2 - R_1]$$

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

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

$$= 0$$

30. $$\Delta = \begin{vmatrix}1 & a & a + b + c \\ 1 & b & a + b + c \\ 1 & c & a + b + c\end{vmatrix}[C_3\rightarrow C_3 + C_2]$$

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

$$= 0$$ because last two columns are equal.

31. Let $$x$$ be the first term and $$d$$ be the common difference of corresponsing A.P.

$$\therefore \frac{1}{a} = x + (p - 1)d; \frac{1}{b} = x + (q - 1)d; \frac{1}{c} = x + (r - 1)d$$

$$\Delta = \begin{vmatrix}c(b - a) & p - q & 0 \\ a(c - b) & q - r & 0 \\ ab & c & 1\end{vmatrix}$$

$$= c(b - a)(q - r) - a(c - b)(p - q)$$

$$= \frac{1}{x + (r - 1)d}\left(\frac{1}{x + (q - 1)d} - \frac{1}{x + (p - 1)d}\right)(q - r) - \frac{1}{x + (p - 1)d}\left(\frac{1}{x + (r - 1)d} - \frac{1}{x + (q - 1)d}\right)(p - q)$$

$$= \frac{(p - q)(q - r)d}{(x + (p - 1)d)(x + (q - 1)d)(x + (r - 1)d)} - \frac{(p - q)(q - r)d}{(x + (p - 1)d)(x + (q - 1)d)(x + (r - 1)d)}$$

$$= 0$$

32. Since $$t$$ is a constant its computation will involve only constant terms of the determinant. What implies is that we can set $$x = 0$$ to form a determinant which will evaluate to $$x$$.

$$\therefore \Delta = \begin{vmatrix}0 & -1 & 3 \\ 1 & 1 & -4 \\ -2 & 4 & 0\end{vmatrix}$$

$$= \frac{1}{3}\begin{vmatrix}0 & 0 & 3 \\ 1 & -1 & -4 \\ -2 & 12 & 0\end{vmatrix}[C_2\rightarrow 3C_2 + C_3]$$

$$= \frac{1}{3}(3(1 * 12 - (-2 * -1))) = 10$$

33. Taking $$a, b, c$$ common from first three columns we have

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

$$= abc\begin{vmatrix}1 & 0 & 0 \\ a & b - a & c - a \\ a^2 & b^2 - a^2 & c^2 - a^2\end{vmatrix}[C_2\rightarrow C_2 - C_1; C_3\rightarrow C_3 - C_1]$$

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

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

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

34. $$\because a, b , c$$ are in A.P. $$b = a + d, c = a + 2d$$ where $$d$$ is the common difference of A.P>

$$\therefore \Delta = \begin{vmatrix}x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + a + d \\ x + 3 & x + 4 & x + a + 2d\end{vmatrix}$$

$$= \begin{vmatrix}x + 1 & x + 2 & x + a \\ 1 & 1 & d \\ 1 & 1 & d\end{vmatrix}[R_3\rightarrow R_3 - R_2;R_2\rightarrow R_2 - R_1]$$

$$= 0$$ because last two rows are equal.

35. $$\Delta = \begin{vmatrix}1 + \omega + \omega^2 & \omega & \omega^2 \\ 1 + \omega + \omega^2 & \omega^2 & 1 \\ 1 + \omega + \omega^2 & 1 & \omega^2\end{vmatrix}[C_1\rightarrow C_1 + C_2 + C_3]$$

$$\because \omega$$ is cube root of unity $$\therefore 1 + \omega + \omega^2 = -1$$

$$\Rightarrow \Delta = \begin{vmatrix}-1 & \omega & \omega^2 \\ -1 &\omega^2 & 1 \\ -1 &1 & \omega^2\end{vmatrix}$$

$$= \begin{vmatrix}-1 & \omega & \omega^2 \\ 0 & \omega^2 - \omega & 1 - \omega^2 \\ 0 & 1 - \omega & 0\end{vmatrix}$$

$$= (1 - \omega)(1 - \omega^2) = 1 + \omega^3 -\omega - \omega^2 = 0$$

36. $$\Delta = \begin{vmatrix}k & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 2 & 3\end{vmatrix}[C_2\rightarrow C_2 - C_1; C_3\rightarrow C_3 - C_2]$$

$$= \begin{vmatrix}k & 0 & 0 \\ 1 & 1 & 0 & 1 & 2 & 1\end{vmatrix}[C_3\rightarrow C_3 - C_2]$$

$$= k$$

37. $$\Delta = \begin{vmatrix}a^2 + b^2 + c^2 + x & b^2 & c^2 \\ a^2 + b^2 + c^2 + x & b^2 + x & c^2 \\ a^2 + b^2 + c^2 + x & b^2 & c^2 + x\end{vmatrix}[C_1\rightarrow C_1 + C_2 + C_3]$$

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

$$= (a^2 + b^2 + c^2 + x)x^2$$

38. $$\Delta = \begin{vmatrix}a + b + c & b + c& a^2 \\ a + b + c & c + a & b^2 \\ a + b + c & a + b & c^2\end{vmatrix}[C_1\rightarrow C_1 + C_2 + C_3]$$

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

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

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

39. $$\Delta = \begin{vmatrix}a + b + c & a - b & a \\ a + b + c & b - c & b \\ a + b + c & c - a & c\end{vmatrix}[C_1 \rightarrow C_1 + C_2 + C_3]$$

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

$$= (a + b + c)\begin{vmatrix}1 & -b & a \\ 1 & -c & b \\ 1 & -a & c\end{vmatrix}[C_2 \rightarrow C_2 - C_3]$$

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

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

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

$$= 3abc - a^3 - b^3 - c^3$$

40. $$\Delta = \begin{vmatrix}2(a + b + c)& b + c & c + a \\ 2(a + b + c) & c + a & a + b \\ 2(a + b + c) & a + b & b + c\end{vmatrix}[C_1\rightarrow C_1 + C_2 + C_3]$$

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

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

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

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

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

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

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

If we take $$x$$ and $$y$$ common from first determinant then its first two rows will become same and thus the value of determinant is $$0$$. Similarly, since two rows of last determinant are same its value is $$0$$

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

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

Following similarly, all four determinants above are $$0$$

42. $$\Delta = -\begin{vmatrix}0 & q - p & r - p \\ p - q & 0 & r - q \\ p - r & q - r & 0\end{vmatrix}$$ Multiplying rows with $$-1$$

$$= -\begin{vmatrix}0 & p - q & p - r \\ q - p & 0 & q - r \\ r - p & r - q & 0\end{vmatrix}$$ Exchanging rows anb columns

$$= -\Delta \Rightarrow 2\Delta = 0 \Rightarrow \Delta = 0$$

43. $$\Delta = \begin{vmatrix}3a + 3b & 3a + 3b & 2a + 3b\\ a + 2b & a & a + b \\ a + b & a + 2b & a\end{vmatrix}[R_1\rightarrow R_1 + R_2 + R_3]$$

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

$$= 3(a + b)\begin{vmatrix}1 & 0 & 0 \\ a + 2b & -2b & -b \\ a + b & b & - b\end{vmatrix}[C_2\rightarrow C_2 - C_1; C_3\rightarrow C_3 - C_1]$$

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

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

44. $$\Delta = \frac{1}{a}\begin{vmatrix}a^2 + b^2 + c^2 & b - c & c + b \\ a^2 + b^2 + c^2 & b & c - a \\ a^2 + b^2 + c^2 \\ b + a & c\end{vmatrix}[C_1\rightarrow aC_1 + bC_2 + cC_3]$$

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

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

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

$$= (a^2 + b^2 + c^2)(a + b + c)$$

$$\because a^2 + b^2 + c^2 > 0$$ the determinant has same sign as $$a + b + c$$

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

Taking out $$a, b, c$$ from columns

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

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

$$= \frac{c^2}{c^2}[-2b^2c^4 + 2b^2c^2(a^2 + b^2) + 2b^2c^2(c^2 + a^2) - 2b^4c^2]$$

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

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

46. $$\Delta = \frac{1}{a^2b^2c^2}\begin{vmatrix}a^2(b + c)^2 & b^2c^2 & b^2c^2 \\ a^2c^2 & b^2(c + a)^2 & c^2a^2 \\ a^2b^2 & b^2a^2 & c^2(a + b)^2\end{vmatrix}[C_1\rightarrow a^2C_1;C_2\rightarrow b^2C_2;C_3\rightarrow c^2C_3]$$

Taking $$\alpha=ab, \beta=bc, \gamma=ca$$

$$= \frac{1}{\alpha\beta\gamma}\begin{vmatrix}(\alpha + \gamma)^2 & \beta^2 & \beta^2 \\ \gamma^2 & (\alpha + \beta)^2 & \gamma^2 \\ \alpha^2 & \alpha^2 & (\beta + \gamma)^2\end{vmatrix}$$

Following like exercise $$15$$

$$= \frac{2\alpha\beta\gamma(\alpha + \beta + \gamma)^3}{\alpha\beta\gamma}$$

$$2(\alpha + \beta + \gamma)^3$$

47. Applying $$R_1\rightarrow cR_1; R_2\rightarrow aR_2; R_3\rightarrow bR_3$$ and then take $$a, b, c$$ common from $$C_2, C_3$$ and $$C_1$$ respectively, we have

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

Following like exercise $$15$$, we get

$$=2abc(a + b + c)^3$$

48. Applyting $$R_1\rightarrow cR_1, R_2\rightarrow aR_2, R_3\rightarrow bR_3$$

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

Following like exercise $$45$$, we get

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

49. $$\Delta = \begin{vmatrix}0 & 0 & x - a & a & a & a \\ b & x & b\end{vmatrix}[R_1\rightarrow R_1 - R_2]$$

$$= (x - a)(ax - ab) = 0\Rightarrow a(x - a)(x - b) = 0$$

$$\therefore x = a, b$$

50. $$\Delta = \begin{vmatrix}x + 13 & x + 14 & x + 15 \\ 6 & x + 4 & 4 \\ 7 & 8 & x + 8\end{vmatrix}[R_1\rightarrow R_1 + R_2 + R_3]$$

$$= \begin{vmatrix}x + 13 & x + 13 & x + 13\\ 6 & x + 4 & 4 \\ 7 & 8 & x + 8\end{vmatrix} + \begin{vmatrix}0 & 1 & 2\\ 6 & x + 4 & 4 \\ 7 & 8 & x + 8\end{vmatrix}$$

Solving these two, we get $$x = 1, \frac{-13 \pm \sqrt{249}}{2}$$