70. Determinants Solutions Part 1ΒΆ

  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 \(x = 1, \frac{-13 \pm \sqrt{249}}{2}\)