76. Determinants Solutions Part 4ΒΆ

  1. \(\Delta = \begin{vmatrix}1 & \frac{\log y}{\log x} & \frac{\log z}{\log x} \\ \frac{\log x}{\log y} & 1 & \frac{\log y}{\log z} \\ \frac{\log x}{\log z} & \frac{\log y}{\log z} & 1\end{vmatrix}\)

    \(= \frac{1}{\log x\log y\log z}\begin{vmatrix}\log x & \log y & \log z \\ \log x & \log y & \log z \\ \log x & \log y & \log x\end{vmatrix}\)

    \(= 0\) because all three rows are identical.

  2. \(\Delta = \begin{vmatrix}a^{2x} + a^{-2x} + 2 & a^{2x} + a^{-2x} - 2 & 1 \\ b^{2x} + b^{-2x} + 2 & b^{2x} + b^{-2x} - 2 & 1 \\ c^{2x} + c^{-2x} + 2 & c^{2x} + c^{-2x} - 2 & 1\end{vmatrix}\)

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

    \(= 0\) because first two columns are identical.

  3. Considering first determinant only:

    \(\Delta = \begin{vmatrix}115 & 114 & 103 \\ 108 & 106 & 111 \\ 113 & 116 & 104\end{vmatrix}[C_1\leftrightarrow C_2; C_2\leftrightarrow C_3]\)

    Performing \(R_1\leftrightarrow R_3\)

    \(\Delta = -\begin{vmatrix}113 & 116 & 104 \\ 108 & 106 & 111 \\ 115 & 114 & 103\end{vmatrix}\)

    Thus, given condition is satisfied.

  4. \(\sum{n = 1}^N U_n = \begin{vmatrix}\sum{n = 1}^N n & 1 & 5 \\ \sum{n = 1}^N n^2 & 2N + 1 & 2N + 1 \\ \sum{n = 1}^N n^3 & 3N^2 & 3N\end{vmatrix}\)

    \(= \begin{vmatrix}\frac{n(n + 1)}{2} & 1 & 5 \\ \frac{n(n + 1)(2n + 1)}{6} & 2N + 1 & 2N + 1 \\ \left\{\frac{n(n + 1)}{2}\right\}^2 & 3N^2 & 3N\end{vmatrix}\)

    Taking \(\frac{N(N + 1)}{2}\) common from first column and then performing \(C_1\rightarrow C_1 - \frac{1}{6}(C_2 + C_3)\)

    \(= \frac{N(N + 1)}{2}\begin{vmatrix}0 & 1 & 5 \\ 0 & 2N + 1 & 2N + 1 \\ 0 & 3N^2 & 3N\end{vmatrix}\)

    Since first column has only \(0\) as element the sum of determinants is zero.

  5. \(\because A, B, C\) are angles of a triangle, therefore \(A + B + C == \pi; \sin(A + B + C) = 0; \cos(A + B) = -\cos C\)

    \(\therefore \Delta = \begin{vmatrix}0 & \sin B & \cos C \\ -\sin B & 0 & \tan A \\ -\cos C & -\tan A & 0 \end{vmatrix}\)

    Changing rows into corresponding columns

    \(= \begin{vmatrix}0 & -\sin B & -\cos C \\ \sin B & 0 & -\tan A \\ \cos C & \tan A & 0\end{vmatrix}\)

    Taking \(-1\) common from second and third columns, we have

    \(= \begin{vmatrix}0 & \sin B & \cos C \\ \sin B & 0 & \tan A \\ \cos C & -\tan A & 0\end{vmatrix} = -\Delta\)

    \(\Rightarrow 2\Delta = 0 \Rightarrow \Delta = 0\)

  6. Taking \(b - a\) common from first and third columns

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

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

    Since the first two columns are same the determinant is zero.

  7. We can rewrite it as \(\sum_{j = 0}^{n - 1}\Delta_j = \begin{vmatrix}\sum_{j = 0}^{n - 1} j & n & 6 \\ \sum_{j = 0}^{n - 1} j^2 & 2n^2 & 4n - 2 \\ \sum_{j = 0}^{n - 1}j^3 & 3n^3 & 3n^2 - 3n\end{vmatrix}\)

    \(= \begin{vmatrix}\frac{n(n - 1)}{2} & n & 6 \\ \frac{n(n - 1)(2n - 1)}{6} & 2n^2 & 4n - 2 \\ \left\{\frac{n(n - 1)}{2}\right\}^2 & 3n^3 & 3n^2 - 3n\end{vmatrix}\)

    \(= \frac{n(n - 1)}{2}\begin{vmatrix}1 & n & 6 \\ \frac{2n - 1}{3} & 2n^2 & 4n - 2 \\ \frac{n(n - 1)}{2} & 3n^3 & 3n^2 - 3n\end{vmatrix}\)

    \(= \frac{n(n - 1)}{2}\begin{vmatrix}0 & n & 6 \\ 0 & 2n^2 & 4n - 2 \\ 0 & 3n^3 & 3n^2 -3n\end{vmatrix}[C_1\rightarrow C_1 - \frac{C_3}{6}]\)

    Since first column is entirely made up of zeros the value of determinant is zero, which is a constant as desired.

  8. \(\sum_{r = 0}^m(2r - 1) = \frac{1}{2}(m + 1)(2m - 1 - 1) = m^2 - 1\)

    \(\sum_{r = 0}^m({}^nC_r) = 2^m\)

    \(\sum_{r = 0}^m1 = m + 1\)

    Thus, first two rows of determinant become zero leading the desired sum to be \(0.\)

  9. \(= \begin{vmatrix}{}^xC_r & {}^{x + 1}C_{r + 1} & {}^{x + 1}C_{r + 2} \\ {}^yC_r & {}^{y + 1}C_{r + 1} & {}^{y + 1}C_{r + 2} \\ {}^zC_r & {}^{z + 1}C_{r + 1} & {}^{z + 1}C_{r + 2}\end{vmatrix} [C_3 \rightarrow C_3 + C_2; C_2\rightarrow C_2 + C_1]\)

    Performing \(C_3\rightarrow C_3 + C_2\) we get the determinant on R.H.S.

  10. \(\sum_{r = 1}^n \Delta_r = \begin{vmatrix}\sum_{r = 1}^nr & n + 1 & 1 \\ \sum_{r = 1}^nr^2 & 2n - 1 & \frac{2n + 1}{3} \\ \sum_{r = 1}^nr^3 & 3n + 2 \\ \frac{n(n + 1)}{2}\end{vmatrix}\)

    \(= \begin{vmatrix}\frac{n(n + 1)}{2} & n + 1 & 1 \\ \frac{n(n + 1)(2n + 1)}{6} & 2n - 1 & \frac{2n + 1}{3} \\ \left\{\frac{n(n + 1)}{2}\right\}^2 & 3n + 2 & \frac{n(n + 1)}{2}\end{vmatrix}\)

    If we take \(\frac{n(n + 1)}{2}\) common from first column then first and third column becomes same. Thus, \(\sum_{r = 1}^n \Delta_r = 0\)

  11. \(\sum_{r = 1}^n 2^{r - 1} = 1 + 2 + \ldots + 2^{n - 1} = \frac{2^n - 1}{2 - 1} = 2^n - 1\)

    \(\sum_{r = 1}^n 2.3^{r - 1} = 2.\frac{3^n - 1}{3 - 1} = 3^n - 1\)

    \(\sum_{r = 1}^n 4.5^{r - 1} = 4.\frac{35^n - 1}{5 - 1} = 5^n - 1\)

    Thus, we see that first row and third rows are equal leading the sum of the determinants to zero.

  12. \(\Delta = \begin{vmatrix}2x - 1 & 2x - 3 & x^2 - 4x + 4 \\ 2x - 3 & 2x - 5 & x^2 - 6x + 9 \\ 2x -5 & 2x -7 & x^2 - 8x + 16\end{vmatrix} [C_1 \rightarrow C_1 - C_1; C_2\rightarrow C_2 - C_3]\)

    \(= \begin{vmatrix}2x - 1 & 2x - 3 & x^2 \\ 2x - 3 & 2x - 5 & x^2 \\ 2x - 5 & 2x - 7 & x^2\end{vmatrix} + \begin{vmatrix}2x - 1 & 2x - 3 & -4x \\ 2x - 3 & 2x - 5 & -6x \\ 2x - 5 & 2x - 7 & -8x\end{vmatrix} + \begin{vmatrix}2x - 1 & 2x - 3 & 4 \\ 2x - 3 & 2x - 5 & 9 \\ 2x - 5 & 2x - 7 & 16\end{vmatrix}\)

    Clearly, if we perform \(R_1\rightarrow R_1- R_2; R_2\rightarrow R_2 - R_3\) will make \(R_1\) and \(R_3\) same in the first determinant.

    This is also true for second determinant.

    \(= \begin{vmatrix}2 & 2 & -5 \\ 2 & 2 & -7 \\ 2x - 5 & 2x - 7 & 16\end{vmatrix}\)

    Clearly, the determinant is independent of \(x\)

  13. \(\Delta = \begin{vmatrix}2 & 1 + i & 3 \\ 1 - i & 0 & 2 + i \\ 3 & 2 - i & 1\end{vmatrix}\)

    Taking complex conjugate and exchanging rows into corresponding columns

    \(\overline{\Delta} = \begin{vmatrix}2 & 1 + i & 3 \\ 1 - i & 0 & 2 + i \\ 3 & 2 - i & 1\end{vmatrix} = \Delta\)

    Since \(\overline{\Delta} = \Delta,\) the determinant is purely real.

  14. \(\Delta = \begin{vmatrix}x - 3 & 2x & 2 \\ 3x + 2 & x & 1 \\ 5x + 1 & 5x & 5\end{vmatrix} + \begin{vmatrix}x - 3 & 1 & 2 \\ 3x + 2 & 2 & 1 \\ 5x + 1 & 4 & 5\end{vmatrix}\)

    If we take out \(x\) common from second column of first determinant then second and third columns are same, making it zero. Now expandng second determinant

    \(= \begin{vmatrix}x & 1 & 2 \\ 3x & 2 & 1 \\ 5x & 4 & 5 \end{vmatrix} +\) a determinant of constants(say \(k\))

    \(= x \begin{vmatrix}0 & 1 & 2 \\ 1 & 2 & 1 \\ 1 & 4 & 5\end{vmatrix} [C_1\rightarrow C_1 - C_2] + k\)

    \(= x \begin{vmatrix}0 & 1 & 2 \\ 1 & 2 & 1 \\ 0 & 2 & 4\end{vmatrix} [R_3\rightarrow R_3 - R_2] + k\)

    \(= x\begin{vmatrix}0 & 1 & 2 \\ 1 & 2 & 1 \\ 0 & 0 & 0\end{vmatrix} [C_3\rightarrow C_3 - 2C_1] + k\)

    \(= k\)

  15. \(\Delta = \begin{vmatrix}a^n - x & a^n(a - 1) & a^{n + 1}(a - 1) \\ a^{n + 3} - x & a^{n + 3}(a - 1) & a^{n + 4}(a - 1) \\ a^{n + 6} - x & a^{n + 6}(a - 1) & a^{n + 7}(a - 1)\end{vmatrix}[R_2\rightarrow R_2 - R_1; R_3\rightarrow R_3 - R_2]\)

    \(=a^{n(n + 1)}(a - 1)^2\begin{vmatrix}a^n - x & 1 & 1 \\ a^{n + 3} - x & a^3 & a^3 \\ a^{n + 6} - x & a^6 & a^6\end{vmatrix} = 0\)

    Since second and third columns are same, the edterminant is zero.

  16. \(\Delta = \sum_{r = 2}^n (-2)^r \begin{vmatrix}{}^nC_r & {}^{n - 2}C_{r - 1} & {}^{n - 2}C_r \\ 0 & 1 & 1 \\ 0 & -1 & 9\end{vmatrix} [C_1\rightarrow C_1 + 2C_2 + C_3]\)

    \(= \sum_{r = 2}^n (-2)^r{}^nC_r\)

    \(= \sum_{r = 0}^n (-2)^r{}^nC_r - ({}^nC_0 -2{}^nC_1)\)

    \(= 2n - 1 + (-1)^n\)

  17. Performing \(R_1\rightarrow aR-1, R_2\rightarrow bR_2, R_3\rightarrow cR_3\) and then taking out \(abc\) out from first two columns,

    \(\Delta = abc\begin{vmatrix}bc & 1 & a(b + c) \\ ca & 1 & b(c + a) \\ ab & 1 & c(a + b)\end{vmatrix}\)

    Performing \(C_3\rightarrow C_3 + C_1\) and then taking \(ab + bc + ca\) out

    \(= abc(ab + bc + ca)\begin{vmatrix}bc & 1 & 1 \\ ca & 1 & 1 \\ ab & 1 & 1\end{vmatrix}\)

    Since last two columns are same, the determinant is zero.

Rest of the problems are left as exercises.