Maths Determinants

$A_{3*3}$

$det\left(A\right)=42A=\left[\begin{array}{ccc}\hfill 2a_1\hfill & \hfill 2a_2\hfill & \hfill 2a_3\hfill \\ \hfill 2b_1\hfill & \hfill 2b_2\hfill & \hfill 2b_3\hfill \\ \hfill 2c_1\hfill & \hfill 2c_2\hfill & \hfill 2c_3\hfill \end{array}\right]$

For R₂ →2R₂ + 5R₃

$det\left(B\right)=\left|\begin{array}{ccc}\hfill 2a_1\hfill & \hfill 2a_2\hfill & \hfill 2a_3\hfill \\ \hfill 4b_1+10c_1\hfill & \hfill 4b_2+10c_2\hfill & \hfill 4b_3+10c_3\hfill \\ \hfill 2c_1\hfill & \hfill 2c_2\hfill & \hfill 2c_3\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill 2a_1\hfill & \hfill 2a_2\hfill & \hfill 2a_3\hfill \\ \hfill 4b_1\hfill & \hfill 4b_2\hfill & \hfill 4b_3\hfill \\ \hfill 2c_1\hfill & \hfill 2c_2\hfill & \hfill 2c_3\hfill \end{array}\right|=16det\left(A\right)=64$

Yes. Determinants can be calculated for any square matrix of n-order, and it is done by expansion of rows and columns. Even in higher dimensions, their job is to define hyper volumes and transformations.

In such cases, the value of determinant turns out to be zero. this is because swapping the values changes the magnitude into the opposite sign (fundamental property of determinants), which results in the final answer being zero.

Yes. Determinants can be fractions and irrational numbers depending on the values of the matrix. This is because the values are calculated as sums and products of the numbers in the matrix which can turn out to be any type of integer.

$A=\left[\begin{array}{ccc}\hfill 2\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 2\hfill \end{array}\right]\Rightarrow \left|A\right|=4\Rightarrow \left|3adj\left(2A^{-1}\right)\right|=\left|3.2^{2}adj\left(A^{-1}\right)\right|$

$\Rightarrow 12^3\left|adj\left(A^{-1}\right)\right|=12^3{\left|A^{-1}\right|}^2=\frac{12^3}{{\left|A\right|}^2}=\frac{12*12*12}{16}=108$

$?a_r=e^{i\frac{2r\pi }{9}}$

$\therefore a_1,a_2,a_3,...........$ are in G.P.

$\left|\begin{array}{ccc}\hfill a_1\hfill & \hfill a_{1}r\hfill & \hfill a_1{r}^2\hfill \\ \hfill a_1{r}^3\hfill & \hfill a_1{r}^4\hfill & \hfill a_1{r}^5\hfill \\ \hfill a_1{r}^6\hfill & \hfill a_1{r}^7\hfill & \hfill a_1{r}^8\hfill \end{array}\right|={a_1}^3{r}^9\left|\begin{array}{ccc}\hfill 1\hfill & \hfill r\hfill & \hfill r^2\hfill \\ \hfill 1\hfill & \hfill r\hfill & \hfill r^2\hfill \\ \hfill 1\hfill & \hfill r\hfill & \hfill r^2\hfill \end{array}\right|=0$

$\left(i\right)a_2{a}_6-a_4{a}_8=a_{1}r.a_1{r}^5-a_1{r}^6{a}_1{r}^7={a_1}^2{r}^3-{a_1}^2{r}^{10}\ne 0$

$\left(ii\right)a_9=a_1{r}^8\ne 0$

$\left(iii\right)a_1.a_1{r}^8-a_1{r}^2.a_1{r}^6=0$

$\left(iv\right)a_5\ne 0$

$\Delta =0$

$\left|\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 5\hfill & \hfill \alpha \hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \end{array}\right|=0$

15 -2α +α- 6 – 1 = 0

α = 8

For = 8, equations are

x + y + 3 = 6

2x + 5y + 8z =β

x + 2y + 3z = 14

$\left(2,5,8\right)=\mathcal{l}\left(1,1,1\right)+m\left(1,2,3\right)$

$\begin{array}{cc}\hfill 2=\mathcal{l}+m\hfill & \hfill 5=\mathcal{l}+2m\hfill \end{array}]\to 3=m,\mathcal{l}=-1$

8 = $\mathcal{l}+3m$

$\beta =6\mathcal{l}+14m$

=- 6 + 42 = 36

α + β = 8 + 36 = 44

  $\Delta =\left|\begin{array}{ccc}\hfill 3sin3\theta \hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 3cos2\theta \hfill & \hfill 4\hfill & \hfill 3\hfill \\ \hfill 6\hfill & \hfill 7\hfill & \hfill 7\hfill \end{array}\right|$

= 3 sin3θ (28 – 21) + (21 cos 2θ - 18) + 1 (21 cos 2θ - 24)

$\Delta =21sin3\theta +42cos2\theta -42$          

for no solution

sin 3θ + 2 cos 2θ = 2

$\theta =\pi ,2\pi ,3\pi ,\frac{\pi }{6},\frac{5\pi }{6},\frac{13\pi }{6},\frac{17\pi }{6}$

$\left(\overrightarrow{a}*\overrightarrow{b}\right)*\widehat{i}=\left(\overrightarrow{a}.\widehat{i}\right)\overrightarrow{b}-\left(\overrightarrow{b}.\widehat{i}\right)\overrightarrow{a}$

$\left(\left(\overrightarrow{a}*\overrightarrow{b}\right)*\widehat{i}\right).\widehat{k}=\left(\overrightarrow{a}.\widehat{i}\right)\left(\overrightarrow{b}.\widehat{k}\right)-\left(\overrightarrow{b}.\widehat{i}\right)\left(\overrightarrow{a}.\widehat{k}\right)$