Determinants

For inconsistent, $\Delta =0,\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill \alpha \hfill & \hfill 2a\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 3\alpha \hfill & \hfill 5\hfill \end{array}\right|=0$

a = 1

$\therefore {\Delta }_1=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 4\hfill & \hfill 3\hfill & \hfill 5\hfill \end{array}\right|=1\left(1\right)+1\left(12+5\right)+1\left(-3-8\right)=7\ne 0$              

ax + by + cz = d

2a + 3b – 5c = d               2a + 3b – 5c = d …………….(v)

Putting (i) & (iv) in (v) we get a = -9d.

In conditions

$\left(2\widehat{i}+\widehat{j}-5\widehat{k}\right),\left(a\widehat{i}+b\widehat{j}+c\widehat{k}\right)=0$

2b = d ………….(i) d > 0

2a + b – 5c = 0                  2a + 3b – 5c = d   $\left|a\right|,\left|b\right|,\left|c\right|,d\to g.c.d$           

=   $\frac{\alpha }{2}$

$\therefore \frac{\alpha }{2}=1\Rightarrow \alpha =2$

$\left(3\widehat{i}+5\widehat{j}-7\widehat{k}\right).\left(a\widehat{i}+b\widehat{j}+c\widehat{k}\right)=0$

3a + 5b – 7c = 0) *   $\frac{4}{7}...................\left(iii\right)$

$\therefore a=-18,b=1$

c = -7, d = 2

(iii)…(ii) ® 2a + 3b   $\frac{-33c}{7}=0$

2a + 3b =  $\frac{33}{7}c$ $c=\frac{-7}{2}d.................\left(iv\right)$

Putting values

a + 7b + c + 20d = 22

$\overrightarrow{v}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}$

$\overrightarrow{v}=\lambda \left(1,1,2\right)+\mu \left(2,-3,1\right)$

$\overrightarrow{v}.\widehat{j}=7\overrightarrow{v}.\frac{\overrightarrow{c}}{\left|\overrightarrow{c}\right|}=\frac{2}{\sqrt[]{3}}$

$\lambda +2\mu -\lambda +3\mu +2\lambda +\mu =2$

$\lambda -3\mu =7$

$\begin{array}{l}2\lambda =8\\ \lambda =4\end{array}$ $\begin{array}{l}\mu =-1\\ \overrightarrow{v}=\left(2,7,7\right)\end{array}$

$2si{n}^2\theta =cos2\theta =1-2si{n}^2\theta$

$\Rightarrow 4si{n}^2\theta =1\Rightarrow sin\theta =\pm \frac{1}{2}$

$\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}$

$Sum\frac{7\pi }{6}+\frac{11\pi }{6}=3\pi \Rightarrow k=3$

$D=\left|\begin{array}{ccc}1& -2& 3\\ 2& 1& 1\\ 1& -7& a\end{array}\right|=0\Rightarrow a=8$

also,  $D_1=\left|\begin{array}{ccc}9& -2& 3\\ b& 1& 1\\ 24& -7& 8\end{array}\right|=0\Rightarrow b=3$

hence,  $a-b=8-3=5$

Let TV (r) denotes truth value of a statement $r$ .

$\text{Now, if}TV\left(p\right)=TV\left(q\right)=T$

$\Rightarrow \mathrm{T}\mathrm{V}\left({\mathrm{S}}_1\right)=\mathrm{F}$

Also, if $TV\left(p\right)=T$ and $TV\left(q\right)=F$

$\Rightarrow \mathrm{T}\mathrm{V}\left({\mathrm{S}}_2\right)=\mathrm{T}$

$\bar{x}=10$

$\Rightarrow \stackrel{?}{\mathrm{x}}=\frac{63+\mathrm{a}+\mathrm{b}}{8}=10$

$\Rightarrow a+b=17$

Since, variance is independent of origin.

So, we subtract 10 from each observation.

So,  ${\sigma }^2=13.5=\frac{79+(a-10{)}^2+(b-10{)}^2}{8}$

$\Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2-20(\mathrm{a}+\mathrm{b})=-171$

$\Rightarrow a^2+b^2=169$

From (1) and (2) ; $a=12$ and $b=5$

Some of the most frequently tested topics for high scores are finding the adjoint and inverse of a matrix, solving equations using Cramer's Rule, expansion using cofactors, and properties of determinants. If students practice questions based on these topics, they can get high scores in the CBSE Board exam and entrance exams.