Determinants Test - 17

$$\displaystyle A=\begin{bmatrix} 5 & 5\alpha  & \alpha  \\ 0 & \alpha  & 5\alpha  \\ 0 & 0 & 5 \end{bmatrix}$$

$$\cos^3 x\begin{vmatrix} \ 1& tanx & tanx\\ tanx & \ 1 & tanx\\ tanx & tanx & \ 1\end{vmatrix} = 0$$

$$\begin{vmatrix} 0 & \dfrac { 8 }{ 3 }  & 1 \\ 1 & 3 & 1 \\ 82 & 30 & 1 \end{vmatrix}$$

$$=216-216=0$$

Hence, the given points are collinear.

It is given that  $$A =\begin{bmatrix} 2 & -3\\  -4 & 1 \end{bmatrix}$$

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

Given, $$[P]=a_{ij}$$ is a order 3 matrix.

Let $$Q=\begin{bmatrix} { 2 }^{ 2 }{ a }_{ 11 } & { 2 }^{ 3 }{ a }_{ 12 } & { 2 }^{ 4 }{ a }_{ 13 } \\ { 2 }^{ 3 }{ a }_{ 21 } & { 2 }^{ 4 }{ a }_{ 22 } & { 2 }^{ 5 }{ a }_{ 23 } \\ { 2 }^{ 4 }{ a }_{ 31 } & { 2 }^{ 5 }{ a }_{ 32 } & { 2 }^{ 6 }{ a }_{ 33 } \end{bmatrix}$$

$$|Q|=\begin{vmatrix} { 2 }^{ 2 }{ a }_{ 11 } & { 2 }^{ 3 }{ a }_{ 12 } & { 2 }^{ 4 }{ a }_{ 13 } \\ { 2 }^{ 3 }{ a }_{ 21 } & { 2 }^{ 4 }{ a }_{ 22 } & { 2 }^{ 5 }{ a }_{ 23 } \\ { 2 }^{ 4 }{ a }_{ 31 } & { 2 }^{ 5 }{ a }_{ 32 } & { 2 }^{ 6 }{ a }_{ 33 } \end{vmatrix}$$

$$\Rightarrow Q={ 2 }^{ 2 }{ 2 }^{ 3 }{ 2 }^{ 4 }\begin{bmatrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { 2 }{ a }_{ 21 } & { 2 }{ a }_{ 22 } & { 2 }{ a }_{ 23 } \\ { 2 }^{ 2 }{ a }_{ 31 } & { 2 }^{ 2 }{ a }_{ 32 } & { 2 }^{ 2 }{ a }_{ 33 } \end{bmatrix}$$

$$\Rightarrow |Q|={ 2 }^{ 9 }{ \times 2 }\times { 2 }^{ 2 }\begin{vmatrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{vmatrix}$$

$$\Rightarrow |Q|=2^{12}|P|$$
$$\Rightarrow |Q|=2^{13}$$

$$\mathrm{P}\equiv(-\sin(\beta-\alpha), -\cos\beta)\equiv(\mathrm{x}_{1}, \mathrm{y}_{1})$$
$$\mathrm{Q}\equiv(\cos(\beta-\alpha), \sin\beta)\equiv(\mathrm{x}_{2}, \mathrm{y}_{2})$$
and $$\mathrm{R}\equiv(\mathrm{x}_{2}\cos\theta+\mathrm{x}_{1}\sin\theta, \mathrm{y}_{2}\cos\theta+\mathrm{y}_{1}\sin\theta)$$
We see that $$\displaystyle \mathrm{T}\equiv(\frac{\mathrm{x}_{2}\cos\theta+\mathrm{x}_{1}\sin\theta}{\cos\theta+\sin\theta}, \frac{\mathrm{y}_{2}\cos\theta+\mathrm{y}_{1}\sin\theta}{\cos\theta+\sin\theta})$$ and $$\mathrm{P},\ \mathrm{Q},\ \mathrm{T}$$ are collinear
$$\Rightarrow \mathrm{P},\ \mathrm{Q},\ \mathrm{R}$$ are non-collinear.

Given $$A=\begin{bmatrix} a\quad  & b \\ c & a \end{bmatrix},a,b,c\in \left\{ 0,1,2,...,p-1 \right\} $$
If A is skew -symmetric matrix, then a =0, b=-c
$$\therefore \left| A \right| =-{ \left| b \right|  }^{ 2 }$$
Thus, p divides $$\left| A \right| $$ only when $$b=0$$   ...(1)
Again, if A is symmetric matrix, then $$v=c$$ and $$\left| A \right| ={ a }^{ 2 }-{ b }^{ 2 }$$
Thus p divides $$\left| A \right| $$ is either p divides $$(a-b)$$ or o divides $$(a+b)$$
p divides $$(a-b)$$ only when $$a=b$$
i.e $$a=b\in \left\{ 0,1,2,...,\left( p-1 \right)  \right\} $$
i.e p choice
p divides $$(a+b)$$ $$\Rightarrow $$ p choice including $$a=b=0$$ in (1)
Therefore total number of choices are $$\left( p+p+1 \right) =2p-1$$

Hence $$U=\begin{bmatrix} 1 & 2 & 2 \\ -2 & -1 & -1 \\ 1 & -4 & -3 \end{bmatrix}\Rightarrow \left| U \right| =3$$