Find the total number of distinct $x \in \mathbb{R}$ for which

$\left|\begin{array}{lll}x & x^2 & 1+x^3 \\ 2 x & 4 x^2 & 1+8 x^3 \\ 3 x & 9 x^2 & 1+27 x^3\end{array}\right|=10$

Ans:-

$\left|\begin{array}{ccc}x & x^2 & 1 \\ 2 x & 4 x^2 & 1 \\ 3 x & 9 x^2 & 1\end{array}\right|+\left|\begin{array}{ccc} x & x^2 &x^3 \\ 2 x & 4 x^2 & 8 x^3 \\ 3 x & 9 x^2 & 27 x^3 \end{array}\right|=10$

$x \times x^2\left|\begin{array}{lll}1 & 1 & 1 \\ 2 & 4 & 1 \\ 3 & 9 & 1\end{array}\right|+x \times x^2 \times x^3\left|\begin{array}{lll}1 & 1 & 1 \\ 2 & 4 & 8 \\ 3 & 9 & 27\end{array}\right|=10$

$\Rightarrow$ $x^3(1(4-9)-1(2-3)+1(18-12))$ + $x^6\times2\times 3$ $\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 3 & 9\end{array}\right|=10$

$2 x^3+12 x^6=10$

$\Rightarrow 6 x^6+x^3-5=0$

$\Rightarrow \quad 6 x^6+6 x^3-5 x^3-5=0$

$\Rightarrow \quad 6 x^3\left(x^3+1\right)-5\left(x^3+1\right)=0$

$\Rightarrow \quad\left(x^3+1\right)\left(6 x^3-5\right)=0$

$\begin{array}{cc}\left(x^3+1\right)=0 \quad \text { or } & 6 x^3-5=0 \\ x^3+a^3=(x+a)\left(x^2-a x+a^2\right)=0 & x^3-5 / 6=0\end{array}$

$\Rightarrow x=-a, x=a \pm \sqrt{a^2-4 a^2} \text { are roots of } \\ x=\frac{a+\sqrt{3}^2 i}{2} \quad x^3+a^3=0$

Let $A=\left[\begin{array}{ccc}(1+\alpha)^2 & (1+2 \alpha)^2 & (1+3-\alpha)^2 \\ (2+\alpha)^2 & (2+2 \alpha)^2 & (2+3 \alpha)^2 \\ (3+\alpha)^2 & (3+2 \alpha)^2 & (3+3 \alpha)^2\end{array}\right]$

be $3 \times 3$ matrix such that $\operatorname{det} A=-648 \alpha$

Then, what will be the value of $\alpha$ ?

$\operatorname{det} A=\left|\begin{array}{lll}1+2 \alpha+\alpha^2 & 1+4 \alpha+4 \alpha^2 & 1+6 \alpha+4 \alpha^2 \\ 4+4 \alpha+\alpha^2 & 4+8 \alpha+4 \alpha^2 & 4+12 \alpha+\alpha^2 \\ 9+6 \alpha+\alpha^2 & 9+12 \alpha+4 \alpha^2 & 9+18 \alpha+9 \alpha^2\end{array}\right|$

$R_2 \rightarrow R_2-R_1$, $\quad R_3 \rightarrow R_3-R_1$

$=\left|\begin{array}{ccc}1+2 \alpha+\alpha^2 & 1+4 \alpha+4 \alpha^2 & 1+6 \alpha+9 \alpha^2 \\ 3+2 \alpha & 3+4 \alpha & 3+6 \alpha \\ 8+4 \alpha & 8+8 \alpha & 8+12 \alpha\end{array}\right|$

$R_3 \rightarrow R_3-2 R_2$

$\operatorname{det} A=\left|\begin{array}{ccc}1+2 \alpha+\alpha^2 & 1+4 \alpha+4 \alpha^2 & 1+6 \alpha+9 \alpha^2 \\ 3+2 \alpha & 3+4 \alpha & 3+6 \alpha \\ 2 & 2 & 2\end{array}\right|$

$c_2 \rightarrow c_2-c_1, c_3 \rightarrow c_3-c_1$

$=\left|\begin{array}{ccc}1+2 \alpha+\alpha^2 & 2 \alpha+3 \alpha^2 & 4 \alpha+2 \alpha^2 \\ 3+2 \alpha & 2 \alpha & 4 \alpha \\ 2 & 0 & 0\end{array}\right|$

$=2\left(4 \alpha\left(2 \alpha+3 \alpha^2\right)-2 \alpha\left(4 \alpha+8 \alpha^2\right)\right)$

let $M$ and $N$ be two $3 \times 3$ matrices which that $M N=N M$. Further if $M \neq N^2$ and $M^2=N^4$, show that

i) $\operatorname{det}\left(M^2+M N^2\right)=0$

ii) There is a $3 \times 3$ nonzero matrix $U$ which that $\left(M^2+M N^2\right) U$ is zero matrix.

$M N =N M \\ M N^2 =N M N \\ M N^2 =N N M=N^2 M \\ M N^2 =N^2 M$

Take

$M^2=N^4$

$\begin{array}{ll}\Rightarrow & M^2-N^4=0 \\ \Rightarrow & M^2-M N^2+M N^2-N^4=0 \\ \Rightarrow & M\left(M-N^2\right)+N^2 M-N^4=0 \\ \Rightarrow & M\left(M-N^2\right)+N^2\left(M-N^2\right)=0\end{array}$

$\left.\left(M+N^2\right) ( M-N^2\right)=0$

Case I $\quad \operatorname{det}\left(M+N^2\right)=0$

$\Rightarrow \operatorname{det}\left(M^2+M N^2\right)$ $=\operatorname{det}\left(M\left(M+N^2\right)\right)$

$=\operatorname{det} M \cdot \operatorname{det}\left(M+N^2\right) =0$

Case II

Case II $\operatorname{det}\left(M+N^2\right) \neq 0$

$\Rightarrow\left(M+N^2\right)^{-1} \text { exists. }$

$\Rightarrow\left(M+N^2\right)^{-1}\left(M+N^2\right)\left(M-N^2\right)=0$

$\Rightarrow \quad M-N^2=0 \quad \quad \Rightarrow \text {which is not possible. }$

$\Rightarrow$ Case II cannot occur.

$\Rightarrow \quad \operatorname{det}\left(M^2+M N^2\right)=0$

ii) Need to show $\left(M^2+M N^2\right) U=0 \rightarrow$ zero matrix for some non - zero matrix $U$

from (1) $\quad\left(M+N^2\right)\left(M-N^2\right)=0$

$\Rightarrow \quad M(M+N^2)(M-N^2)=0 \\ \Rightarrow \quad(M^2+M N^2)(M-N^2)=0$

$U=M-N^2 \neq 0 \Rightarrow(M^2+M N^2) U=0$

Let

$M=\left[\begin{array}{lll}0 & 1 & a \\1 & 2 & 3 \\3 & b & 1\end{array}\right]$

adj $M=\left[\begin{array}{ccc}-1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1\end{array}\right]$, where $a$ and $b$ are real number

Then, show that

i) $a+b=3$

ii) $(\operatorname{adj} M)^{-1}+\operatorname{adj} M^{-1}=-M$

iii) if $M\left[\begin{array}{c}\alpha \\ \beta \\ \gamma\end{array}\right]=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$, then $\alpha-\beta+\gamma=3$.

i) $M=\left[\begin{array}{lll}0 & 1 & a \\ 1 & 2 & 3 \\3 & b & 1\end{array}\right]$

(1,1) co-factor of $M = \left|\begin{array}{ll}2 & 3 \\ 3 & 1\end{array}\right|$ $= 2-3b$

$\Rightarrow 2-3b=-1$ $\Rightarrow 3b=3$ $\rightarrow b=1$

(3,1) co-factor of $M = \left|\begin{array}{ll}1 & a \\ 2 & 3\end{array}\right|$ $= 3-2a$

$\Rightarrow 3-2a = -1 \quad \Rightarrow 2a=4 \quad \Rightarrow a=2$

ii) $(\operatorname{adj} M)^{-1}+\operatorname{adj} M^{-1}=-M$

$M=\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$

$det M = -1(1-9)+2(1-6)= 8-10=-2$

$\operatorname{det}(\operatorname{adj} M)=(\operatorname{det} M)^2=4 \neq 0$

$\Rightarrow(\operatorname{adj} M)^{-1}+\operatorname{adj} M^{-1}=-M$

iii) If $M \left[\begin{array}{l}\alpha \\ \beta \\ \gamma\end{array}\right]=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$, then

$\alpha-\beta+\gamma=3$

$\operatorname{det} M=-2, \quad M^{-1}=\frac{\operatorname{adj} M}{\operatorname{det} M}$ $M^{-1}=\left[\begin{array}{ccc}1/2 & -1/2 & 1/2 \\ -4 & 3 & -1 \\ 5/2 & -3/2 & 1/2 \end{array}\right]$

let $x \in \mathbb{R}$ and let

$P=\left[\begin{array}{lll}1 & 1 & 1 \\0 & 2 & 2 \\0 & 0 & 3\end{array}\right], \sum=\left[\begin{array}{lll}2 & x & x \\0 & 4 & 0 \\x & x & 6\end{array}\right]$ & $\quad R=P$ & $P^{-1}$, then show that

i) $\operatorname{det} R=\operatorname{det}\left[\begin{array}{lll}2 & x & x \\ 0 & 4 & 0 \\ x & x & 5\end{array}\right]+8$

ii) For $x=0$, if $R\left[\begin{array}{c}1 \\ a \\ b\end{array}\right]=6\left[\begin{array}{l}1 \\ a \\ b\end{array}\right]$, then $a+b=5$.

$R=P Q P^{-1}$

$\operatorname{det} R=\operatorname{det} P \operatorname{det} Q \cdot \operatorname{det} P^{-1}$

$=\operatorname{det} P \operatorname{det} 2 \frac{1}{\operatorname{det} P}$

$=\operatorname{det} Q$

$\operatorname{det} R =\operatorname{det}\left[\begin{array}{lll}2 & x & x+0 \\0 & 4 & 0+0 \\x & x & 5+1\end{array}\right] \\ =\operatorname{det}\left[\begin{array}{ccc}2 & x & x \\0 & 4 & 0 \\x & x & 5\end{array}\right]+\operatorname{det}\left[\begin{array}{lll}2 & x & 0 \\0 & 4 & 0 \\x & x & 1\end{array}\right]$

$\operatorname{det} R=\operatorname{det}\left[\begin{array}{ccc}2 & x & x \\0 & 4 & 0 \\x & x & 5\end{array}\right]+8$

ii) If $x=0$

$\begin{array}{ll}R=\operatorname{PQP}^{-1} Q=\left[\begin{array}{ccc}2 & 0 & 0 \\0 & 4 & 0 \\0 & 0 & 6\end{array}\right] \\P=\left[\begin{array}{lll}1 & 1 & 1 \\0 & 2 & 2 \\0 & 0 & 3\end{array}\right], \quad P^{-1}=\left[\begin{array}{ccc}1 & -1/2 & 0 \\0 & 1/2 & -1/3 \\0 & 0 & 1/3 \end{array}\right]\end{array}$

$\operatorname{det} P=6$

$\left.\begin{array}{c}a+\frac{2}{3} b=4 \\ 2 a-\frac{4}{3} b=0\end{array}\right\rangle \rightarrow \begin{aligned} & 4 a=8 \\ & a=2 \\ & b=3\end{aligned}$

$R\left[\begin{array}{l}1 \\ a \\ b \end{array}\right]=6\left[\begin{array}{l}1 \\ a \\ b \end{array}\right] \\ 2+a+\frac{2}{3} b=6 \\ 4 a + 4/3 b=6 a \\ 6 b=6 b \\ a+b=2+3=5$