mathematics-class-12-unit-03-chapter-09-matrix-and-determinant-l-3-5-pwvqcvyz6em

### <Success style="font-size:25px"> Matrix And Determinant L-3 </Success>
### <primary style="font-size:24px"> Problem </primary>
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-  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:-
- $\left|\begin{array}{ccc}x & x^2 & 1 \\\ 2 x & 4 x^2 & 1 \\\ 3 x & 9 x^2 & 1\end{array}\right|+\left|\begin{array}{ccc} x & x^2 &x^3 \\\ 2 x & 4 x^2 & 8 x^3 \\\ 3 x & 9 x^2 & 27 x^3 \end{array}\right|=10$

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### <Success style="font-size:25px"> Matrix And Determinant L-3 </Success>
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- $ 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 $
 
 - $\Rightarrow$  $x^3(1(4-9)-1(2-3)+1(18-12))$ + $x^6\times2\times 3$  $\left|\begin{array}{ccc}1 & 1 & 1 \\\ 1 & 2 & 4 \\\ 1 &  3 &  9\end{array}\right|=10$
 
- $ \Rightarrow x^3(-5+1+6)+6x^6[1(18-12)-1(9-4) +1(3-2)]=10 \\\\ \Rightarrow  2 x^3+6 x^6(6-5+1)=10$

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### <Success style="font-size:25px"> Matrix And Determinant L-3 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $  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 $

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<ul>
<li>$x^3+a^3=0$ has only ore real root which is $x=-a$</li>
<li>Similarly, $x^3-a^3=0$ has only one real root given by $x=a$ $\Rightarrow x^3+1=0$ has $x=-1$ one real root</li>
<li>$4 x^3-5 / 6 \Rightarrow$ has real root $x=(5 / 2)^{1 / 3}$</li>
<li>$\Rightarrow$ we get two real roots for</li>
<li>$\left|\begin{array}{ccc}x & x^2 & 1+x^3 \\ 2 x & 4 x^2 & 1+8 x^3 \\ 3 x & 9 x^2 & 1+27 x^3\end{array}\right|=10$</li>
<li>and they an given by</li>
<li>$x=-1 \quad \ \quad x=(5 / 6)^{1 / 3}$</li>
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### <Success style="font-size:25px"> Matrix And Determinant L-3 </Success>
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- $  \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|$
 
- $R_2 \rightarrow R_2-R_1$, $\quad R_3 \rightarrow R_3-R_1$
 
- $=\left|\begin{array}{ccc}1+2 \alpha+\alpha^2 & 1+4 \alpha+4 \alpha^2 & 1+6 \alpha+9 \alpha^2 \\\ 3+2 \alpha & 3+4 \alpha & 3+6 \alpha \\\ 8+4 \alpha & 8+8 \alpha & 8+12 \alpha\end{array}\right|$

- $R_3 \rightarrow R_3-2 R_2$

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- $ \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) $
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### <Success style="font-size:25px"> Matrix And Determinant L-3 </Success>
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- $\left.\left(M+N^2\right) ( M-N^2\right)=0$

- 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 $

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

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### <Success style="font-size:25px"> Matrix And Determinant L-3 </Success>
### <primary style="font-size:24px"> Problem </primary>
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-  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 $\alpha-\beta+\gamma=3$.

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<li>
<p>i) $M=\left[\begin{array}{lll}0 & 1 & a \\ 1 & 2 & 3 \\3 & b & 1\end{array}\right]$</p>
</li>
<li>
<p>(1,1) co-factor of  $M = \left|\begin{array}{ll}2 & 3 \\ 3 & 1\end{array}\right|$ $ = 2-3b$</p>
</li>
<li>
<p>$\Rightarrow 2-3b=-1 $ $\Rightarrow 3b=3$ $\rightarrow b=1$</p>
</li>
<li>
<p>(3,1) co-factor of $M = \left|\begin{array}{ll}1 & a \\ 2 & 3\end{array}\right|$ $ = 3-2a$</p>
</li>
<li>
<p>$\Rightarrow 3-2a = -1 \quad \Rightarrow 2a=4 \quad \Rightarrow a=2 $</p>
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<li>
<p>ii) $(\operatorname{adj} M)^{-1}+\operatorname{adj} M^{-1}=-M$</p>
</li>
<li>
<p>$  M=\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$</p>
</li>
<li>
<p>$det M = -1(1-9)+2(1-6)= 8-10=-2 $</p>
</li>
<li>
<p>$\operatorname{det}(\operatorname{adj} M)=(\operatorname{det} M)^2=4 \neq 0$</p>
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- $(\operatorname{adj} M)(\operatorname{adj} M)^{-1}  =I \\\ \operatorname{det} M \cdot M^{-1}(\operatorname{adj} M)^{-1}  =I \\\ $
- $ \Rightarrow(\operatorname{adj} M)^{-1}  =\frac{M}{\operatorname{det} M} \\\ $ 
- $ \quad =\operatorname{det} M^{-1}\left(M^{-1}\right)^{-1} \\\  =\operatorname{adj} M^{-1} \\\ $ 
- $ (\operatorname{adj} M)^{-1}=\frac{M}{\operatorname{det} M}  =\operatorname{adj} M^{-1} \\\ $
- $ \Rightarrow(\operatorname{adj} M)^{-1}+\operatorname{adj} M^{-1}  =\frac{2 M}{\operatorname{det} M}=-M$

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<ul>
<li>
<p>$ \Rightarrow(\operatorname{adj} M)^{-1}+\operatorname{adj} M^{-1}=-M$</p>
</li>
<li>
<p>iii) If $M  \left[\begin{array}{l}\alpha \\ \beta \\ \gamma\end{array}\right]=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$, then</p>
</li>
<li>
<p>$\alpha-\beta+\gamma=3$</p>
</li>
<li>
<p>$ \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]$</p>
</li>
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<ul>
<li>$ \begin{aligned} & M\left[\begin{array}{l}\alpha \\ B \\ \gamma\end{array}\right]=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right] \\ & \Rightarrow \begin{array}{l}\alpha=1 \\ \beta=-1 \\ \gamma=1\end{array} \quad \Rightarrow\left[\begin{array}{l}\alpha \\ \beta \\ \gamma\end{array}\right]=M^{-1}\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right] \\ & \beta=-1, r=1, \Rightarrow \alpha-\beta+\gamma=3 \\ & =\left[\begin{array}{ccc} 1/2 & -1/2 & 1/2 \\ -4 & 3 & -1 \\ 5/2 & -3/2 & 1/2 \end{array}\right]\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right] \\ & {\left[\begin{array}{l}\alpha \\ \beta \\ \gamma\end{array}\right]=\left[\begin{array}{c}1 \\ -1 \\ 1\end{array}\right]} \\ & \end{aligned}$</li>
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### <Success style="font-size:25px"> Matrix And Determinant L-3 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- 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 $a+b=5$.

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![image]( /subject-images/iitpal/image/mathematics-class-12-unit-03-chapter-09-matrix-and-determinant-l-3-5-pwvqcvyz6em-22.jpg)

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

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

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

- $=\operatorname{det} Q$

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

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![image]( /subject-images/iitpal/image/mathematics-class-12-unit-03-chapter-09-matrix-and-determinant-l-3-5-pwvqcvyz6em-23.jpg)

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

- ii) If $x=0$

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

- $\operatorname{det} P=6$

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<li>$\begin{aligned}R & =\left[\begin{array}{lll}1 & 1 & 1 \\0 & 2 & 2 \\0 & 0 & 3\end{array}\right]\left[\begin{array}{lll}2 & 0 & 0 \\0 & 4 & 0 \\0 & 0 & 6\end{array}\right]\left[\begin{array}{ccc}1 & -1/2 & 0 \\ 0 & 1/2 & -1/3 \\ 0 & 0 & 1/3 \end{array}\right] \\ & =\left[\begin{array}{lll}1 & 1 & 1 \\0 & 2 & 2 \\ 0 & 0 & 3\end{array}\right]\left[\begin{array}{ccc}2 & -1 & 0 \\ 0 & 2 & -4 / 3 \\ 0 & 0 & 2\end{array}\right] \\ R & =\left[\begin{array}{lll}2 & 1 & 2/3 \\ 0 & 4 & 4 / 3 \\ 0 & 0 & 6\end{array}\right]\left.\right.\end{aligned}$</li>
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