mathematics-class-12-unit-04-chapter-10-determinants-l-10-11-fbz9jlntlt0

<h3 id="success-stylefont-size25px-determinants-l-10-success-1"><Success style="font-size:25px"> Determinants L-10 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-1-primary"><primary style="font-size:24px"> Problem 1 </primary></h3>
<hr>
<main>
<div style='float: left; width:99%;font-size:30px; text-align:left;'>
<ul>
<li>
<p>What is the determinant of</p>
</li>
<li>
<p>$A=\left(\begin{array}{lll}a & a^2 & a^3-1 \\ a^w & a^{2 \omega} & a^{3 \omega}-1 \\ a^{\omega^2} & a^{2 \omega^2} & a^{3 \omega z}-1\end{array}\right)$</p>
</li>
<li>
<p>What $\omega$ is the cube root of unity</p>
</li>
<li>
<p>Soln.  And we known that</p>
</li>
<li>
<p>$ 1 + \omega +  {\omega}^2 = 0 $</p>
</li>
</ul>
</div>
<div style='float: right; width:1%;'>
	
</div>
</main>
<hr>
<footer style = "font-size: 18px"> $\rightarrow$ Determinants Lecture - 10 $\rightarrow$ <span style = "color:blue"> Problem 1 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Determinants L-10 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

- $ \therefore|A| $

- $ = \left|\begin{array}{lll}a & a^2 & a^3 \\\ a^\omega & a^{2 \omega} & a^{3 \omega} \\\a^{\omega^2} & a^{2 \omega^2} & a^{3 \omega^2}\end{array}\right| -  \left|\begin{array}{lll}a & a^2 & 1 \\\a^{\omega} & a^{2 \omega} & 1 \\\a^{\omega^2} & a^{2 \omega^2} & 1\end{array}\right| $

- $ \left|A_1\right|=a \cdot a^\omega \cdot a^{\omega^2}\left|\begin{array}{ccc}1 & a & a^2 \\\1 & a^\omega & a^{2 \omega} \\\1 & a^{\omega^2} & a^{2 \omega^2}\end{array}\right| $

- $ =a^{{1+\omega+\omega^2}{=0}}\left|\begin{array}{lll}1 & a & a^2 \\\1 & a^\omega & a^2 \omega \\\1 & a^{\omega^2} & a^{2 \omega^2}\end{array}\right| $
</div>
<div style='float: right; width:0%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px">Determinants Lecture - 10 $\rightarrow$ Problem 1 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Determinants L-10 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

- $ \therefore\left|A_1\right|=\left|\begin{array}{lll}1 & a & a^2 \\\1 & a^\omega & a^{2 \omega} \\\1 & a^{\omega^2} &a^{2\omega^2}\end{array}\right| $

- Now $ |A_2| = \left|\begin{array}{lll} a & a^2 & 1 \\\a^{\omega} & a^{2\omega} & 1 \\\a^{\omega^2} & a^{2\omega^2} & 1\end{array}\right| $

- We can get $ |A_2| $ from $ |A_1| $  by first interchanging column with column , then interchanging column 1 and column 2
</div>
<div style='float: right; width:1%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px">Problem 1 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Remarks </footer>

<h3 id="success-stylefont-size25px-determinants-l-10-success-2"><Success style="font-size:25px"> Determinants L-10 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary"><primary style="font-size:24px"> Solution </primary></h3>
<hr>
<main>
<div style='float: left; width:99%;font-size:30px; text-align:left;'>
<ul>
<li>
<p>We know that $A(\operatorname{adj} A)=|A| I_n$</p>
</li>
<li>
<p>$
\therefore A \frac{(\operatorname{adi} A)}{|A|}=I_n
$</p>
</li>
<li>
<p>$\therefore$ We obtain a matrix $\frac{(\operatorname{adi} A)}{|A|} ∃ \frac{(\operatorname{adj} A)}{|A|}=I$.</p>
</li>
<li>
<p>and parrallely $ \frac{(\operatorname{adj} A)}{|A|} \cdot A=I_n$.</p>
</li>
</ul>
</div>
<div style='float: right; width:1%;'>
		
</div>
</main>
<hr>
<footer style = "font-size: 18px">Note $\rightarrow$ Problem 2 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Remarks $\rightarrow$ Result </footer>

### <Success style="font-size:25px"> Determinants L-10 </Success>
### <primary style="font-size:24px"> Result </primary>
<hr>
<main>

- Result: If $A \\& B$ are non singular matrix then (adj $A B$ )
- $ =(\operatorname{adj} B)(\operatorname{adj} A) \text {. }$
- Proof consider $(A B)(\operatorname{adj} B)(\operatorname{adj} A)$

- $ =A(B(a d j B))(a d j A) $

- $ =A(|B| I)(\operatorname{adj} A) $

- $ =|B|(A \cdot I \cdot(\operatorname{adj} A)) $

- $ =|B|(A \cdot(\operatorname{adj} A \mid)=|B||A| I $

</div>
</main>
<hr>
<footer style = "font-size: 18px">Remarks $\rightarrow$ Result $\rightarrow$ <span style = "color:blue"> Result </span> $\rightarrow$ Proof $\rightarrow$ Problem 3 </footer>

### <Success style="font-size:25px"> Determinants L-10 </Success>
### <primary style="font-size:24px"> Proof </primary>
<hr>
<main>

- Since (adj $A B$ ) in such that

- $(A B)($ adj $A B)=|A B| I$
- $ =|A||B| I $
- $\therefore$ We find that

- $ (A B)(\operatorname{adj} A B)=(A B)(\operatorname{adj} B)(\operatorname{adj} A) $

- $ =|A||B| I .$

- $\therefore$ By Multiplying both sides with $(A B)^{-1}$

- we get $( adj A B) $ = $ (\operatorname{adj} B)(\operatorname{adj} A)$
</div>
<div style='float: right; width:0%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px">Result $\rightarrow$ Result $\rightarrow$ <span style = "color:blue"> Proof </span> $\rightarrow$ Problem 3 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Determinants L-10 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

- $ A=\left(\begin{array}{ccc}1 & -1 & 2 \\\\ 0 & 2 & -3 \\\\ 3 & -2 & 4\end{array}\right)$

- $\therefore A_{11}=4.2-(-3)(-2)-8-6=2 \\\\  A_{12}=(-1)^{1+2}(0.4-(-3) .3)=-9 $

- $A_{13}=0 .(-2)-3.2=-6 $

- $A_{21}=-4-2(-2)=0 $

- $A_{22}=4.1-3.2=-2 $

- $A_{23}=(-1)^{2+3}\left(1 \cdot(-2)-(-1)(3)\right)=-(-2+3) =-1 $

- $A_ {31}=-1, ~ A_ {32}  =3 ~ $ and $~ A_{33}=2  $
</div>
<div style='float: right; width:1%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px">Proof $\rightarrow$ Problem 3 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem 4 </footer>

### <Success style="font-size:25px"> Determinants L-10 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

$
\begin{aligned}
& \therefore(\operatorname{adj} A)=\left(\begin{array}{ccc}
 2 & 0 & -1 \\\\
-9 & -2 & 3 \\\\
-6 & -1 & 2
\end{array}\right) \\\\
& \therefore A^{-1}=\left(\begin{array}{ccc}
-2 & 0 & 1 \\\\
 9 & 2 & -3 \\\\
 6 & 1 & -2
\end{array}\right) \\
&
\end{aligned}
  $

- Verify that $A \cdot A^{-1}=I_3$ $A^{-1} \cdot A=I_3$

</div>
</main>
<hr>
<footer style = "font-size: 18px">Problem 3 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem 4 $\rightarrow$ Problem 5 </footer>

### <Success style="font-size:25px"> Determinants L-10 </Success>
### <primary style="font-size:24px"> Problem 5 </primary>
<hr>
<main>

- Ex what is the inverse of
- $A=\left(\begin{array}{ccc}1 & 0 & 0 \\\0 & \cos \alpha & \sin \alpha \\\0 & \sin \alpha & -\cos \alpha\end{array}\right)$

- so we first compute the cofactor A.

- $A_{11}=\cos \alpha(-\cos \alpha)-\sin \alpha \cdot \sin \alpha $

- $=-\cos ^2 \alpha-\sin ^2 \alpha =0 \cdot 1 $

- $A_{12}=0 \cdot(-\cos \alpha)-0(\sin \alpha)=0 $

- $A_{13}=0 \cdot \sin \alpha-0 \cos \alpha=0 $

- $A_{21}=0(-\cos \alpha)-0(\sin \alpha)=0 $

- $A_{22}=-\cos \alpha-0=-\cos \alpha$

</div>
</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Problem 4 $\rightarrow$ <span style = "color:blue"> Problem 5 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-determinants-l-10-success-3"><Success style="font-size:25px"> Determinants L-10 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-1"><primary style="font-size:24px"> Solution </primary></h3>
<hr>
<main>
<div style='float: left; width:99%;font-size:30px; text-align:left;'>
<ul>
<li>
<p>In a similar way:</p>
</li>
<li>
<p>$A_{23}  =(-1)^{2+3}(1 \cdot \sin \alpha-0) = $</p>
</li>
<li>
<p>$-\sin \alpha $</p>
</li>
<li>
<p>$A_{31}  =0 \cdot \sin \alpha-0 \sin \alpha=0 $</p>
</li>
<li>
<p>$A_{32}  =(-1)(1 \cdot \sin \alpha-0)= -\sin \alpha $</p>
</li>
<li>
<p>$A_{33}  = 1 - \cos \alpha - 0 $</p>
</li>
<li>
<p>$ =\cos \alpha .$</p>
</li>
</ul>
</div>
<div style='float: right; width:1%;'>
		
</div>
</main>
<hr>
<footer style = "font-size: 18px">Problem 4 $\rightarrow$ Problem 5 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Determinants L-10 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

- $ \therefore(\operatorname{adj} A) $ $ =\left(\begin{array}{ccc}-1 & 0 & 0 \\\0 & -\cos \alpha & -\sin \alpha \\\0 & -\sin \alpha & \cos \alpha\end{array}\right) $

- $ \therefore A \text { (adj } A) $

- $ =\left[\begin{array}{ccc}1 & 0 & 0 \\\0 & \operatorname{cos} \alpha & \sin \alpha \\\0 & \sin \alpha & -\cos \alpha\end{array}\right]\left[\begin{array}{ccc}-1 & 0 & 0 \\\0 & -\cos \lambda & -\sin \alpha \\\0 & -\sin \alpha & \cos \alpha\end{array}\right] $

- $ =\left[\begin{array}{ccc}-1 & 0 & 0 \\\0 & -\cos ^2 \alpha-\sin^2 \alpha & -\cos \alpha \cdot \sin \alpha +\cos \alpha \cdot \sin \alpha \\\0 & -\cos \alpha \cdot \sin \alpha + \cos\alpha \cdot \sin \alpha & -\sin^2 \alpha - \cos^2 \alpha \end{array}\right] $ 
</div>
<div style='float: right; width:1%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px">Problem 5 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem 6 </footer>

<h3 id="success-stylefont-size25px-determinants-l-10-success-4"><Success style="font-size:25px"> Determinants L-10 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-2"><primary style="font-size:24px"> Solution </primary></h3>
<hr>
<main>
<div style='float: left; width:99%;font-size:30px; text-align:left;'>
<ul>
<li>
<p>$\begin{aligned}\therefore A(\operatorname{adj} A) & =\left[\begin{array}{ccc}-1 & 0 & 0 \\0 & -1 & 0 \\0 & 0 & -1\end{array}\right]
\text { i.e }(-1) I_3 & =|A| I_3 .\end{aligned}$</p>
</li>
<li>
<p>$\therefore$ The inverse of $A$ is</p>
</li>
<li>
<p>$\begin{aligned}& \frac{(\operatorname{adj} A)}{|A|}=\left(\begin{array}{ccc}-1 & 0 & 0 \\0 & -\cos \alpha & -\sin \alpha \\0 & -\sin \alpha & \cos \alpha\end{array}\right) \end{aligned}$</p>
</li>
<li>
<p>$\begin{aligned}& =\left(\begin{array}{ccc}1 & 0 & 0 \\0 & \cos \alpha & \sin \alpha \\0 & \sin \alpha & -\cos \alpha
\end{array}\right)^{-1} \&\end{aligned}$</p>
</li>
</ul>
</div>
<div style='float: right; width:1%;'>
		
</div>
</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem 6 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Determinants L-10 </Success>
### <primary style="font-size:24px"> Problem 6 </primary>
<hr>
<main>

- Ex : Show that the matrix $A=\left[\begin{array}{lll}1 & 2 & 2 \\\ 2 & 1 & 2 \\\ 2 & 2 & 1\end{array}\right]$ satisfies the equation $A^2-4 A-5 I_3=C$ where $O$ in the 0 -mamix of orden $3 \times 3$.
- Also, find out the inverse 8 the above matrix.

- since
- $A=\left[\begin{array}{lll}1 & 2 & 2 \\\2 & 1 & 2 \\\2 & 2 & 1\end{array}\right]$

- $A^2=A \cdot A  =\left[\begin{array}{lll}1 & 2 & 2 \\\2 & 1 & 2 \\\2 & 2 & 1\end{array}\right]\left[\begin{array}{lll}1 & 2 & 2 \\\2 & 1 & 2 \\\2 & 2 & 1\end{array}\right]  =\left[\begin{array}{lll}9 & 8 & 8 \\\8 & 9 & 8 \\\8 & 8 & 9\end{array}\right] $

</div>
<div style='float: right; width:1%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem 6 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Determinants L-10 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

- $\begin{aligned} & \therefore A^2-4 A \\\ & =\left[\begin{array}{lll}9 & 8 & 8 \\\ 8 & 9 & 8 \\\ 8 & 8 & 9\end{array}\right]-4\left[\begin{array}{lll}1 & 2 & 2 \\\ 2 & 1 & 2 \\\\ 2 & 2 & 1\end{array}\right] \\\\ & =\left[\begin{array}{ccc}5 & 0 & 0 \\\ 0 & 5 & 0 \\\ 0 & 0 & 5\end{array}\right]=5 I_3 \\\ & \therefore A^2-4 A=5 I \\\ & \therefore A^2-4 A-5 I=0_{3 \times 3}\end{aligned}$

</div>
</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Problem 6 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem 7 </footer>

### <Success style="font-size:25px"> Determinants L-10 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

- $\therefore \quad A^2-4 A=5 I$

- Pre-multiplying by $A^{-1}$

- $ A^{-1}\left(A^2-4 A\right)=5 \cdot A^{-1} \cdot I \quad \text { or } \quad A-4 I=5 \cdot A^{-1} $

- $ \therefore A^{-1} \cdot \frac{A-4 \cdot I}{5} $

- $ = \frac{1}{5}\left(\begin{array}{lll}1 & 2 & 2 \\\2 & 1 & 2 \\\2 & 2 & 1\end{array}\right) -\left(\begin{array}{lll}4 & 0 & 0 \\\0 & 4 & 0 \\\0 & 0 & 4\end{array}\right)$

- $ =\frac{1}{5}\left(\begin{array}{ccc}-3 & 2 & 2 \\\2 & -3 & 2 \\\2 & 2 & -3\end{array}\right) $
- $=\left(\begin{array}{ccc}-3 / 5 & 2 / 5 & 2 / 5 \\\2 / 5 & -3 / 5 & 2 / 5 \\\2 / 5 & 2 / 5 & -3 / 5\end{array}\right) $

- Verify that $A \cdot A^{-1}=A^{-1} A$ $=I$.
</div>

</div>
</main>
<hr>
<footer style = "font-size: 18px">Problem 6 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem 7 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Determinants L-10 </Success>
### <primary style="font-size:24px"> Problem 7 </primary>
<hr>
<main>

- What is determinant if (adj $(\operatorname{adj} A))$
- when $|A|$ is given

- Soln.  we know $|A(\operatorname{adj} A)|$

- $\begin{aligned}& =|A||(\operatorname{ad} j A)|=|| A\left|I_n\right| \\& =\left|\begin{array}{cccc}
|A| & & & \\\\& |A| & & \\\\& 0 & & |A|\end{array}\right|_{\operatorname{n × n}} \\&\end{aligned} $

- $=|A|^n $

- $ \therefore|A||(\operatorname{adi} A)|=|A|^n $

- $ \therefore \mid \operatorname{adj} A)\left.|=| A\right|^{n-1} $

- $\therefore|\operatorname{adj}(\operatorname{adj} A)|=\left(|A|^{n-1}\right)^{n-1}$

</div>
<div style='float: right; width:1%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem 7 </span> $\rightarrow$ Solution $\rightarrow$ Thank you </footer>