mathematics-class-12-unit-04-chapter-11-determinants-l-11-11-8tb38zluiqc

<h3 id="success-stylefont-size25px-determinants-l-11-success-1"><Success style="font-size:25px"> Determinants L-11 </Success></h3>
<h3 id="primary-stylefont-size24px-system-of-linear-equations-primary"><primary style="font-size:24px"> System of linear equations </primary></h3>
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<p>An equation is said to he linear if the underlying variables have exponent 1.</p>
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<p>For example $a x+b y=c$</p>
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<p>$2 x+3 y=5$</p>
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<footer style = "font-size: 18px"> $\rightarrow$ Determinant lecture 11 $\rightarrow$ <span style = "color:blue"> System of linear equations </span> $\rightarrow$  System of linear equations $\rightarrow$ Example 1 </footer>

<h3 id="success-stylefont-size25px-determinants-l-11-success-2"><Success style="font-size:25px"> Determinants L-11 </Success></h3>
<h3 id="primary-stylefont-size24px--system-of-linear-equations-primary"><primary style="font-size:24px">  System of linear equations </primary></h3>
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<li>Systems of linear equations we consider multiple equation with multiple variable, and we like to solve for the variables such that the obtained values satisfy all the equations simultaneously.</li>
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<footer style = "font-size: 18px">Determinant lecture 11 $\rightarrow$ System of linear equations $\rightarrow$ <span style = "color:blue">  System of linear equations </span> $\rightarrow$ Example 1 $\rightarrow$ Example 1 </footer>

### <Success style="font-size:25px"> Determinants L-11 </Success>
### <primary style="font-size:24px"> Systems of equation </primary>
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- We apply matrix & matrix inverse based techniques:

- Let the systems of equation be

- $ a_{11} x_1+a_{12} x_2+\cdots+a_{1 n} x_n=b_1$

- $ a_{21} x_1+a_{22} x_2+\cdots+a_{2 n} x_n=b_2 $

- $a_{n 1} x_1+a_{n 2} x_2+\cdots+a_{n n} x_n=b_1$   
- We are looking at $x$ equation in $n$ unknowns, $x_1 \cdots x_n \text {. }$

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<footer style = "font-size: 18px">Example 1 $\rightarrow$ Example 1 $\rightarrow$ <span style = "color:blue"> Systems of equation </span> $\rightarrow$ AX = B $\rightarrow$ AX = B </footer>

### <Success style="font-size:25px"> Determinants L-11 </Success>
### <primary style="font-size:24px"> AX = B </primary>
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- Note that we can represent such a system of equation in a matrix form:

- $A_{n \times n} X_{n \times 1}=B_{n \times 1}$

- $or \left[\begin{array}{ccc}a_{11} & \cdots & a_{1 n} \\\ a_{21} & \cdots & a_{2 n} \\\ \vdots & & \\\ a_{n 1} & \cdots & a_{n n}\end{array}\right] \cdot\left[\begin{array}{c}x_1 \\\ \vdots \\ x_n\end{array}\right]=\left[\begin{array}{c}b_1 \\\ b_2 \\ \vdots \\\ b_n\end{array}\right]$
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<footer style = "font-size: 18px">Example 1 $\rightarrow$ Systems of equation $\rightarrow$ <span style = "color:blue"> AX = B </span> $\rightarrow$ AX = B $\rightarrow$ Example 2 </footer>

### <Success style="font-size:25px"> Determinants L-11 </Success>
### <primary style="font-size:24px"> Example 2 </primary>
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- consider 2 x+3 y=8

- 3 x-y = 1

- In matrix notation:

- $ \left[\begin{array}{cc}2 & 3 \\\ 3 & -1\end{array}\right]\left[\begin{array}{l}x \\\ y
\end{array}\right]=\left[\begin{array}{l}8 \\\ 1\end{array}\right]$

- $ Now |A|=\left|\begin{array}{cc}2 & 3 \\\ 3 & -1\end{array}\right|=-2-9=-11\neq 0$

- $ \therefore A^{-1} exists.$

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<footer style = "font-size: 18px">AX = B $\rightarrow$ AX = B $\rightarrow$ <span style = "color:blue"> Example 2 </span> $\rightarrow$ Example 2 $\rightarrow$ Example 3 </footer>

### <Success style="font-size:25px"> Determinants L-11 </Success>
### <primary style="font-size:24px"> Example 2 </primary>
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- We know:

- $ A^{-1}=\frac{(a d j A)}{|A|}$

- $ Now \left[\begin{array}{cc}2 & 3 \\\  3 & -1\end{array}\right] \quad \therefore(\operatorname{adj} A)=\left[\begin{array}{ll}-1 & -3 \\\ -3 & 2\end{array}\right].$

- $\therefore A^{-1}=\frac{\left[\begin{array}{cc}-1 & -3 \\\ -3 & 2\end{array}\right]}{-11}=\left(\begin{array}{cc}1 / 11 & 3 / 11 \\\ 3 / 11 & -2 / 11\end{array}\right) .$

- $\therefore A^{-1} B=\left(\begin{array}{cc}1 / 11 & 3 / 11 \\\ 3 / 11 & -2 / 11\end{array}\right)\left(\begin{array}{l}8 \\\ 1\end{array}\right)$

- $ =\left(\begin{array}{l}\frac{8}{11}+\frac{3}{11} \\\ \frac{24}{11}-\frac{2}{11}\end{array}\right)=\left(\begin{array}{l}1 \\\ 2\end{array}\right) \text {. }$

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<footer style = "font-size: 18px">AX = B $\rightarrow$ Example 2 $\rightarrow$ <span style = "color:blue"> Example 2 </span> $\rightarrow$ Example 3 $\rightarrow$ Example 3 </footer>

### <Success style="font-size:25px"> Determinants L-11 </Success>
### <primary style="font-size:24px"> Example 3 </primary>
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- We represent the system of equation as:

- $ \left[\begin{array}{lll}4 & 3 & 2 \\\ 2 & n & 6 \\\ 6 & 2 & 3\end{array}\right]\left[\begin{array}{l}x \\\ y \\\ z \end{array}\right]=\left[\begin{array}{l}60 \\\ 90 \\\ 70\end{array}\right]$

- where,

- $x$ in price of per kg of $x$ is price of onion.

- $y$ is price of  wheat.

- $z$ is price  of rice.   
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<footer style = "font-size: 18px">Example 2 $\rightarrow$ Example 3 $\rightarrow$ <span style = "color:blue"> Example 3 </span> $\rightarrow$ Example 3 $\rightarrow$ Example 3 </footer>

### <Success style="font-size:25px"> Determinants L-11 </Success>
### <primary style="font-size:24px"> Example 3 </primary>
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- $\therefore$ We need to compute $A^{-1}=\frac{(\operatorname{adj} A)}{|A|}$ we compute (adj $A$ ).

- $  A_{11}=4.3-6.2=0$

- $ A_{12}=(-1)(2.3-6.6)=30 $

- $ A_{13}=2.2-6.4=-20 $

- $ A_{21}=(-1)(3.3-2.2)=-5 $

- $ A_{22}=4.3-6.2=0 $

- $ A_{23}=(-1)(4.2-6.3)=10 $

- $ A_{31}=3.6-4.2=10$

- $ A_{32}=(-1)(4.6-2.2)=-20 $

- $ A_{33}=4.4-3.2=+10$

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<footer style = "font-size: 18px">Example 3 $\rightarrow$ Example 3 $\rightarrow$ <span style = "color:blue"> Example 3 </span> $\rightarrow$ Example 3 $\rightarrow$ Example 3 </footer>

### <Success style="font-size:25px"> Determinants L-11 </Success>
### <primary style="font-size:24px"> Example 3 </primary>
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- $ |A|=4.0+3.30+2(-20)$      
 
- $ =0+90-40=50 $

- $ \therefore A^{-1}=\frac{\left(\begin{array}{ccc}0 & -5 & 10 \\\ 30 & 0 & -20 \\\ -20 & 10 & 10\end{array}\right)}{50}$

- $ =\left(\begin{array}{ccc}0 & -\frac{1}{10} & \frac{1}{5} \\\ 3 / 5 & 0 & -2 / 5 \\\ -2 / 5 & \frac{1}{5} & 1 / 5\end{array}\right) \cdot\left[\begin{array}{l}60 \\\ 90 \\\ 70\end{array}\right]$

- $ \therefore\left(\begin{array}{l}x \\\ y \\\ 3\end{array}\right)=\left(\begin{array}{l}\ 5 \\\ 8 \\\ 8\end{array}\right)$

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- Three equations :

-  x + y + z = 6   
 
-  x + y + 2 z = 7   
 
-  3 x + y + z = 12   
 
- $\text { or }\left[\begin{array}{lll}1 & 1 & 1 \\\ 1 & 0 & 2 \\\ 3 & 1 & 1\end{array}\right]\left[\begin{array}{l}x \\\ y \\\ z \end{array}\right]=\left[\begin{array}{l}6 \\\ 7 \\\ 12\end{array}\right]$

- Determinant of $A$

- $=1 \cdot(0-2)-1 \cdot(1-6)+1(1-0)$

- $=-2+5+1=4 \neq 0$

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<footer style = "font-size: 18px">Example 3 $\rightarrow$ Example 4 $\rightarrow$ <span style = "color:blue"> Example 4 </span> $\rightarrow$ Example 4 $\rightarrow$ Example 4 </footer>

### <Success style="font-size:25px"> Determinants L-11 </Success>
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- $\therefore(\operatorname{adj} A)=\left(\begin{array}{ccc}-2 & 0 & 2 \\\ 5 & -2 & -1 \\\ 1 & 2 & -1\end{array}\right)$

- $\therefore A^{-1}=\frac{1}{4}\left(\begin{array}{ccc}-2 & 0 & 2 \\\ 5 & -2 & -1 \\\ 1 & 2 & -1\end{array}\right)$.

- $\therefore$ Solution of the equation

- $ =A^{-1} B=\frac{1}{4}\left(\begin{array}{ccc}-2 & 0 & 2 \\\ 5 & -2 & -1 \\\ 1 & 2 & -1\end{array}\right)\left(\begin{array}{l}6 \\\ 7 \\\ 12 \end{array}\right)$

- $= \frac{1}{4}\left(\begin{array}{cc}-12+24 \\\ 30-14 & -12 \\\ 6+14 & -12\end{array}\right)=\frac{1}{4}\left(\begin{array}{c}12 \\\ 4 \\\ 8 \end{array}\right)=\left(\begin{array}{l}3 \\\ 1 \\\ 2\end{array}\right)$

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<footer style = "font-size: 18px">Example 4 $\rightarrow$ Example 4 $\rightarrow$ <span style = "color:blue"> Example 4 </span> $\rightarrow$ Cramer rule $\rightarrow$ Cramer Rule </footer>

### <Success style="font-size:25px"> Determinants L-11 </Success>
### <primary style="font-size:24px"> Cramer rule </primary>
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- If the given system of equation are $\underset{\tiny n \times n}{A} \underset{~ ~ \tiny n \times 1}{X}=\underset{\tiny n \times 1}{B}$

- such that $A$ in non-Singular

-  $B \neq 0_{n \times 1}$ (ie a $n \times 1$ matrix where all values are 0 ) then the solution of the  equation can be compute as follow.
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<footer style = "font-size: 18px">Example 4 $\rightarrow$ Example 4 $\rightarrow$ <span style = "color:blue"> Cramer rule </span> $\rightarrow$ Cramer Rule $\rightarrow$ Cramer Rule </footer>

### <Success style="font-size:25px"> Determinants L-11 </Success>
### <primary style="font-size:24px"> Cramer Rule </primary>
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- similarly Let $D_2$ be determinant of $\left|\begin{array}{lll}a_{11} & b_1 & a_{13} \\\  a_{21} & b_2 & a_{23} \\\  a_{31} & b_3 & a_{33}\end{array}\right|$ 
- similarly campute $D_3$ as determinant  of

- $\left|\begin{array}{lll}a_{11} & a_{12} & b_1 \\\ a_{21} & a_{22} & b_2 \\\ a_{31} & a_{32} & b_3\end{array}\right|$

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<footer style = "font-size: 18px">Cramer rule $\rightarrow$ Cramer Rule $\rightarrow$ <span style = "color:blue"> Cramer Rule </span> $\rightarrow$ Cramer Rule $\rightarrow$ Solution by Cramer'Rule </footer>

### <Success style="font-size:25px"> Determinants L-11 </Success>
### <primary style="font-size:24px"> Cramer Rule </primary>
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- Then $x=\frac{D_1}{D} \quad y=\frac{D_2}{D} \quad \\& z=\frac{D_3}{D}$

- Verification

- A=$\left[\begin{array}{lll}1 & 1 & 1 \\\ 1 & 0 & 2 \\\ 3 & 1 & 1\end{array}\right]\left[\begin{array}{l}x \\\ y \\\ z\end{array}\right]=\left[\begin{array}{c}6 \\\ 7 \\\ 12\end{array}\right]$

- We know the answers in

- $\left(\begin{array}{l}3 \\\ 1 \\\ 2\end{array}\right)$

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<footer style = "font-size: 18px">Cramer Rule $\rightarrow$ Cramer Rule $\rightarrow$ <span style = "color:blue"> Cramer Rule </span> $\rightarrow$ Solution by Cramer'Rule $\rightarrow$ Solution by Cramer'Rule </footer>

### <Success style="font-size:25px"> Determinants L-11 </Success>
### <primary style="font-size:24px"> Example 5 </primary>
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- Consider,

- 2 x+3 y=5 and 4 x+6 y=10

- $ A=\left[\begin{array}{ll}2 & 3 \\\ 4 & 6\end{array}\right] \therefore|A|=0$

- $\therefore$ We compute $($ adj $A$ ) B

- $= \left(\begin{array}{cc}6 & -3 \\\ -4 & 2\end{array}\right)\left(\begin{array}{c}5 \\\ 10\end{array}\right)$

- $= \left(\begin{array}{c}30-30 \\\ -20+20\end{array}\right)=\left(\begin{array}{l}0 \\\ 0\end{array}\right)=0$

- There are infinitely many solution.
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<footer style = "font-size: 18px">Solution by Cramer'Rule $\rightarrow$ If $A$ is singular $\rightarrow$ <span style = "color:blue"> Example 5 </span> $\rightarrow$ Example 6 $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Determinants L-11 </Success>
### <primary style="font-size:24px"> Example 6 </primary>
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- 2 x+3 y=5

- 4 x+6 y=15

- $\therefore|A|=0 $

- (adj A)$ \left(\begin{array}{c}5 \\\ 15\end{array}\right)=\left(\begin{array}{cc}6 & -3 \\\ -4 & 2\end{array}\right)\left(\begin{array}{c}5 \\\ 15 \end{array}\right)$

- $= 30 - 45  $

- $-20 + 30 =\left(\begin{array}{c}-15 \\\ 10\end{array}\right)$

- Not a zero matrix  of size $2 \times 1$
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<footer style = "font-size: 18px">If $A$ is singular $\rightarrow$ Example 5 $\rightarrow$ <span style = "color:blue"> Example 6 </span> $\rightarrow$ Thank you $\rightarrow$  </footer>