mathematics-class-12-unit-04-chapter-03-determinants-l-3-11-enuieuavt1y

<h3 id="success-stylefont-size25px-determinants-l-3-success-1"><Success style="font-size:25px"> Determinants L-3 </Success></h3>
<h3 id="primary-stylefont-size24px-inverse-of-a-matrix-primary"><primary style="font-size:24px"> Inverse of a matrix </primary></h3>
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<p>Inverse of a matrix A is defined by</p>
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<p>$AA^{-1} =A^{-1}A = I$</p>
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<p>$A^{-1}$ =notation for inverse</p>
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<p>“idea” of an inverse ?</p>
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<p>$ 2 x=1 $</p>
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<p>(2 : $1\times 1$ matrix or scalar)</p>
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<p>Is there a way to find inverse of ‘2’?</p>
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<footer style = "font-size: 18px"> $\rightarrow$ Determinants lecture - 3 $\rightarrow$ <span style = "color:blue"> Inverse of a matrix </span> $\rightarrow$ Inverse of a matrix $\rightarrow$ Solution of the system of equations </footer>

### <Success style="font-size:25px"> Determinants L-3 </Success>
### <primary style="font-size:24px"> Inverse of a matrix </primary>
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- Notion of multiplicative inverse

- $ 2^{-1} \times[2 x=1 $

- $\Rightarrow(\underbrace{2^{-1} \cdot 2}_{1}) x=2^{-1} $

- $\Rightarrow   x=2^{-1}=1 / 2$

- multiplicative inverse
- $\longrightarrow$ matrix inverse may be used for matrix equation

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<footer style = "font-size: 18px">Determinants lecture - 3 $\rightarrow$ Inverse of a matrix $\rightarrow$ <span style = "color:blue"> Inverse of a matrix </span> $\rightarrow$ Solution of the system of equations $\rightarrow$ Proof of $A(\operatorname{adj}(A))  =\operatorname{det}(A) I $ </footer>

<h3 id="success-stylefont-size25px-determinants-l-3-success-2"><Success style="font-size:25px"> Determinants L-3 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-of-the-system-of-equations-primary"><primary style="font-size:24px"> Solution of the system of equations </primary></h3>
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<p>Consider $A x=b$</p>
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<p>If we could find $A^{-1}$,</p>
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<p>$A^{-1} \times A x  =b$</p>
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<p>$\Rightarrow A^{-1} A x  =A^{-1} b $</p>
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<p>$\Rightarrow \quad x  =A^{-1} b$</p>
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<p>Goal: Show the use of determinants in checking for invertibity of in a matrix and to actually compute it.</p>
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<footer style = "font-size: 18px">Inverse of a matrix $\rightarrow$ Inverse of a matrix $\rightarrow$ <span style = "color:blue"> Solution of the system of equations </span> $\rightarrow$ Proof of $A(\operatorname{adj}(A))  =\operatorname{det}(A) I $ $\rightarrow$ Proof of $A(\operatorname{adj}(A))  =\operatorname{det}(A) I $ </footer>

<h3 id="success-stylefont-size25px-determinants-l-3-success-3"><Success style="font-size:25px"> Determinants L-3 </Success></h3>
<h3 id="primary-stylefont-size24px-proof-of-aoperatornameadja--operatornamedeta-i--primary"><primary style="font-size:24px"> Proof of $A(\operatorname{adj}(A))  =\operatorname{det}(A) I $ </primary></h3>
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<p>Consider a $3 \times 3$ matrix</p>
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<p>$\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33}\end{array}\right]$</p>
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<p>denote $A_{i j}$ as the cofactor of $a_{i j}$ define adjoint of this matrix</p>
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<p>$\left[\begin{array}{lll}x_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33}\end{array}\right]^{\top}$</p>
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<p>adjoint of a matrix is obtained by taking transpose of matrix where each element is replaced by cofactor</p>
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<footer style = "font-size: 18px">Inverse of a matrix $\rightarrow$ Solution of the system of equations $\rightarrow$ <span style = "color:blue"> Proof of $A(\operatorname{adj}(A))  =\operatorname{det}(A) I $ </span> $\rightarrow$ Proof of $A(\operatorname{adj}(A))  =\operatorname{det}(A) I $ $\rightarrow$ Proof of $A(\operatorname{adj}(A))  =\operatorname{det}(A) I $ </footer>

<h3 id="success-stylefont-size25px-determinants-l-3-success-4"><Success style="font-size:25px"> Determinants L-3 </Success></h3>
<h3 id="primary-stylefont-size24px-proof-of-aoperatornameadja--operatornamedeta-i--primary-1"><primary style="font-size:24px"> Proof of $A(\operatorname{adj}(A))  =\operatorname{det}(A) I $ </primary></h3>
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<p>matrix (A) adjacent</p>
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<p>$\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]_{3 \times 3}$</p>
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<p>$\times\left[\begin{array}{lll}A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33}\end{array}\right]_{3 \times 3}$</p>
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<p>$= \left[\begin{array}{lll}{a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13}} & a_{11} A_{21}+a_{12} A_{22}+a_{13} A_{23} &  \\ a_{21} A_{11}+a_{22} A_{12}+a_{23} A_{13} &  &  \\  &  &  \end{array}\right]_{3 \times 3}$</p>
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<p>$\left[\begin{array}{lll}\operatorname{det}(A) & 0 & 0 \\ 0 & \operatorname{det}(A) & 0 \\ 0 & 0 & \operatorname{det}(A)\end{array}\right]$</p>
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<p>=$\operatorname{det}(A) \cdot I_{3 \times 3}$</p>
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<footer style = "font-size: 18px">Solution of the system of equations $\rightarrow$ Proof of $A(\operatorname{adj}(A))  =\operatorname{det}(A) I $ $\rightarrow$ <span style = "color:blue"> Proof of $A(\operatorname{adj}(A))  =\operatorname{det}(A) I $ </span> $\rightarrow$ Proof of $A(\operatorname{adj}(A))  =\operatorname{det}(A) I $ $\rightarrow$ Note </footer>

### <Success style="font-size:25px"> Determinants L-3 </Success>
### <primary style="font-size:24px"> Proof of $A(\operatorname{adj}(A))  =\operatorname{det}(A) I $ </primary>
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- $A(\operatorname{adj}(A))  =\operatorname{det}(A) I $

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

- Combining these,

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

- $\Rightarrow A \frac{\operatorname{adj}(A)}{\operatorname{det}(A)}=\frac{\operatorname{adj}(A)}{\operatorname{det}(A)} \cdot A=I \quad(\text { when } \operatorname{det}(A) \neq 0)$

- compare with $A A^{-1}=A^{\prime} A=I$

- $\operatorname{det}(\Delta) \neq 0)$

- $\Rightarrow A^{-1}=\frac{\operatorname{adj}(A)}{\operatorname{det}(A)}, \operatorname{det}(A) \neq 0 $

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<footer style = "font-size: 18px">Proof of $A(\operatorname{adj}(A))  =\operatorname{det}(A) I $ $\rightarrow$ Proof of $A(\operatorname{adj}(A))  =\operatorname{det}(A) I $ $\rightarrow$ <span style = "color:blue"> Proof of $A(\operatorname{adj}(A))  =\operatorname{det}(A) I $ </span> $\rightarrow$ Note $\rightarrow$ Note </footer>

### <Success style="font-size:25px"> Determinants L-3 </Success>
### <primary style="font-size:24px"> Note </primary>
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- Note : What is $\operatorname{det}(\operatorname{adj}(A))$ ?

- We use the property:

- $\operatorname{det}(A B)=\operatorname{det}(A) \cdot \operatorname{det}(B)$

- ( A and B are square matrices) Example. det

- $ A=\left[\begin{array}{ll}1 & 2 \\\2 & 1\end{array}\right]\rightarrow-3 $

-  B=$\left[\begin{array}{ll}2 & 1 \\\1 & 2\end{array}\right] \rightarrow 3 $
 
-  A B=$\left[\begin{array}{ll}4 & 5 \\\5 & 4\end{array}\right] \rightarrow-9 $
 
- $ \therefore \operatorname{det}(A B)=\operatorname{det}(A) \operatorname{det}(B)$

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<footer style = "font-size: 18px">Proof of $A(\operatorname{adj}(A))  =\operatorname{det}(A) I $ $\rightarrow$ Proof of $A(\operatorname{adj}(A))  =\operatorname{det}(A) I $ $\rightarrow$ <span style = "color:blue"> Note </span> $\rightarrow$ Note $\rightarrow$ Note </footer>

### <Success style="font-size:25px"> Determinants L-3 </Success>
### <primary style="font-size:24px"> Note </primary>
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- $A \operatorname{adj}(A)=\operatorname{det}(A) \cdot I$

- Total determinant

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

- using property

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

- $=[\operatorname{det}(A)]^3(\operatorname{for} 3 \times 3)$

- $=[\operatorname{det}(A)]^n(\operatorname{for} n \times n)$

- $ \operatorname{det}(A) \neq 0$

- $ \operatorname{det}(A) \neq 0 $

- $ \Rightarrow \operatorname{det}(\operatorname{adj}(A))=[\operatorname{det}(A)]^2(\operatorname{for} 3 \times 2) $

- $=[\operatorname{det}(A)]^{n-1}(\operatorname{for} n \times n)$

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<footer style = "font-size: 18px">Proof of $A(\operatorname{adj}(A))  =\operatorname{det}(A) I $ $\rightarrow$ Note $\rightarrow$ <span style = "color:blue"> Note </span> $\rightarrow$ Note $\rightarrow$ Theorem </footer>

### <Success style="font-size:25px"> Determinants L-3 </Success>
### <primary style="font-size:24px"> Theorem </primary>
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- $A$ is invertible $\Longleftrightarrow A$ is non-singular

- $\Leftrightarrow$ ) A is invertible

- $\Rightarrow \exists B$ such that $A B=B A=I$

- taking determinant $\Rightarrow \operatorname{det}(A B)=\operatorname{det}(I)$

- $\Rightarrow \operatorname{det}(A) \cdot \operatorname{det}(B)=1$

- $\Rightarrow \operatorname{det}(A) \neq 0 \Rightarrow A$ is non-singular
- ( $A$ is non-singular

- $\Rightarrow \operatorname{det}(A) \neq 0$

- $\Rightarrow A^{\prime}=\frac{\operatorname{adj}(A)}{\operatorname{det}(A)}$ satisfies $A A^{\prime}=A^{\prime} A=I$

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

### <Success style="font-size:25px"> Determinants L-3 </Success>
### <primary style="font-size:24px"> Example </primary>
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- Determinants help in checking invertibility of matrix, computing the inverse and significance of theorem.

- Example: Consider
- $\left[\begin{array}{lll}x & 0 & 0 \\\0 & y & 0 \\\0 & 0 & z\end{array}\right]$

- $\xrightarrow{I} \left[\begin{array}{ccc}1 / x & 0 & 0 \\\0 & 1/ y & 0 \\\0 & 0 & 1 / z\end{array}\right]$

- $\operatorname{det}=x y z $
- is non-singular  $xyz \neq 0 \begin{aligned} & x \neq 0 \\ & y \neq 0 \\ & z \neq 0\end{aligned}$

- $\left.\operatorname{adj}=\left[\begin{array}{ccc}yz & 0 & 0 \\\ 0 & xz & 0 \\\ 0 & 0 &xy\end{array}\right]^{\top}=\left[\begin{array}{ccc}y z & 0 & 0 \\\ 0 & xz & 0 \\\ 0 & 0 & xy\end{array}\right]\right]$

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

### <Success style="font-size:25px"> Determinants L-3 </Success>
### <primary style="font-size:24px"> Example </primary>
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- Example: $A=\left[\begin{array}{ll}2 & 3 \\\\ 1 & 2\end{array}\right]$

- calculate $A^{-1}$.
 
- Does inverse exit?

- $\operatorname{det}(A)=1$

- The inverse exists as $\operatorname{det}(A) \neq 0$.

- What is $A^{\prime}=\operatorname{adj}(A) \quad(\operatorname{as} \operatorname{det}(A)=1$

- $=\left[\begin{array}{cc}2 & -1 \\\ -3 & 2\end{array}\right]^{\top}=\left[\begin{array}{cc}2 & -3 \\\ -1 & 2\end{array}\right]$

- Check: $A A^{-1}=\left[\begin{array}{ll}1 & 0 \\\ 0 & 1\end{array}\right]$

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