Inverse of a matrix A is defined by

$AA^{-1} =A^{-1}A = I$

$A^{-1}$ =notation for inverse

“idea” of an inverse ?

$2 x=1$

(2 : $1\times 1$ matrix or scalar)

Is there a way to find inverse of ‘2’?

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

Consider $A x=b$

If we could find $A^{-1}$,

$A^{-1} \times A x =b$

$\Rightarrow A^{-1} A x =A^{-1} b$

$\Rightarrow \quad x =A^{-1} b$

Goal: Show the use of determinants in checking for invertibity of in a matrix and to actually compute it.

Consider a $3 \times 3$ matrix

$\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33}\end{array}\right]$

denote $A_{i j}$ as the cofactor of $a_{i j}$ define adjoint of this matrix

$\left[\begin{array}{lll}x_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33}\end{array}\right]^{\top}$

adjoint of a matrix is obtained by taking transpose of matrix where each element is replaced by cofactor

matrix (A) adjacent

$\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]_{3 \times 3}$

$\times\left[\begin{array}{lll}A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33}\end{array}\right]_{3 \times 3}$

$= \left[\begin{array}{lll}{a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13}} & a_{11} A_{21}+a_{12} A_{22}+a_{13} A_{23} & \\ a_{21} A_{11}+a_{22} A_{12}+a_{23} A_{13} & & \\ & & \end{array}\right]_{3 \times 3}$

$\left[\begin{array}{lll}\operatorname{det}(A) & 0 & 0 \\ 0 & \operatorname{det}(A) & 0 \\ 0 & 0 & \operatorname{det}(A)\end{array}\right]$

=$\operatorname{det}(A) \cdot I_{3 \times 3}$

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

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

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

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

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

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=$\left[\begin{array}{ll}4 & 5 \\5 & 4\end{array}\right] \rightarrow-9$

$\therefore \operatorname{det}(A B)=\operatorname{det}(A) \operatorname{det}(B)$

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

Total determinant

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

using property

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

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

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

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

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

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

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

Define a singular matrix as a matrix with a zero determinant .

Define a non-singular matrix as a matrix with a non-zero determinant

$A$ 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 $A A^{\prime}=A^{\prime} A=I$

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 $xyz \neq 0 \begin{aligned} & x \neq 0 \ & y \neq 0 \ & z \neq 0\end{aligned}$

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

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

Calculating inverses?

inspection for some simple matrices

using determinant and adjoint (bonus: provides a conduction to check for existemes of an inverse?

Example (continued):

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

Satusfies $\quad A^2-4 A+I=0$ (can be checked)

How to use this to calculate $A^{-1}$ ?

multiply by $A^{-1} \Rightarrow A^{-1}\left(A^A A\right)-4 A^{-1} A+A^{-1} I=0$

$\Rightarrow I A-4 I+A^{-1}=0$

$\Rightarrow A^{-1}=4 I-A= \left[\begin{array}{ll}4 & 0 \\0 & 4\end{array}\right]$-A

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

Another way to get $A^{-1}$.

$\leftrightarrow|\lambda I-A|=\left|\begin{array}{cc}\lambda-2 & -3 \\ -1 & \lambda-2\end{array}\right|$