Question

if A=a1 b1 c1 a2 b2 c2 a3 b3 c3 is non singular matrix then show that Ais invertible and A-1=adjA/det A​

Answer 1

Answer:

this is help

Step-by-step explanation:

Answer 2

Given:

A is non-singular matrix.

To find:

The show that A is invertible.

Step-by-step explanation:

$A=\left[\begin{array}{ccc}a_{1} &b_{1} &c_{1} \\a_{2} &b_{2} &c_{2} \\a_{3} &b_{3} &c_{3} \end{array}\right]$

$c_{11} =|\eft\begin{array}{ccc}b_{2} &c_{2} \\b_{3} &c_{3} \\\end{array}|\right|$,$c_{12} =|\eft\begin{array}{ccc}a_{2} &c_{2} \\a_{3} &c_{3} \\\end{array}|\right|$,$c_{13} =|\eft\begin{array}{ccc}a_{2} &b_{2} \\a_{3} &b_{3} \\\end{array}|\right|$

$A.adjA=A=\left[\begin{array}{ccc}a_{1} &b_{1} &c_{1} \\a_{2} &b_{2} &c_{2} \\a_{3} &b_{3} &c_{3} \end{array}\right] \left[\begin{array}{ccc}c_{11} &c_{21} &c_{31} \\c_{12} &c_{22} &c_{32} \\c_{13} &c_{23} &c_{33} \end{array}\right]$

$A[\left adjA ] =\left\left[\begin{array}{ccc}detA&0&0\\0&detA&0\\0&0&detA\end{array}\right]$$=detAI$

because A is invertible, det A$\neq$0,

$\frac{1}{detA}$$A(adjA)=I$

$A\left[\frac{1}{detA} ] =I$

$\frac{1}{detA} adjA=A^{-1}$

$A^{-1} =\frac{adjA}{detA}$