Question

Hey mate please answer my question with step by step explanation​

Answer 1

Question:-

Find the inverse of the non singular matrix

$\rm \: A= \begin{vmatrix}2& 1& 1\\3 & 2 & 1\\2& 1 & 2\end{vmatrix}$

and reduce it to the identity by elminatory transmittion

Solution:-

Property of A inverse is

$\rm \: A {}^{ - 1} = \dfrac{1}{ |A| } adjA$

Now we find co - factor, we get

$\rm \: A= \begin{vmatrix}3& 4& - 1\\1 & 2 & 0\\ - 1& - 1 & 1\end{vmatrix}$

$\rm \: A= \begin{vmatrix} 3& - 4& - 1\\ - 1 & 2 & 0\\ - 1& 1 & 1\end{vmatrix}$

Now we will find

$\rm \: adjA {}^{T}$

we get

$\rm \: adjA= \begin{vmatrix} 3& - 1& - 1\\ - 4 & 2 & 1\\ - 1& 0& 1\end{vmatrix}$

Now find

$\rm \: |A|$

Take determinant wrt a₁₁ we get

$\rm \: |A| = 3$

Put the value on property, we get

$\rm \: A {}^{ - 1} = \frac{1}{3} \begin{vmatrix} 3& - 1& - 1\\ - 4 & 2 & 1\\ - 1& 0& 1\end{vmatrix}$

Answer 2

Property of A inverse is

\rm \: A {}^{ - 1} = \dfrac{1}{ |A| } adjAA

−1

=

∣A∣

1

adjA

Now we find co - factor, we get

\begin{gathered}\rm \: A= \begin{vmatrix}3& 4& - 1\\1 & 2 & 0\\ - 1& - 1 & 1\end{vmatrix}\end{gathered}

A=

∣

∣

∣

∣

∣

∣

∣

3

1

−1

4

2

−1

−1

0

1

∣

∣

∣

∣

∣

∣

∣

\begin{gathered}\rm \: A= \begin{vmatrix} 3& - 4& - 1\\ - 1 & 2 & 0\\ - 1& 1 & 1\end{vmatrix}\end{gathered}

A=

∣

∣

∣

∣

∣

∣

∣

3

−1

−1

−4

2

1

−1

0

1

∣

∣

∣

∣

∣

∣

∣

Now we will find

\rm \: adjA {}^{T}adjA

T

we get

\begin{gathered}\rm \: adjA= \begin{vmatrix} 3& - 1& - 1\\ - 4 & 2 & 1\\ - 1& 0& 1\end{vmatrix}\end{gathered}

adjA=

∣

∣

∣

∣

∣

∣

∣

3

−4

−1

−1

2

0

−1

1

1

∣

∣

∣

∣

∣

∣

∣

Now find

\rm \: |A|∣A∣

Take determinant wrt a₁₁ we get

\rm \: |A| = 3∣A∣=3

Put the value on property, we get

\begin{gathered}\rm \: A {}^{ - 1} = \frac{1}{3} \begin{vmatrix} 3& - 1& - 1\\ - 4 & 2 & 1\\ - 1& 0& 1\end{vmatrix}\end{gathered}

A

−1

=

3

1

∣

∣

∣

∣

∣

∣

∣

3

−4

−1

−1

2

0

−1

1

1

∣

∣

∣

∣

∣

∣

∣