Find Inverse

Question

Inverse of this matrix

$\begin{bmatrix} 2 &1 &1 \\ 3 &2 &1 \\ 2& 1 &2 \end{bmatrix}$

will be

A)$\begin{bmatrix} 3 &-4 &-1 \\ -1 &2 &0 \\ -1 &1 &1 \end{bmatrix}$

B)$\begin{bmatrix} 3 &-1 &-1 \\ -4 &2 &1 \\ -1& 0 &1 \end{bmatrix}$

Comments

@Magma 

see options,it is ok?

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Lakshman edited

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@Mk Utkarsh

thanks 

Answer 1

Let us suppose $A=\begin{bmatrix} 2& 1 &1 \\ 3& 2 &1 \\ 2 &1 &2 \end{bmatrix}$

let another matrix $B=\begin{bmatrix} a _{11}&a _{12} &a _{13} \\ a _{21}& a _{22} &a _{23} \\ a _{31} &a _{32} &a _{33} \end{bmatrix}$

find the cofactor of:

$a_{11}=(-1)^{1+1}\begin{vmatrix} 2 & 1\\ 1 &2 \end{vmatrix}=1(4-1)=3$

$a_{12}=(-1)^{1+2}\begin{vmatrix} 3 & 2\\ 1 &2 \end{vmatrix}=(-1)(6-2)=(-1)(4)=-4$

$a_{13}=(-1)^{1+3}\begin{vmatrix} 3 & 2\\ 2 &1 \end{vmatrix}=1(3-4)=-1$

$a_{21}=(-1)^{2+1}\begin{vmatrix} 1 & 1\\ 1 &2 \end{vmatrix}=(-1)(2-1)=-1$

$a_{22}=(-1)^{2+2}\begin{vmatrix} 2 & 2\\ 1 &2 \end{vmatrix}=1(4-2)=2$

$a_{23}=(-1)^{2+3}\begin{vmatrix} 2 & 2\\ 1 &1 \end{vmatrix}=(-1)(2-2)=0$

$a_{31}=(-1)^{3+1}\begin{vmatrix} 1 & 2\\ 1 &1 \end{vmatrix}=1(1-2)=-1$

$a_{32}=(-1)^{3+2}\begin{vmatrix} 2 & 3\\ 1 &1 \end{vmatrix}=(-1)(2-3)=(-1)(-1)=1$

$a_{33}=(-1)^{3+3}\begin{vmatrix} 2 & 3\\ 1 &2 \end{vmatrix}=1(4-3)=1$

Now $B=\begin{bmatrix} 3 &-4 &-1 \\ -1 &2 &0 \\ -1 &1 &1 \end{bmatrix}$

$Adj(A)=B^{T}=\begin{bmatrix} 3 &-1 &-1 \\ -4 &2 &1 \\ -1 &0 &1 \end{bmatrix}$

Now we find $|A|=\begin{vmatrix} 2 &1 &1 \\ 3 &2 &1 \\ 2 &1 &2 \end{vmatrix}$

$|A|=(2)\begin{vmatrix} 2&1 \\ 1&2 \end{vmatrix}-(1)\begin{vmatrix} 3&2 \\ 1&2 \end{vmatrix}+(1)\begin{vmatrix} 3&2 \\2& 1\end{vmatrix}$

$|A|=2(4-1)-1(6-2)+1(3-4)$

$|A|=2(3)-1(4)+1(-1)$

$|A|=6-4-1=6-5=1$

$So,$$|A|=1$

Now$,A^{-1}=\frac{Adj(A)}{|A|}$

we can simply put the value,

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

$A^{-1}=\begin{bmatrix} 3 &-1 &-1 \\ -4 &2 &1 \\ -1 &0 &1 \end{bmatrix}$

So,this is the correct answer.