Question

Let A be a 3 × 3 matrix such that A $\left( \begin{matrix}
   1 & 2 & 3 \\
   0 & 2 & 3 \\
   0 & 1 & 1 \\
\end{matrix} \right)$ = $\left( \begin{matrix}
   0 & 0 & 1 \\
   1 & 0 & 0 \\
   0 & 1 & 0 \\
\end{matrix} \right)$
Then ${{A}^{-1}}$ =

Answer 1

Hint: When two matrices are multiplied to get a new matrix , we can also find one of two matrices by multiplying inverse of the other given matrix on both sides. Now you can find the inverse of the matrix by finding adjacent matrices and then multiplying the inverse matrix with the product to get that value of the unknown matrix.

Complete step by step answer:
Let us assume that the matrix $\left( \begin{matrix}
   1 & 2 & 3 \\
   0 & 2 & 3 \\
   0 & 1 & 1 \\
\end{matrix} \right)$ is B.
B = $\left( \begin{matrix}
   1 & 2 & 3 \\
   0 & 2 & 3 \\
   0 & 1 & 1 \\
\end{matrix} \right)$
$\Rightarrow $ (AB) = $\left( \begin{matrix}
   0 & 0 & 1 \\
   1 & 0 & 0 \\
   0 & 1 & 0 \\
\end{matrix} \right)$
By multiplying ${{B}^{-1}}$on both sides on the right side of the above equation, we get
(AB)${{B}^{-1}}$ = $\left( \begin{matrix}
   0 & 0 & 1 \\
   1 & 0 & 0 \\
   0 & 1 & 0 \\
\end{matrix} \right)$ ${{B}^{-1}}$
By associative law of matrices we have, (AB)C = A(BC)
$A(B{{B}^{-1}})$) = $\left( \begin{matrix}
   0 & 0 & 1 \\
   1 & 0 & 0 \\
   0 & 1 & 0 \\
\end{matrix} \right) {{B}^{-1}}$
$\Rightarrow $ A = $\left( \begin{matrix}
   0 & 0 & 1 \\
   1 & 0 & 0 \\
   0 & 1 & 0 \\
\end{matrix} \right)$ ${{B}^{-1}}$ (because B${{B}^{-1}}$= I and A I = A , where I is the identity matrix)
Now we need to find the value of B inverse and then multiply it with $\left( \begin{matrix}
   0 & 0 & 1 \\
   1 & 0 & 0 \\
   0 & 1 & 0 \\
\end{matrix} \right)$ to get the value of A.
We know that \[{{B}^{-1}}\] = $\dfrac{1}{\det B}(AdjB)$
The determinant of matrix $\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right)$ will be $\det \left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right)={{a}_{11}}({{a}_{22}}{{a}_{33}}-{{a}_{32}}{{a}_{23}})-{{a}_{12}}({{a}_{21}}{{a}_{33}}-{{a}_{31}}{{a}_{23}})+{{a}_{13}}({{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}})$
Comparing $\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right)$ with $\left( \begin{matrix}
   1 & 2 & 3 \\
   0 & 2 & 3 \\
   0 & 1 & 1 \\
\end{matrix} \right)$
$\left( \begin{matrix}
   {{a}_{11}}=1 & {{a}_{12}}=2 & {{a}_{13}}=3 \\
   {{a}_{21}}=0 & {{a}_{22}}=2 & {{a}_{23}}=3 \\
   {{a}_{31}}=0 & {{a}_{32}}=1 & {{a}_{33}}=1 \\
\end{matrix} \right)$
\[detB\text{ }=\text{ }1\left( 2-3 \right)2\left( 0-0 \right)+3\left( 0-0 \right)=-1\]
Now we need the value of Adj B
For finding the adjoint matrix we will first find the minors of each element and then writing the minors matrix by taking determinant values of each minor and then convert it into cofactor matrix and then finally taking the transpose of cofactor matrix.
The minors of the matrix $\left( \begin{matrix}
   {{a}_{11}}=1 & {{a}_{12}}=2 & {{a}_{13}}=3 \\
   {{a}_{21}}=0 & {{a}_{22}}=2 & {{a}_{23}}=3 \\
   {{a}_{31}}=0 & {{a}_{32}}=1 & {{a}_{33}}=1 \\
\end{matrix} \right)$ will be
$\begin{align}
  & {{M}_{11}}=\text{minor of }{{\text{a}}_{11}}=\det \left( \begin{matrix}
   2 & 3 \\
   1 & 1 \\
\end{matrix} \right)=2(1)-3(1)=-1 \\
 & {{M}_{12}}=\text{minor of }{{\text{a}}_{12}}=\det \left( \begin{matrix}
   0 & 3 \\
   0 & 1 \\
\end{matrix} \right)=0-0=0 \\
 & {{M}_{13}}=\text{minor of }{{\text{a}}_{13}}=\det \left( \begin{matrix}
   0 & 2 \\
   0 & 1 \\
\end{matrix} \right)=0-0=0 \\
 & {{M}_{21}}=\text{minor of }{{\text{a}}_{21}}=\det \left( \begin{matrix}
   2 & 3 \\
   1 & 1 \\
\end{matrix} \right)=2-3=-1 \\
 & {{M}_{22}}=\text{minor of }{{\text{a}}_{22}}=\det \left( \begin{matrix}
   1 & 3 \\
   0 & 1 \\
\end{matrix} \right)=1-0=1 \\
 & {{M}_{23}}=\text{minor of }{{\text{a}}_{23}}=\det \left( \begin{matrix}
   1 & 2 \\
   0 & 1 \\
\end{matrix} \right)=1-0=1 \\
 & {{M}_{31}}=\text{minor of }{{\text{a}}_{31}}=\det \left( \begin{matrix}
   2 & 3 \\
   2 & 3 \\
\end{matrix} \right)=6-6=0 \\
 & {{M}_{32}}=\text{minor of }{{\text{a}}_{32}}=\det \left( \begin{matrix}
   1 & 3 \\
   0 & 3 \\
\end{matrix} \right)=3-0=3 \\
 & {{M}_{33}}=\text{minor of }{{\text{a}}_{33}}=\det \left( \begin{matrix}
   1 & 2 \\
   0 & 2 \\
\end{matrix} \right)=2-0=2 \\
\end{align}$
The minor matrix will be
$\left( \begin{matrix}
   {{M}_{11}} & {{M}_{12}} & {{M}_{13}} \\
   {{M}_{21}} & {{M}_{22}} & {{M}_{23}} \\
   {{M}_{31}} & {{M}_{32}} & {{M}_{33}} \\
\end{matrix} \right)=$ $\left( \begin{matrix}
   -1 & 0 & 0 \\
   -1 & 1 & 1 \\
   0 & 3 & 2 \\
\end{matrix} \right)$
now the co-factor matrix will be${{\left( -1 \right)}^{i+j}}{{a}_{ij}}$
co-factor matrix= $\left( \begin{matrix}
   -1 & -(0) & 0 \\
   -(-1) & 1 & -(1) \\
   0 & -(3) & 2 \\
\end{matrix} \right)=$ $\left( \begin{matrix}
   -1 & 0 & 0 \\
   1 & 1 & -1 \\
   0 & -3 & 2 \\
\end{matrix} \right)$
 the transpose of cofactor matrix that is AdjB will be
${{\left( \text{co-factor matrix} \right)}^{T}}={{\left( \begin{matrix}
   -1 & 0 & 0 \\
   1 & 1 & -1 \\
   0 & -3 & 2 \\
\end{matrix} \right)}^{T}}=$ $\left( \begin{matrix}
   -1 & 1 & 0 \\
   0 & 1 & -3 \\
   0 & -1 & 2 \\
\end{matrix} \right)$
now ${{B}^{-1}}$ will be $\dfrac{1}{\det B}(AdjB)$
that implies
${{B}^{-1}}$= $\dfrac{1}{(-1)}\left( \begin{matrix}
   -1 & 1 & 0 \\
   0 & 1 & -3 \\
   0 & -1 & 2 \\
\end{matrix} \right)$
${{B}^{-1}}$=$\left( \begin{matrix}
   1 & -1 & 0 \\
   0 & -1 & 3 \\
   0 & 1 & -2 \\
\end{matrix} \right)$
Now by multiplying $\left( \begin{matrix}
   0 & 0 & 1 \\
   1 & 0 & 0 \\
   0 & 1 & 0 \\
\end{matrix} \right)$ with ${{B}^{-1}}$ =$\left( \begin{matrix}
   1 & -1 & 0 \\
   0 & -1 & 3 \\
   0 & 1 & -2 \\
\end{matrix} \right)$we get A
A=$\left( \begin{matrix}
   0 & 0 & 1 \\
   1 & 0 & 0 \\
   0 & 1 & 0 \\
\end{matrix} \right)\times \left( \begin{matrix}
   1 & -1 & 0 \\
   0 & -1 & 3 \\
   0 & 1 & -2 \\
\end{matrix} \right)=\left( \begin{matrix}
   0 & 1 & -2 \\
   1 & -1 & 0 \\
   0 & -1 & 3 \\
\end{matrix} \right)$
A = $\left( \begin{matrix}
   0 & 1 & -2 \\
   1 & -1 & 0 \\
   0 & -1 & 3 \\
\end{matrix} \right)$
In the similar let us find \[{{A}^{-1}}\],
\[{{A}^{-1}}\]=$\dfrac{1}{\det A}(adjA)$
$\begin{align}
  & \det \left( \begin{matrix}
   0 & 1 & -2 \\
   1 & -1 & 0 \\
   0 & -1 & 3 \\
\end{matrix} \right)=0(-3)-1(3)-2(-1)=-1 \\
 & \det A=-1 \\
\end{align}$
First let us find the minor matrix of A,
Comparing $\left( \begin{matrix}
   0 & 1 & -2 \\
   1 & -1 & 0 \\
   0 & -1 & 3 \\
\end{matrix} \right)$,$\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right)$we have $\left( \begin{matrix}
   {{a}_{11}}=0 & {{a}_{12}}=1 & {{a}_{13}}=-2 \\
   {{a}_{21}}=1 & {{a}_{22}}=-1 & {{a}_{23}}=0 \\
   {{a}_{31}}=0 & {{a}_{32}}=-1 & {{a}_{33}}=3 \\
\end{matrix} \right)$



$\begin{align}
  & {{M}_{11}}=\text{minor of }{{\text{a}}_{11}}=\det \left( \begin{matrix}
   -1 & 0 \\
   -1 & 3 \\
\end{matrix} \right)=-1(3)-0(-1)=-3 \\
 & {{M}_{12}}=\text{minor of }{{\text{a}}_{12}}=\det \left( \begin{matrix}
   1 & 0 \\
   0 & 3 \\
\end{matrix} \right)=3-0=3 \\
 & {{M}_{13}}=\text{minor of }{{\text{a}}_{13}}=\det \left( \begin{matrix}
   1 & -1 \\
   0 & -1 \\
\end{matrix} \right)=-1-0=-1 \\
 & {{M}_{21}}=\text{minor of }{{\text{a}}_{21}}=\det \left( \begin{matrix}
   1 & -2 \\
   -1 & 3 \\
\end{matrix} \right)=3-2=1 \\
 & {{M}_{22}}=\text{minor of }{{\text{a}}_{22}}=\det \left( \begin{matrix}
   0 & -2 \\
   0 & 3 \\
\end{matrix} \right)=0-0=0 \\
 & {{M}_{23}}=\text{minor of }{{\text{a}}_{23}}=\det \left( \begin{matrix}
   0 & 1 \\
   0 & -1 \\
\end{matrix} \right)=0-0=0 \\
 & {{M}_{31}}=\text{minor of }{{\text{a}}_{31}}=\det \left( \begin{matrix}
   1 & -2 \\
   -1 & 0 \\
\end{matrix} \right)=0-2=-2 \\
 & {{M}_{32}}=\text{minor of }{{\text{a}}_{32}}=\det \left( \begin{matrix}
   0 & -2 \\
   1 & 0 \\
\end{matrix} \right)=0+2=2 \\
 & {{M}_{33}}=\text{minor of }{{\text{a}}_{33}}=\det \left( \begin{matrix}
   0 & 1 \\
   1 & -1 \\
\end{matrix} \right)=0-1=-1 \\
\end{align}$
Minor matrix would be ,
$\left( \begin{matrix}
   {{M}_{11}} & {{M}_{12}} & {{M}_{13}} \\
   {{M}_{21}} & {{M}_{22}} & {{M}_{23}} \\
   {{M}_{31}} & {{M}_{32}} & {{M}_{33}} \\
\end{matrix} \right)=$ $\left( \begin{matrix}
   -3 & 3 & -1 \\
   1 & 0 & 0 \\
   -2 & 2 & -1 \\
\end{matrix} \right)$
now the co-factor matrix will be${{\left( -1 \right)}^{i+j}}{{a}_{ij}}$
co-factor matrix= $\left( \begin{matrix}
   -3 & -(3) & -1 \\
   -(1) & 0 & -(0) \\
   -2 & -(2) & -1 \\
\end{matrix} \right)=\left( \begin{matrix}
   -3 & -3 & -1 \\
   -1 & 0 & 0 \\
   -2 & -2 & -1 \\
\end{matrix} \right)$
the transpose of cofactor matrix that is AdjA will be
${{\left( \text{co-factor matrix} \right)}^{T}}={{\left( \begin{matrix}
   -3 & -3 & -1 \\
   -1 & 0 & 0 \\
   -2 & -2 & -1 \\
\end{matrix} \right)}^{T}}=\left( \begin{matrix}
   -3 & -1 & -2 \\
   -3 & 0 & -2 \\
   -1 & 0 & -1 \\
\end{matrix} \right)=adjA$
\[{{A}^{-1}}\]=$\dfrac{1}{\det A}(adjA)$

\[{{A}^{-1}}\]=$\dfrac{1}{(-1)}\left( \begin{matrix}
   -3 & -1 & -2 \\
   -3 & 0 & -2 \\
   -1 & 0 & -1 \\
\end{matrix} \right)=\left( \begin{matrix}
   3 & 1 & 2 \\
   3 & 0 & 2 \\
   1 & 0 & 1 \\
\end{matrix} \right)$
\[{{A}^{-1}}\] = $\left( \begin{matrix}
   3 & 1 & 2 \\
   3 & 0 & 2 \\
   1 & 0 & 1 \\
\end{matrix} \right)$

Note: Make sure that the determinant of the matrix for which you are finding the inverse is not 0. Because if it is zero , then the matrix will not be invertible , hence we cannot find the inverse of that particular matrix. Make sure there are no calculation mistakes as there are many calculation processes.