The element of second row and third column in the inverse of $\left( \begin{matrix}
   1 & 2 & 1 \\
   2 & 1 & 0 \\
   -1 & 0 & 1 \\
\end{matrix} \right)$ is
A. $-2$
B. $-1$
C. $1$
D. $2$

Hint: To find the element of second row and third column in the inverse of the matrix of $\left( \begin{matrix}
   1 & 2 & 1 \\
   2 & 1 & 0 \\
   -1 & 0 & 1 \\
\end{matrix} \right)$, we will find the inverse of this matrix and select the element ${{a}_{23}}$. We will first calculate the determinant and the adjoint of the matrix $\left( \begin{matrix}
   1 & 2 & 1 \\
   2 & 1 & 0 \\
   -1 & 0 & 1 \\
\end{matrix} \right)$ . Then we will substitute the values in the formula of the inverse.

Formula Used: \[{{A}^{-1}}=\frac{1}{|A|}\,adj(A)\]
If $A=\left( \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right)\,$ then
 $\begin{align}
  & |A|=a\left( \begin{matrix}
   e & f \\
   h & i \\
\end{matrix} \right)-b\left( \begin{matrix}
   d & f \\
   g & i \\
\end{matrix} \right)+c\left( \begin{matrix}
   d & e \\
   g & h \\
\end{matrix} \right) \\
 & =a(ei-fh)-b(di-fg)+c(dh-eg)
\end{align}$.


Complete step by step solution: We are given a matrix $\left( \begin{matrix}
   1 & 2 & 1 \\
   2 & 1 & 0 \\
   -1 & 0 & 1 \\
\end{matrix} \right)$ and we have to find the element of second row and third column in the inverse of this matrix.
We will find the inverse of this matrix. To do that first we will calculate the determinant and adjoint of the matrix.
Let us assume this matrix be $A=\left( \begin{matrix}
   1 & 2 & 1 \\
   2 & 1 & 0 \\
   -1 & 0 & 1 \\
\end{matrix} \right)$ so the determinant will be,
$|A|=1(1-0)-2(2-0)+1(0+1)$
$\begin{align}
  & |A|=1-4+1 \\
 & \,\,\,\,\,\,\,=-2 \\
\end{align}$
Now we will find adjoint of the matrix.
The adjoint of the matrix can be defined as the transpose of the cofactor of the matrix.
The cofactor of the matrix is,
$A=\left( \begin{matrix}
   +{{A}_{11}} & -{{A}_{12}} & +{{A}_{13}} \\
   -{{A}_{21}} & +{{A}_{22}} & -{{A}_{23}} \\
   +{{A}_{31}} & -{{A}_{32}} & +{{A}_{33}} \\
\end{matrix} \right)$
The transpose of the cofactor of the matrix will be,
${{A}^{T}}=\left( \begin{matrix}
   +{{A}_{11}} & -{{A}_{21}} & +{{A}_{31}} \\
   -{{A}_{12}} & +{{A}_{22}} & -{{A}_{32}} \\
   +{{A}_{13}} & -{{A}_{23}} & +{{A}_{33}} \\
\end{matrix} \right)$
We will now find the inverse of the matrix,
\[{{A}^{-1}}=\frac{1}{-2}\,\left( \begin{matrix}
   +{{A}_{11}} & -{{A}_{21}} & +{{A}_{31}} \\
   -{{A}_{12}} & +{{A}_{22}} & -{{A}_{32}} \\
   +{{A}_{13}} & -{{A}_{23}} & +{{A}_{33}} \\
\end{matrix} \right)\]
We have to find the element of the second row and third column so we will select that element from the cofactor,
\[{{({{A}^{-1}})}_{23}}=\frac{-{{A}_{32}}}{-2}\,\]
\[{{({{A}^{-1}})}_{23}}=\frac{-\left( \begin{matrix}
   1 & 1 \\
   2 & 0 \\
\end{matrix} \right)}{-2}\,\]
\[{{({{A}^{-1}})}_{23}}=\frac{-\left[ 0-2 \right]}{-2}\,\]
\[{{({{A}^{-1}})}_{23}}=\frac{2}{-2}\,\]
\[{{({{A}^{-1}})}_{23}}=-1\]
The value of the element of second row and third column in the inverse of the matrix $\left( \begin{matrix}
   1 & 2 & 1 \\
   2 & 1 & 0 \\
   -1 & 0 & 1 \\
\end{matrix} \right)$ is \[{{({{A}^{-1}})}_{23}}=-1\]

Option ‘B’ is correct

Note:The transpose of the matrix is calculated by interchanging the element of the principal diagonal and only changing the sign of the other diagonal.