The element in the first row and third column of the inverse of the matrix $\left[ \begin{matrix}
   1 & 2 & 3 \\
   0 & 1 & 2 \\
   0 & 0 & 1 \\
\end{matrix} \right]$ is
(a) $-2$
(b) 0
(c) 1
(d) None of these

Hint: First, we have to find minors of all the 9 elements in the matrix present i.e. determinant. Then we have to apply sign to find cofactor using the concept of $\left[ \begin{matrix}
   + & - & + \\
   - & + & - \\
   + & - & + \\
\end{matrix} \right]$ . Then we have to find adjoint of matrix which is called transpose of matrix and at last we have to find determinant of only first row elements and add using the formula $a-b+c=\left| A \right|$ . Thus having all the values we will put in the formula of inverse ${{A}^{-1}}=\dfrac{Adj\left( A \right)}{\left| A \right|}$ and get the answer.

Complete step by step solution:
Here, we should know the steps how to find inverse of $3\times 3$ matrix which is given to us
$\left[ \begin{matrix}
   1 & 2 & 3 \\
   0 & 1 & 2 \\
   0 & 0 & 1 \\
\end{matrix} \right]$
So, there are 4 steps for finding inverse matrix:
Step 1: Here, we have to create minors of the matrix usually known as determinant of all 9 elements. We have to ignore the value of current row and column and calculate the determinant of the remaining $2\times 2$ matrix.
So, we will calculate here for all the 9 elements in the matrix.
$1=\left[ \begin{matrix}
   1 & 2 \\
   0 & 1 \\
\end{matrix} \right]=\left( 1\times 1-\left( 2\times 0 \right) \right)=1-0=1$ So, the minor of first element 1 is 1.
Now, finding the second element in the first row.
$2=\left[ \begin{matrix}
   0 & 2 \\
   0 & 1 \\
\end{matrix} \right]=\left( 0\times 1-\left( 2\times 0 \right) \right)=0$ So, minor for 2 is 0.
Similarly, we have to calculate for the other 7 elements. After calculating, we get matrix of minors as
$\left[ \begin{matrix}
   1 & 0 & 0 \\
   2 & 1 & 0 \\
   1 & 2 & 1 \\
\end{matrix} \right]$ ………………………………(1)
Step 2: We have to find cofactors i.e. we have to change sign of every alternative elements in the format as
$\left[ \begin{matrix}
   + & - & + \\
   - & + & - \\
   + & - & + \\
\end{matrix} \right]$ . So, applying this sign in our minor matrix of equation (1), we get
$\left[ \begin{matrix}
   1 & 0 & 0 \\
   -2 & 1 & 0 \\
   1 & -2 & 1 \\
\end{matrix} \right]$ ……………………….(2)
Step 3: We have to find the adjoint of the matrix $Adj\left( A \right)$ which is basically the transpose of the matrix. So, here we will interchange elements of row with column and vice versa. So, on doing that we will get matrix as
$\left[ \begin{matrix}
   1 & -2 & 1 \\
   0 & 1 & -2 \\
   0 & 0 & 1 \\
\end{matrix} \right]$ ………………………….(4)
Step 4: Now, we will multiply the matrix with determinant $\left| A \right|$ . Here, we will find the determinant of elements of the first row only. We have already calculated in step 1 in equation (1) so, we get as $a-b+c=\left| A \right|$ where a, b and c are values of the determinant of the first three elements of the first row respectively.
Thus, we get $\left| A \right|=1-0+0=1$ ………………..(5)
So, formula of inverse matrix is ${{A}^{-1}}=\dfrac{Adj\left( A \right)}{\left| A \right|}$
Now, substituting the values we get
${{A}^{-1}}=\dfrac{1}{1}\left[ \begin{matrix}
   1 & -2 & 1 \\
   0 & 1 & -2 \\
   0 & 0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & -2 & 1 \\
   0 & 1 & -2 \\
   0 & 0 & 1 \\
\end{matrix} \right]$
Thus, element in the first row and third column of the inverse of the matrix $\left[ \begin{matrix}
   1 & 2 & 3 \\
   0 & 1 & 2 \\
   0 & 0 & 1 \\
\end{matrix} \right]$ is 1.
Hence, option (c) is the correct answer.

Note: Be careful while finding cofactor of minor matrix. Mostly, students make sign mistake which affects the whole answer i.e. instead of using $\left[ \begin{matrix}
   + & - & + \\
   - & + & - \\
   + & - & + \\
\end{matrix} \right]$ they use $\left[ \begin{matrix}
   - & + & - \\
   + & - & + \\
   - & + & - \\
\end{matrix} \right]$ . So, do not make this mistake. Also, in finding determinant $a-b+c=\left| A \right|$ this minus is applied to second elements while finding determinant. Also sometimes students make mistakes and incorrect answers are selected. So, do not make these silly mistakes.