Question

The element in the first row and third column of the below inverse matrix is:
\[\left[ \begin{matrix}
   1 & 2 & -3 \\
   0 & 1 & 2 \\
   0 & 0 & 1 \\
\end{matrix} \right]\]
(a) -2
(b) 0
(c) 1
(d) 7

Answer 1

Hint: We solve this problem by finding the inverse matrix of the given matrix
We have the condition that the inverse of a matrix is given as
\[{{A}^{-1}}=\dfrac{adjA}{\left| A \right|}\]
We have the result that the adjoint of a matrix is given as the transpose of a matric which is obtained by replacing the elements of the given matrix by its cofactors.
A cofactor of an element of a matrix is obtained by using the minors which the determinant of matrix eliminating the column and row of the element. Then the cofactor is given by applying the alternative negative to minors.
By using the above conditions we find the inverse of a given matrix

Complete step by step answer:
Let us assume that the given matrix as
\[\Rightarrow A=\left[ \begin{matrix}
   1 & 2 & -3 \\
   0 & 1 & 2 \\
   0 & 0 & 1 \\
\end{matrix} \right]\]
Now, let us find the minor matrix of the given matrix
Let us assume that the minor matrix of the given matrix as\[M\]
We know that minors of each element are obtained by the determinant of matrix eliminating the column and row of the element
By using the above condition to the element of first row first column then we get
\[\begin{align}
  & \Rightarrow {{m}_{1}}=\left| \begin{matrix}
   1 & 2 \\
   0 & 1 \\
\end{matrix} \right| \\
 & \Rightarrow {{m}_{1}}=1\times 1-0\times 2 \\
 & \Rightarrow {{m}_{1}}=1 \\
\end{align}\]
Similarly by using the above condition to all elements of the given matrix we get
\[\begin{align}
  & \Rightarrow M=\left[ \begin{matrix}
   \left( 1\times 1-0\times 1 \right) & \left( 0\times 1-0\times 2 \right) & \left( 0\times 0-0\times 1 \right) \\
   \left( 2\times 1-0\times \left( -3 \right) \right) & 1\times 1-0\times \left( -3 \right) & \left( 1\times 0-0\times 2 \right) \\
   \left( 2\times 2-1\times \left( -3 \right) \right) & 2\times 1-0\times \left( -3 \right) & \left( 1\times 1-0\times 2 \right) \\
\end{matrix} \right] \\
 & \Rightarrow M=\left[ \begin{matrix}
   1 & 0 & 0 \\
   2 & 1 & 0 \\
   7 & 2 & 1 \\
\end{matrix} \right] \\
\end{align}\]
Now, let us find the cofactor matrix of the above minor matrix
Let us assume that the cofactor matrix as \[C\]
We know that the cofactor matrix is obtained by giving the alternative negative to minor matrix then we get
\[\begin{align}
  & \Rightarrow C=\left[ \begin{matrix}
   1 & -0 & 0 \\
   -2 & 1 & -0 \\
   7 & -2 & 1 \\
\end{matrix} \right] \\
 & \Rightarrow C=\left[ \begin{matrix}
   1 & 0 & 0 \\
   -2 & 1 & 0 \\
   7 & -2 & 1 \\
\end{matrix} \right] \\
\end{align}\]
Now, let us find the adjoint matrix
We know that the adjoint matrix is given as the transpose of the cofactor matrix
By using the above condition we get
\[\begin{align}
  & \Rightarrow adjA={{\left[ \begin{matrix}
   1 & 0 & 0 \\
   -2 & 1 & 0 \\
   7 & -2 & 1 \\
\end{matrix} \right]}^{T}} \\
 & \Rightarrow adjA=\left[ \begin{matrix}
   1 & -2 & 7 \\
   0 & 1 & -2 \\
   0 & 0 & 1 \\
\end{matrix} \right] \\
\end{align}\]
Now, let us find the determinant of the given matrix
\[\Rightarrow \left| A \right|=\left| \begin{matrix}
   1 & 2 & -3 \\
   0 & 1 & 2 \\
   0 & 0 & 1 \\
\end{matrix} \right|\]
By expanding the determinant along first column then we get
\[\begin{align}
  & \Rightarrow \left| A \right|=1\left( 1\times 1-0\times 2 \right)-0\left( 2\times 1-0\times \left( -3 \right) \right)+0\left( 2\times 2-1\times \left( -3 \right) \right) \\
 & \Rightarrow \left| A \right|=1 \\
\end{align}\]
We know that the condition that inverse of matrix is given as
\[{{A}^{-1}}=\dfrac{adjA}{\left| A \right|}\]
By using the above formula we get the inverse of the given matrix as
\[\begin{align}
  & \Rightarrow {{A}^{-1}}=\dfrac{1}{1}\left[ \begin{matrix}
   1 & -2 & 7 \\
   0 & 1 & -2 \\
   0 & 0 & 1 \\
\end{matrix} \right] \\
 & \Rightarrow {{A}^{-1}}=\left[ \begin{matrix}
   1 & -2 & 7 \\
   0 & 1 & -2 \\
   0 & 0 & 1 \\
\end{matrix} \right] \\
\end{align}\]
We are asked to find the element in the first row and third column of the inverse matrix of given matrix.
Here, we can see that the element in the first row and the second column is 7
So, option (d) is correct answer.


Note:
 We have a shortcut for this problem.
We are asked to find the element in the first row and third column of the inverse of given matrix.
By using the definition of inverse matrix the element in first row and third column of inverse matrix is the cofactor of element of third row and first column of given matrix because we use the transpose of the cofactor matrix.
By using the definition of cofactor we get the value of cofactor of element of third row and first column we get
\[\begin{align}
  & \Rightarrow {{a}_{31}}=\left| \begin{matrix}
   2 & -3 \\
   1 & 2 \\
\end{matrix} \right| \\
 & \Rightarrow {{a}_{31}}=2\times 2-1\times \left( -3 \right) \\
 & \Rightarrow {{a}_{31}}=7 \\
\end{align}\]
Therefore we can conclude that the element in the first row third column of inverse of given matrix is 7
So, option (d) is correct answer.