Question

The value of determinant
\[\left| \begin{matrix}
   1+{{a}^{2}}-{{b}^{2}} & 2ab & -2b \\
   2ab & 1-{{a}^{2}}+{{b}^{2}} & 2a \\
   2b & -2a & 1-{{a}^{2}}-{{b}^{2}} \\
\end{matrix} \right|\]
(a) 0
(b) \[\left( 1+{{a}^{2}}+{{b}^{2}} \right)\]
(c) \[{{\left( 1+{{a}^{2}}+{{b}^{2}} \right)}^{2}}\]
(d) \[{{\left( 1+{{a}^{2}}+{{b}^{2}} \right)}^{3}}\]

Answer 1

Hint: We solve this problem by using the determinant properties by taking the given determinant. That is we assume that
\[A=\left| \begin{matrix}
   1+{{a}^{2}}-{{b}^{2}} & 2ab & -2b \\
   2ab & 1-{{a}^{2}}+{{b}^{2}} & 2a \\
   2b & -2a & 1-{{a}^{2}}-{{b}^{2}} \\
\end{matrix} \right|\]
The properties of determinant are simple, that is nothing but changing the rows and columns by subtracting or adding the columns or rows with respect to other columns and rows. By using the suitable transformations we get RHS from LHS we need to get two values as ‘0’ either in one row or column so that we can expand the determinant easily.

Complete step by step answer:
Let us assume that the given determinant as
\[\Rightarrow A=\left| \begin{matrix}
   1+{{a}^{2}}-{{b}^{2}} & 2ab & -2b \\
   2ab & 1-{{a}^{2}}+{{b}^{2}} & 2a \\
   2b & -2a & 1-{{a}^{2}}-{{b}^{2}} \\
\end{matrix} \right|\]
Now let us use the row transformation for first row that is by applying the transformation
\[{{R}_{1}}\to {{R}_{1}}+b\times {{R}_{3}}\] then we get
\[\begin{align}
  & \Rightarrow A=\left| \begin{matrix}
   1+{{a}^{2}}-{{b}^{2}}+2{{b}^{2}} & 2ab-2ab & -2b+b-{{a}^{2}}b-{{b}^{3}} \\
   2ab & 1-{{a}^{2}}+{{b}^{2}} & 2a \\
   2b & -2a & 1-{{a}^{2}}-{{b}^{2}} \\
\end{matrix} \right| \\
 & \Rightarrow A=\left| \begin{matrix}
   1+{{a}^{2}}+{{b}^{2}} & 0 & -b\left( 1+{{a}^{2}}+{{b}^{2}} \right) \\
   2ab & 1-{{a}^{2}}+{{b}^{2}} & 2a \\
   2b & -2a & 1-{{a}^{2}}-{{b}^{2}} \\
\end{matrix} \right| \\
\end{align}\]
Now, by taking the common term out from the first row we get
\[\Rightarrow A=\left( 1+{{a}^{2}}+{{b}^{2}} \right)\left| \begin{matrix}
   1 & 0 & -b \\
   2ab & 1-{{a}^{2}}+{{b}^{2}} & 2a \\
   2b & -2a & 1-{{a}^{2}}-{{b}^{2}} \\
\end{matrix} \right|\]
Now let us use the row transformation for second row that is by applying the transformation
\[{{R}_{2}}\to {{R}_{2}}-a\times {{R}_{3}}\] then we get
\[\begin{align}
  & \Rightarrow A=\left( 1+{{a}^{2}}+{{b}^{2}} \right)\left| \begin{matrix}
   1 & 0 & -b \\
   2ab-2ab & 1-{{a}^{2}}+{{b}^{2}}-2{{a}^{2}} & 2a-a+{{a}^{3}}+{{b}^{2}}a \\
   2b & -2a & 1-{{a}^{2}}-{{b}^{2}} \\
\end{matrix} \right| \\
 & \Rightarrow A=\left( 1+{{a}^{2}}+{{b}^{2}} \right)\left| \begin{matrix}
   1 & 0 & -b \\
   0 & 1+{{a}^{2}}+{{b}^{2}} & a\left( 1+{{a}^{2}}+{{b}^{2}} \right) \\
   2b & -2a & 1-{{a}^{2}}-{{b}^{2}} \\
\end{matrix} \right| \\
\end{align}\]
Now, by taking the common term out from the second row we get
\[\Rightarrow A={{\left( 1+{{a}^{2}}+{{b}^{2}} \right)}^{2}}\left| \begin{matrix}
   1 & 0 & -b \\
   0 & 1 & a \\
   2b & -2a & 1-{{a}^{2}}-{{b}^{2}} \\
\end{matrix} \right|\]
Now let us use the column transformation for third row that is by applying the transformation
\[{{C}_{3}}\to {{C}_{3}}+b\times {{C}_{1}}-a\times {{C}_{2}}\] then we get
\[\Rightarrow A={{\left( 1+{{a}^{2}}+{{b}^{2}} \right)}^{2}}\left| \begin{matrix}
   1 & 0 & -b+b-0 \\
   0 & 1 & a+0-a \\
   2b & -2a & 1-{{a}^{2}}-{{b}^{2}}+2{{b}^{2}}+2{{a}^{2}} \\
\end{matrix} \right|\]
\[\Rightarrow A={{\left( 1+{{a}^{2}}+{{b}^{2}} \right)}^{2}}\left| \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   2b & -2a & 1+{{a}^{2}}+{{b}^{2}} \\
\end{matrix} \right|\]
Here, we can see that there are two zeros in third column.
Now, by expanding the determinant along third column we get
\[\begin{align}
  & \Rightarrow A={{\left( 1+{{a}^{2}}+{{b}^{2}} \right)}^{2}}\left[ \left( 1+{{a}^{2}}+{{b}^{2}} \right)\left( 1-0 \right) \right] \\
 & \Rightarrow A={{\left( 1+{{a}^{2}}+{{b}^{2}} \right)}^{3}} \\
\end{align}\]
Therefore the value of determinant given is \[{{\left( 1+{{a}^{2}}+{{b}^{2}} \right)}^{3}}\]

So, the correct answer is “Option d”.

Note: We have a shortcut for using the transformations.
Here, we have the determinant of order 3 so, we can apply 2 transformations at a time but those 2 transformations must be either row or column transformations.
Here, we have the determinant
\[\Rightarrow A=\left| \begin{matrix}
   1+{{a}^{2}}-{{b}^{2}} & 2ab & -2b \\
   2ab & 1-{{a}^{2}}+{{b}^{2}} & 2a \\
   2b & -2a & 1-{{a}^{2}}-{{b}^{2}} \\
\end{matrix} \right|\]
Now, we can apply 2 transformations at a time that is by applying \[{{R}_{1}}\to {{R}_{1}}+b\times {{R}_{3}}\] and \[{{R}_{2}}\to {{R}_{2}}-a\times {{R}_{3}}\] we can get directly
\[\Rightarrow A={{\left( 1+{{a}^{2}}+{{b}^{2}} \right)}^{2}}\left| \begin{matrix}
   1 & 0 & -b \\
   0 & 1 & a \\
   2b & -2a & 1-{{a}^{2}}-{{b}^{2}} \\
\end{matrix} \right|\]
But we should not apply the transformations \[{{R}_{2}}\to {{R}_{2}}-a\times {{R}_{3}}\] and \[{{C}_{3}}\to {{C}_{3}}+b\times {{C}_{1}}-a\times {{C}_{2}}\] at a time because one is row and other is column.