The value of the determinant \[\Delta =\left| \begin{matrix}
2{{a}_{1}}{{b}_{1}} & {{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}} & {{a}_{1}}{{b}_{3}}+{{a}_{3}}{{b}_{1}} \\
{{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}} & 2{{a}_{2}}{{b}_{2}} & {{a}_{2}}{{b}_{3}}+{{b}_{2}}{{a}_{3}} \\
{{a}_{1}}{{b}_{3}}+{{a}_{3}}{{b}_{1}} & {{a}_{3}}{{b}_{2}}+{{a}_{2}}{{b}_{3}} & 2{{a}_{3}}{{b}_{3}} \\
\end{matrix} \right|\] is:
A. 1
B. -1
C. 0
D. ${{a}_{1}}{{a}_{2}}{{a}_{3}}{{b}_{1}}{{b}_{2}}{{b}_{3}}$
Answer
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Hint: We have been given the determinant \[\Delta =\left| \begin{matrix}
2{{a}_{1}}{{b}_{1}} & {{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}} & {{a}_{1}}{{b}_{3}}+{{a}_{3}}{{b}_{1}} \\
{{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}} & 2{{a}_{2}}{{b}_{2}} & {{a}_{2}}{{b}_{3}}+{{b}_{2}}{{a}_{3}} \\
{{a}_{1}}{{b}_{3}}+{{a}_{3}}{{b}_{1}} & {{a}_{3}}{{b}_{2}}+{{a}_{2}}{{b}_{3}} & 2{{a}_{3}}{{b}_{3}} \\
\end{matrix} \right|\] and we need to find its value. From observing this given determinant, we can see the terms are in the form of sum of product of two numbers. Thus, they can be written in the form of a product of two determinants. So, we will try to write the given determinant in the same way and find their product and equate that with the given determinant. This will give us the value of our determinants and then we will try to solve the then obtained two determinants separately. Once we obtain their values, we can multiply them and we will get the value of the required determinant.
Complete step-by-step answer:
Now, we have to find the value of the determinant \[\Delta =\left| \begin{matrix}
2{{a}_{1}}{{b}_{1}} & {{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}} & {{a}_{1}}{{b}_{3}}+{{a}_{3}}{{b}_{1}} \\
{{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}} & 2{{a}_{2}}{{b}_{2}} & {{a}_{2}}{{b}_{3}}+{{b}_{2}}{{a}_{3}} \\
{{a}_{1}}{{b}_{3}}+{{a}_{3}}{{b}_{1}} & {{a}_{3}}{{b}_{2}}+{{a}_{2}}{{b}_{3}} & 2{{a}_{3}}{{b}_{3}} \\
\end{matrix} \right|\].
Now, we can see that except for the diagonal elements, all the elements of the determinant are written in the form of sum of product of two elements. So we will try making the diagonal elements into the same form.
Now, if we write $2{{a}_{1}}{{b}_{1}},2{{a}_{2}}{{b}_{2}}$ and $2{{a}_{3}}{{b}_{3}}$ as the sum of ${{a}_{1}}{{b}_{1}},{{a}_{2}}{{b}_{2}}$ and ${{a}_{3}}{{b}_{3}}$, we will get all the elements in the same form. Thus, our determinant will become:
\[\begin{align}
& \Delta =\left| \begin{matrix}
2{{a}_{1}}{{b}_{1}} & {{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}} & {{a}_{1}}{{b}_{3}}+{{a}_{3}}{{b}_{1}} \\
{{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}} & 2{{a}_{2}}{{b}_{2}} & {{a}_{2}}{{b}_{3}}+{{b}_{2}}{{a}_{3}} \\
{{a}_{1}}{{b}_{3}}+{{a}_{3}}{{b}_{1}} & {{a}_{3}}{{b}_{2}}+{{a}_{2}}{{b}_{3}} & 2{{a}_{3}}{{b}_{3}} \\
\end{matrix} \right| \\
& \Rightarrow \Delta =\left| \begin{matrix}
{{a}_{1}}{{b}_{1}}+{{a}_{1}}{{b}_{1}} & {{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}} & {{a}_{1}}{{b}_{3}}+{{a}_{3}}{{b}_{1}} \\
{{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}} & {{a}_{2}}{{b}_{2}}+{{a}_{2}}{{b}_{2}} & {{a}_{2}}{{b}_{3}}+{{b}_{2}}{{a}_{3}} \\
{{a}_{1}}{{b}_{3}}+{{a}_{3}}{{b}_{1}} & {{a}_{3}}{{b}_{2}}+{{a}_{2}}{{b}_{3}} & {{a}_{3}}{{b}_{3}}+{{a}_{3}}{{b}_{3}} \\
\end{matrix} \right| \\
\end{align}\]
Now, we can see that all the elements of the determinant are in the form of sum of product of two different numbers.
We know that when we do product of two determinants, every element of the resultant determinant given by ${{r}_{ij}}$ is obtained as:
${{r}_{ij}}=\sum{{{a}_{ij}}.{{b}_{ji}}}$
Where,
i=row number
j=column number
${{a}_{ij}}$=the element of first determinant in the ${{i}^{th}}$ row and ${{j}^{th}}$ column.
${{b}_{ji}}$=the element of the second determinant in the ${{j}^{th}}$ row and ${{i}^{th}}$ column.
Since, in the given determinant, all the elements are written as sum of product of two numbers, we can write $\Delta $ as the product of two determinants.
Let those two determinants be ${{\Delta }_{1}}$ and ${{\Delta }_{2}}$.
In the given determinant $\Delta $, one of the general elements in the product is given in the forms of ‘a’ and the other in the form of ‘b’. Also, every ‘a’ in the determinant is only multiplied with ‘b’ and vice-versa. There is no third variable used in $\Delta $.
Thus, we can say that ${{a}_{1}},{{a}_{2}},{{a}_{3}}$ and ${{b}_{1}},{{b}_{2}},{{b}_{3}}$ are the first and second column of ${{\Delta }_{1}}$ and second and first row of ${{\Delta }_{2}}$ respectively.
Since, there is no other element given in $\Delta $, we can assume the elements of third column of ${{\Delta }_{1}}$ to be in the form of ‘c’ and the third row of ${{\Delta }_{2}}$ in the form of ‘d’.
Thus, we get our two determinants as:
\[\begin{align}
& {{\Delta }_{1}}=\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right| \\
& {{\Delta }_{2}}=\left| \begin{matrix}
{{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
{{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
{{d}_{1}} & {{d}_{2}} & {{d}_{3}} \\
\end{matrix} \right| \\
\end{align}\]
Now, we have already established that $\Delta $ is the product of ${{\Delta }_{1}}$ and ${{\Delta }_{2}}$. So now we will multiply ${{\Delta }_{1}}$ and ${{\Delta }_{2}}$ and place it equal to $\Delta $.
Thus, we have:
\[\begin{align}
& {{\Delta }_{1}}=\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right| \\
& {{\Delta }_{2}}=\left| \begin{matrix}
{{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
{{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
{{d}_{1}} & {{d}_{2}} & {{d}_{3}} \\
\end{matrix} \right| \\
& \Rightarrow {{\Delta }_{1}}\times {{\Delta }_{2}}=\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|\times \left| \begin{matrix}
{{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
{{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
{{d}_{1}} & {{d}_{2}} & {{d}_{3}} \\
\end{matrix} \right| \\
& \Rightarrow {{\Delta }_{1}}\times {{\Delta }_{2}}=\left| \begin{matrix}
{{a}_{1}}{{b}_{1}}+{{b}_{1}}{{a}_{1}}+{{c}_{1}}{{d}_{1}} & {{a}_{1}}{{b}_{2}}+{{b}_{1}}{{a}_{2}}+{{c}_{1}}{{d}_{2}} & {{a}_{1}}{{b}_{3}}+{{b}_{1}}{{a}_{3}}+{{c}_{1}}{{d}_{3}} \\
{{a}_{2}}{{b}_{1}}+{{b}_{2}}{{a}_{1}}+{{c}_{2}}{{d}_{1}} & {{a}_{2}}{{b}_{2}}+{{b}_{2}}{{a}_{2}}+{{c}_{2}}{{d}_{2}} & {{a}_{2}}{{b}_{3}}+{{b}_{2}}{{a}_{3}}+{{c}_{2}}{{d}_{3}} \\
{{a}_{3}}{{b}_{1}}+{{b}_{3}}{{a}_{1}}+{{c}_{3}}{{d}_{1}} & {{a}_{3}}{{b}_{2}}+{{b}_{3}}{{a}_{2}}+{{c}_{3}}{{d}_{2}} & {{a}_{3}}{{b}_{3}}+{{b}_{3}}{{a}_{3}}+{{c}_{3}}{{d}_{3}} \\
\end{matrix} \right| \\
& \Rightarrow {{\Delta }_{1}}\times {{\Delta }_{2}}=\left| \begin{matrix}
2{{a}_{1}}{{b}_{1}}+{{c}_{1}}{{d}_{1}} & {{a}_{1}}{{b}_{2}}+{{b}_{1}}{{a}_{2}}+{{c}_{1}}{{d}_{2}} & {{a}_{1}}{{b}_{3}}+{{b}_{1}}{{a}_{3}}+{{c}_{1}}{{d}_{3}} \\
{{a}_{2}}{{b}_{1}}+{{b}_{2}}{{a}_{1}}+{{c}_{2}}{{d}_{1}} & 2{{a}_{2}}{{b}_{2}}+{{c}_{2}}{{d}_{2}} & {{a}_{2}}{{b}_{3}}+{{b}_{2}}{{a}_{3}}+{{c}_{2}}{{d}_{3}} \\
{{a}_{3}}{{b}_{1}}+{{b}_{3}}{{a}_{1}}+{{c}_{3}}{{d}_{1}} & {{a}_{3}}{{b}_{2}}+{{b}_{3}}{{a}_{2}}+{{c}_{3}}{{d}_{2}} & 2{{a}_{3}}{{b}_{3}}+{{c}_{3}}{{d}_{3}} \\
\end{matrix} \right| \\
\end{align}\]
Now, we have obtained the product of the two determinants ${{\Delta }_{1}}$ and ${{\Delta }_{2}}$. Since, the product is equal to $\Delta $ , we can say that:
$\Delta ={{\Delta }_{1}}\times {{\Delta }_{2}}$
Putting in their values, we get:
\[\left| \begin{matrix}
2{{a}_{1}}{{b}_{1}} & {{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}} & {{a}_{1}}{{b}_{3}}+{{a}_{3}}{{b}_{1}} \\
{{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}} & 2{{a}_{2}}{{b}_{2}} & {{a}_{2}}{{b}_{3}}+{{b}_{2}}{{a}_{3}} \\
{{a}_{1}}{{b}_{3}}+{{a}_{3}}{{b}_{1}} & {{a}_{3}}{{b}_{2}}+{{a}_{2}}{{b}_{3}} & 2{{a}_{3}}{{b}_{3}} \\
\end{matrix} \right|=\left| \begin{matrix}
2{{a}_{1}}{{b}_{1}}+{{c}_{1}}{{d}_{1}} & {{a}_{1}}{{b}_{2}}+{{b}_{1}}{{a}_{2}}+{{c}_{1}}{{d}_{2}} & {{a}_{1}}{{b}_{3}}+{{b}_{1}}{{a}_{3}}+{{c}_{1}}{{d}_{3}} \\
{{a}_{2}}{{b}_{1}}+{{b}_{2}}{{a}_{1}}+{{c}_{2}}{{d}_{1}} & 2{{a}_{2}}{{b}_{2}}+{{c}_{2}}{{d}_{2}} & {{a}_{2}}{{b}_{3}}+{{b}_{2}}{{a}_{3}}+{{c}_{2}}{{d}_{3}} \\
{{a}_{3}}{{b}_{1}}+{{b}_{3}}{{a}_{1}}+{{c}_{3}}{{d}_{1}} & {{a}_{3}}{{b}_{2}}+{{b}_{3}}{{a}_{2}}+{{c}_{3}}{{d}_{2}} & 2{{a}_{3}}{{b}_{3}}+{{c}_{3}}{{d}_{3}} \\
\end{matrix} \right|\]
Now, by equating their corresponding terms, we get:
\[\begin{align}
& {{c}_{1}}{{d}_{1}}=0 \\
& {{c}_{2}}{{d}_{1}}=0 \\
& {{c}_{3}}{{d}_{1}}=0 \\
& {{c}_{1}}{{d}_{2}}=0 \\
\end{align}\]
$\begin{align}
& {{c}_{2}}{{d}_{2}}=0 \\
& {{c}_{3}}{{d}_{2}}=0 \\
& {{c}_{1}}{{d}_{3}}=0 \\
& {{c}_{2}}{{d}_{3}}=0 \\
& {{c}_{3}}{{d}_{3}}=0 \\
\end{align}$
This is only possible if ${{c}_{1}},{{c}_{2}},{{c}_{3}},{{d}_{1}},{{d}_{2}},{{d}_{3}}$ all are equal to 0.
Thus, by putting in the obtained values of ${{c}_{1}},{{c}_{2}},{{c}_{3}},{{d}_{1}},{{d}_{2}},{{d}_{3}}$, we will get the values of ${{\Delta }_{1}}$ and ${{\Delta }_{2}}$.
Thus, we get:
$\begin{align}
& {{\Delta }_{1}}=\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & 0 \\
{{a}_{2}} & {{b}_{2}} & 0 \\
{{a}_{3}} & {{b}_{3}} & 0 \\
\end{matrix} \right| \\
& {{\Delta }_{2}}=\left| \begin{matrix}
{{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
{{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
0 & 0 & 0 \\
\end{matrix} \right| \\
\end{align}$
Now, that we have the determinants ${{\Delta }_{1}}$ and ${{\Delta }_{2}}$, we can obtain their value.
Now, ${{\Delta }_{1}}$ is given as:
\[\begin{align}
& {{\Delta }_{1}}=\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & 0 \\
{{a}_{2}} & {{b}_{2}} & 0 \\
{{a}_{3}} & {{b}_{3}} & 0 \\
\end{matrix} \right| \\
& \text{Now, opening via }{{\text{R}}_{1}}\text{ we get:} \\
& {{\Delta }_{1}}={{a}_{1}}\left( {{b}_{2}}.0-{{b}_{3}}.0 \right)-{{b}_{1}}\left( {{a}_{2}}.0-{{a}_{3}}.0 \right)+0\left( {{a}_{2}}{{b}_{3}}-{{a}_{3}}{{b}_{2}} \right) \\
& \Rightarrow {{\Delta }_{1}}={{a}_{1}}.0-{{b}_{1}}.0+0\left( {{a}_{2}}{{b}_{3}}-{{a}_{3}}{{b}_{2}} \right) \\
& \Rightarrow {{\Delta }_{1}}=0-0+0 \\
& \Rightarrow {{\Delta }_{1}}=0 \\
\end{align}\]
Thus, the value of ${{\Delta }_{1}}$ is 0.
Now, ${{\Delta }_{2}}$ is given as:
$\begin{align}
& {{\Delta }_{2}}=\left| \begin{matrix}
{{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
{{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
0 & 0 & 0 \\
\end{matrix} \right| \\
& \text{Opening via }{{\text{R}}_{1}}\text{ we get:} \\
& {{\Delta }_{2}}={{b}_{1}}\left( {{a}_{2}}.0-{{a}_{3}}.0 \right)-{{b}_{2}}\left( {{a}_{1}}.0-{{a}_{3}}.0 \right)+{{b}_{3}}\left( {{a}_{1}}.0-{{a}_{2}}.0 \right) \\
& \Rightarrow {{\Delta }_{2}}={{b}_{1}}.0-{{b}_{2}}.0+{{b}_{3}}.0 \\
& \Rightarrow {{\Delta }_{2}}=0-0+0 \\
& \Rightarrow {{\Delta }_{2}}=0 \\
\end{align}$
Thus, the value of ${{\Delta }_{2}}$ is also 0.
Now, we know that $\Delta $ is the product of ${{\Delta }_{1}}$ and ${{\Delta }_{2}}$.
Thus, we get:
$\begin{align}
& \Delta ={{\Delta }_{1}}\times {{\Delta }_{2}} \\
& \Rightarrow \Delta =0\times 0 \\
& \Rightarrow \Delta =0 \\
\end{align}$
Thus, the value of the required determinant is 0.
So, the correct answer is “Option C”.
Note: Here we have solved ${{\Delta }_{1}}$ and ${{\Delta }_{2}}$ to obtain their values. But we can straightaway write them equal to 0 because they have all the elements of one row or one column as 0. So, if we were to open the determinant by that row or column, we will obtain the determinant as 0 only. Thus, the value of all the determinants having all the elements of at least one row or column as 0 is 0.
2{{a}_{1}}{{b}_{1}} & {{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}} & {{a}_{1}}{{b}_{3}}+{{a}_{3}}{{b}_{1}} \\
{{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}} & 2{{a}_{2}}{{b}_{2}} & {{a}_{2}}{{b}_{3}}+{{b}_{2}}{{a}_{3}} \\
{{a}_{1}}{{b}_{3}}+{{a}_{3}}{{b}_{1}} & {{a}_{3}}{{b}_{2}}+{{a}_{2}}{{b}_{3}} & 2{{a}_{3}}{{b}_{3}} \\
\end{matrix} \right|\] and we need to find its value. From observing this given determinant, we can see the terms are in the form of sum of product of two numbers. Thus, they can be written in the form of a product of two determinants. So, we will try to write the given determinant in the same way and find their product and equate that with the given determinant. This will give us the value of our determinants and then we will try to solve the then obtained two determinants separately. Once we obtain their values, we can multiply them and we will get the value of the required determinant.
Complete step-by-step answer:
Now, we have to find the value of the determinant \[\Delta =\left| \begin{matrix}
2{{a}_{1}}{{b}_{1}} & {{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}} & {{a}_{1}}{{b}_{3}}+{{a}_{3}}{{b}_{1}} \\
{{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}} & 2{{a}_{2}}{{b}_{2}} & {{a}_{2}}{{b}_{3}}+{{b}_{2}}{{a}_{3}} \\
{{a}_{1}}{{b}_{3}}+{{a}_{3}}{{b}_{1}} & {{a}_{3}}{{b}_{2}}+{{a}_{2}}{{b}_{3}} & 2{{a}_{3}}{{b}_{3}} \\
\end{matrix} \right|\].
Now, we can see that except for the diagonal elements, all the elements of the determinant are written in the form of sum of product of two elements. So we will try making the diagonal elements into the same form.
Now, if we write $2{{a}_{1}}{{b}_{1}},2{{a}_{2}}{{b}_{2}}$ and $2{{a}_{3}}{{b}_{3}}$ as the sum of ${{a}_{1}}{{b}_{1}},{{a}_{2}}{{b}_{2}}$ and ${{a}_{3}}{{b}_{3}}$, we will get all the elements in the same form. Thus, our determinant will become:
\[\begin{align}
& \Delta =\left| \begin{matrix}
2{{a}_{1}}{{b}_{1}} & {{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}} & {{a}_{1}}{{b}_{3}}+{{a}_{3}}{{b}_{1}} \\
{{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}} & 2{{a}_{2}}{{b}_{2}} & {{a}_{2}}{{b}_{3}}+{{b}_{2}}{{a}_{3}} \\
{{a}_{1}}{{b}_{3}}+{{a}_{3}}{{b}_{1}} & {{a}_{3}}{{b}_{2}}+{{a}_{2}}{{b}_{3}} & 2{{a}_{3}}{{b}_{3}} \\
\end{matrix} \right| \\
& \Rightarrow \Delta =\left| \begin{matrix}
{{a}_{1}}{{b}_{1}}+{{a}_{1}}{{b}_{1}} & {{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}} & {{a}_{1}}{{b}_{3}}+{{a}_{3}}{{b}_{1}} \\
{{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}} & {{a}_{2}}{{b}_{2}}+{{a}_{2}}{{b}_{2}} & {{a}_{2}}{{b}_{3}}+{{b}_{2}}{{a}_{3}} \\
{{a}_{1}}{{b}_{3}}+{{a}_{3}}{{b}_{1}} & {{a}_{3}}{{b}_{2}}+{{a}_{2}}{{b}_{3}} & {{a}_{3}}{{b}_{3}}+{{a}_{3}}{{b}_{3}} \\
\end{matrix} \right| \\
\end{align}\]
Now, we can see that all the elements of the determinant are in the form of sum of product of two different numbers.
We know that when we do product of two determinants, every element of the resultant determinant given by ${{r}_{ij}}$ is obtained as:
${{r}_{ij}}=\sum{{{a}_{ij}}.{{b}_{ji}}}$
Where,
i=row number
j=column number
${{a}_{ij}}$=the element of first determinant in the ${{i}^{th}}$ row and ${{j}^{th}}$ column.
${{b}_{ji}}$=the element of the second determinant in the ${{j}^{th}}$ row and ${{i}^{th}}$ column.
Since, in the given determinant, all the elements are written as sum of product of two numbers, we can write $\Delta $ as the product of two determinants.
Let those two determinants be ${{\Delta }_{1}}$ and ${{\Delta }_{2}}$.
In the given determinant $\Delta $, one of the general elements in the product is given in the forms of ‘a’ and the other in the form of ‘b’. Also, every ‘a’ in the determinant is only multiplied with ‘b’ and vice-versa. There is no third variable used in $\Delta $.
Thus, we can say that ${{a}_{1}},{{a}_{2}},{{a}_{3}}$ and ${{b}_{1}},{{b}_{2}},{{b}_{3}}$ are the first and second column of ${{\Delta }_{1}}$ and second and first row of ${{\Delta }_{2}}$ respectively.
Since, there is no other element given in $\Delta $, we can assume the elements of third column of ${{\Delta }_{1}}$ to be in the form of ‘c’ and the third row of ${{\Delta }_{2}}$ in the form of ‘d’.
Thus, we get our two determinants as:
\[\begin{align}
& {{\Delta }_{1}}=\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right| \\
& {{\Delta }_{2}}=\left| \begin{matrix}
{{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
{{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
{{d}_{1}} & {{d}_{2}} & {{d}_{3}} \\
\end{matrix} \right| \\
\end{align}\]
Now, we have already established that $\Delta $ is the product of ${{\Delta }_{1}}$ and ${{\Delta }_{2}}$. So now we will multiply ${{\Delta }_{1}}$ and ${{\Delta }_{2}}$ and place it equal to $\Delta $.
Thus, we have:
\[\begin{align}
& {{\Delta }_{1}}=\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right| \\
& {{\Delta }_{2}}=\left| \begin{matrix}
{{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
{{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
{{d}_{1}} & {{d}_{2}} & {{d}_{3}} \\
\end{matrix} \right| \\
& \Rightarrow {{\Delta }_{1}}\times {{\Delta }_{2}}=\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|\times \left| \begin{matrix}
{{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
{{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
{{d}_{1}} & {{d}_{2}} & {{d}_{3}} \\
\end{matrix} \right| \\
& \Rightarrow {{\Delta }_{1}}\times {{\Delta }_{2}}=\left| \begin{matrix}
{{a}_{1}}{{b}_{1}}+{{b}_{1}}{{a}_{1}}+{{c}_{1}}{{d}_{1}} & {{a}_{1}}{{b}_{2}}+{{b}_{1}}{{a}_{2}}+{{c}_{1}}{{d}_{2}} & {{a}_{1}}{{b}_{3}}+{{b}_{1}}{{a}_{3}}+{{c}_{1}}{{d}_{3}} \\
{{a}_{2}}{{b}_{1}}+{{b}_{2}}{{a}_{1}}+{{c}_{2}}{{d}_{1}} & {{a}_{2}}{{b}_{2}}+{{b}_{2}}{{a}_{2}}+{{c}_{2}}{{d}_{2}} & {{a}_{2}}{{b}_{3}}+{{b}_{2}}{{a}_{3}}+{{c}_{2}}{{d}_{3}} \\
{{a}_{3}}{{b}_{1}}+{{b}_{3}}{{a}_{1}}+{{c}_{3}}{{d}_{1}} & {{a}_{3}}{{b}_{2}}+{{b}_{3}}{{a}_{2}}+{{c}_{3}}{{d}_{2}} & {{a}_{3}}{{b}_{3}}+{{b}_{3}}{{a}_{3}}+{{c}_{3}}{{d}_{3}} \\
\end{matrix} \right| \\
& \Rightarrow {{\Delta }_{1}}\times {{\Delta }_{2}}=\left| \begin{matrix}
2{{a}_{1}}{{b}_{1}}+{{c}_{1}}{{d}_{1}} & {{a}_{1}}{{b}_{2}}+{{b}_{1}}{{a}_{2}}+{{c}_{1}}{{d}_{2}} & {{a}_{1}}{{b}_{3}}+{{b}_{1}}{{a}_{3}}+{{c}_{1}}{{d}_{3}} \\
{{a}_{2}}{{b}_{1}}+{{b}_{2}}{{a}_{1}}+{{c}_{2}}{{d}_{1}} & 2{{a}_{2}}{{b}_{2}}+{{c}_{2}}{{d}_{2}} & {{a}_{2}}{{b}_{3}}+{{b}_{2}}{{a}_{3}}+{{c}_{2}}{{d}_{3}} \\
{{a}_{3}}{{b}_{1}}+{{b}_{3}}{{a}_{1}}+{{c}_{3}}{{d}_{1}} & {{a}_{3}}{{b}_{2}}+{{b}_{3}}{{a}_{2}}+{{c}_{3}}{{d}_{2}} & 2{{a}_{3}}{{b}_{3}}+{{c}_{3}}{{d}_{3}} \\
\end{matrix} \right| \\
\end{align}\]
Now, we have obtained the product of the two determinants ${{\Delta }_{1}}$ and ${{\Delta }_{2}}$. Since, the product is equal to $\Delta $ , we can say that:
$\Delta ={{\Delta }_{1}}\times {{\Delta }_{2}}$
Putting in their values, we get:
\[\left| \begin{matrix}
2{{a}_{1}}{{b}_{1}} & {{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}} & {{a}_{1}}{{b}_{3}}+{{a}_{3}}{{b}_{1}} \\
{{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}} & 2{{a}_{2}}{{b}_{2}} & {{a}_{2}}{{b}_{3}}+{{b}_{2}}{{a}_{3}} \\
{{a}_{1}}{{b}_{3}}+{{a}_{3}}{{b}_{1}} & {{a}_{3}}{{b}_{2}}+{{a}_{2}}{{b}_{3}} & 2{{a}_{3}}{{b}_{3}} \\
\end{matrix} \right|=\left| \begin{matrix}
2{{a}_{1}}{{b}_{1}}+{{c}_{1}}{{d}_{1}} & {{a}_{1}}{{b}_{2}}+{{b}_{1}}{{a}_{2}}+{{c}_{1}}{{d}_{2}} & {{a}_{1}}{{b}_{3}}+{{b}_{1}}{{a}_{3}}+{{c}_{1}}{{d}_{3}} \\
{{a}_{2}}{{b}_{1}}+{{b}_{2}}{{a}_{1}}+{{c}_{2}}{{d}_{1}} & 2{{a}_{2}}{{b}_{2}}+{{c}_{2}}{{d}_{2}} & {{a}_{2}}{{b}_{3}}+{{b}_{2}}{{a}_{3}}+{{c}_{2}}{{d}_{3}} \\
{{a}_{3}}{{b}_{1}}+{{b}_{3}}{{a}_{1}}+{{c}_{3}}{{d}_{1}} & {{a}_{3}}{{b}_{2}}+{{b}_{3}}{{a}_{2}}+{{c}_{3}}{{d}_{2}} & 2{{a}_{3}}{{b}_{3}}+{{c}_{3}}{{d}_{3}} \\
\end{matrix} \right|\]
Now, by equating their corresponding terms, we get:
\[\begin{align}
& {{c}_{1}}{{d}_{1}}=0 \\
& {{c}_{2}}{{d}_{1}}=0 \\
& {{c}_{3}}{{d}_{1}}=0 \\
& {{c}_{1}}{{d}_{2}}=0 \\
\end{align}\]
$\begin{align}
& {{c}_{2}}{{d}_{2}}=0 \\
& {{c}_{3}}{{d}_{2}}=0 \\
& {{c}_{1}}{{d}_{3}}=0 \\
& {{c}_{2}}{{d}_{3}}=0 \\
& {{c}_{3}}{{d}_{3}}=0 \\
\end{align}$
This is only possible if ${{c}_{1}},{{c}_{2}},{{c}_{3}},{{d}_{1}},{{d}_{2}},{{d}_{3}}$ all are equal to 0.
Thus, by putting in the obtained values of ${{c}_{1}},{{c}_{2}},{{c}_{3}},{{d}_{1}},{{d}_{2}},{{d}_{3}}$, we will get the values of ${{\Delta }_{1}}$ and ${{\Delta }_{2}}$.
Thus, we get:
$\begin{align}
& {{\Delta }_{1}}=\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & 0 \\
{{a}_{2}} & {{b}_{2}} & 0 \\
{{a}_{3}} & {{b}_{3}} & 0 \\
\end{matrix} \right| \\
& {{\Delta }_{2}}=\left| \begin{matrix}
{{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
{{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
0 & 0 & 0 \\
\end{matrix} \right| \\
\end{align}$
Now, that we have the determinants ${{\Delta }_{1}}$ and ${{\Delta }_{2}}$, we can obtain their value.
Now, ${{\Delta }_{1}}$ is given as:
\[\begin{align}
& {{\Delta }_{1}}=\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & 0 \\
{{a}_{2}} & {{b}_{2}} & 0 \\
{{a}_{3}} & {{b}_{3}} & 0 \\
\end{matrix} \right| \\
& \text{Now, opening via }{{\text{R}}_{1}}\text{ we get:} \\
& {{\Delta }_{1}}={{a}_{1}}\left( {{b}_{2}}.0-{{b}_{3}}.0 \right)-{{b}_{1}}\left( {{a}_{2}}.0-{{a}_{3}}.0 \right)+0\left( {{a}_{2}}{{b}_{3}}-{{a}_{3}}{{b}_{2}} \right) \\
& \Rightarrow {{\Delta }_{1}}={{a}_{1}}.0-{{b}_{1}}.0+0\left( {{a}_{2}}{{b}_{3}}-{{a}_{3}}{{b}_{2}} \right) \\
& \Rightarrow {{\Delta }_{1}}=0-0+0 \\
& \Rightarrow {{\Delta }_{1}}=0 \\
\end{align}\]
Thus, the value of ${{\Delta }_{1}}$ is 0.
Now, ${{\Delta }_{2}}$ is given as:
$\begin{align}
& {{\Delta }_{2}}=\left| \begin{matrix}
{{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
{{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
0 & 0 & 0 \\
\end{matrix} \right| \\
& \text{Opening via }{{\text{R}}_{1}}\text{ we get:} \\
& {{\Delta }_{2}}={{b}_{1}}\left( {{a}_{2}}.0-{{a}_{3}}.0 \right)-{{b}_{2}}\left( {{a}_{1}}.0-{{a}_{3}}.0 \right)+{{b}_{3}}\left( {{a}_{1}}.0-{{a}_{2}}.0 \right) \\
& \Rightarrow {{\Delta }_{2}}={{b}_{1}}.0-{{b}_{2}}.0+{{b}_{3}}.0 \\
& \Rightarrow {{\Delta }_{2}}=0-0+0 \\
& \Rightarrow {{\Delta }_{2}}=0 \\
\end{align}$
Thus, the value of ${{\Delta }_{2}}$ is also 0.
Now, we know that $\Delta $ is the product of ${{\Delta }_{1}}$ and ${{\Delta }_{2}}$.
Thus, we get:
$\begin{align}
& \Delta ={{\Delta }_{1}}\times {{\Delta }_{2}} \\
& \Rightarrow \Delta =0\times 0 \\
& \Rightarrow \Delta =0 \\
\end{align}$
Thus, the value of the required determinant is 0.
So, the correct answer is “Option C”.
Note: Here we have solved ${{\Delta }_{1}}$ and ${{\Delta }_{2}}$ to obtain their values. But we can straightaway write them equal to 0 because they have all the elements of one row or one column as 0. So, if we were to open the determinant by that row or column, we will obtain the determinant as 0 only. Thus, the value of all the determinants having all the elements of at least one row or column as 0 is 0.
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