Answer
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Hint: Expand the given determinant about any row or column. Expansion of a determinant
$\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|$ is given along the first row of the determinant as
${{a}_{1}}({{b}_{2}}{{c}_{3}}-{{c}_{2}}{{b}_{3}})-{{b}_{1}}({{a}_{2}}{{c}_{3}}-{{a}_{3}}{{c}_{2}})+{{c}_{1}}({{a}_{2}}{{b}_{3}}-{{a}_{3}}{{b}_{2}})$
Complete step-by-step answer:
Let us suppose the value of given determinant be D. So, we get
$D$= $\left| \begin{matrix}
18 & 40 & 89 \\
40 & 89 & 198 \\
89 & 198 & 440 \\
\end{matrix} \right|$ ………………. (i)
As, we know any determinant can be expanded about any of the row and column of that determinant in the following way:-
Let us suppose we have a determinant as
$\vartriangle =\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|$
We can expand this determinant about the first row as
${{a}_{1}}({{b}_{2}}{{c}_{3}}-{{b}_{3}}{{c}_{2}})-{{b}_{1}}({{a}_{2}}{{c}_{3}}-{{a}_{3}}{{c}_{2}})+{{c}_{1}}({{a}_{2}}{{b}_{3}}-{{b}_{2}}{{a}_{3}})$ ………….. (ii)
So, we can expand the given determinant in equation with the help of equation (ii) as (about first row):-
$D=\left| \begin{matrix}
18 & 40 & 89 \\
40 & 89 & 198 \\
89 & 198 & 440 \\
\end{matrix} \right|$
$D=18(39160-39204)-40(17600-17622)+89(7920-7921)$
On further simplifying the above expression, we get
$D=18(-44)-40(-22)+89(-1)$
$D=-1$
Hence, the value of the given determinant is -1.
So, we get $\left| \begin{matrix}
18 & 40 & 89 \\
40 & 89 & 198 \\
89 & 198 & 440 \\
\end{matrix} \right|=-1$
Note: Instead of doing above lengthy calculations we can perform operations on rows or columns of the determinant to convert their elements to smaller numbers and then expand the determinant along the most suitable row or column , This method will actually save a lot of time if the right operations click to you.
$\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|$ is given along the first row of the determinant as
${{a}_{1}}({{b}_{2}}{{c}_{3}}-{{c}_{2}}{{b}_{3}})-{{b}_{1}}({{a}_{2}}{{c}_{3}}-{{a}_{3}}{{c}_{2}})+{{c}_{1}}({{a}_{2}}{{b}_{3}}-{{a}_{3}}{{b}_{2}})$
Complete step-by-step answer:
Let us suppose the value of given determinant be D. So, we get
$D$= $\left| \begin{matrix}
18 & 40 & 89 \\
40 & 89 & 198 \\
89 & 198 & 440 \\
\end{matrix} \right|$ ………………. (i)
As, we know any determinant can be expanded about any of the row and column of that determinant in the following way:-
Let us suppose we have a determinant as
$\vartriangle =\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|$
We can expand this determinant about the first row as
${{a}_{1}}({{b}_{2}}{{c}_{3}}-{{b}_{3}}{{c}_{2}})-{{b}_{1}}({{a}_{2}}{{c}_{3}}-{{a}_{3}}{{c}_{2}})+{{c}_{1}}({{a}_{2}}{{b}_{3}}-{{b}_{2}}{{a}_{3}})$ ………….. (ii)
So, we can expand the given determinant in equation with the help of equation (ii) as (about first row):-
$D=\left| \begin{matrix}
18 & 40 & 89 \\
40 & 89 & 198 \\
89 & 198 & 440 \\
\end{matrix} \right|$
$D=18(39160-39204)-40(17600-17622)+89(7920-7921)$
On further simplifying the above expression, we get
$D=18(-44)-40(-22)+89(-1)$
$D=-1$
Hence, the value of the given determinant is -1.
So, we get $\left| \begin{matrix}
18 & 40 & 89 \\
40 & 89 & 198 \\
89 & 198 & 440 \\
\end{matrix} \right|=-1$
Note: Instead of doing above lengthy calculations we can perform operations on rows or columns of the determinant to convert their elements to smaller numbers and then expand the determinant along the most suitable row or column , This method will actually save a lot of time if the right operations click to you.
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