Answer
Verified
417.6k+ views
Hint: For evaluating the given determinant, we can apply row or column operations to make the expansion easier.
Complete step-by-step answer:
$ \Rightarrow $ Let $\vartriangle = $ $\left| {\begin{array}{*{20}{c}}
1&a&{bc} \\
1&b&{ca} \\
1&c&{ab}
\end{array}} \right|$
Now, here we will apply row operations for solving the above determinant.
Now, applying row operation in ${R_2}$
$ \Rightarrow {R_2} \to {R_2} - {R_1}$
\[ \Rightarrow \vartriangle = \left| {\begin{array}{*{20}{c}}
1&a&{bc} \\
0&{b - a}&{ca - bc} \\
1&c&{ab}
\end{array}} \right|\]
Now, applying row operation in R3
$ \Rightarrow {R_3} \to {R_3} - {R_1}$
\[ \Rightarrow \vartriangle = \left| {\begin{array}{*{20}{c}}
1&a&{bc} \\
0&{b - a}&{ca - bc} \\
0&{c - a}&{ab - bc}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
1&a&{bc} \\
0&{b - a}&{c\left( {a - b} \right)} \\
0&{c - a}&{b\left( {a - c} \right)}
\end{array}} \right|\]
Now, for making the expansion easy we can take common terms out from the determinant. So taking (a – b) and (a – c) common from ${R_2}$ and ${R_3}$ respectively, we get,
\[ \Rightarrow \vartriangle = \left( {a - b} \right)\left( {a - c} \right)\left| {\begin{array}{*{20}{c}}
1&a&{bc} \\
0&{ - 1}&c \\
0&{ - 1}&b
\end{array}} \right|\]
Now we can expand the determinant along any row or column.
So, here it is better to expand the above determinant along the first column as two elements of the first column are 0. So, it will make our calculations easier.
So, expanding along first column, we get,
\[ \Rightarrow \vartriangle = \left( {a - b} \right)\left( {a - c} \right)\left| {\begin{array}{*{20}{c}}
{ - 1}&c \\
{ - 1}&b
\end{array}} \right| = \left( {a - b} \right)\left( {a - c} \right)\left( { - b + c} \right)\]
$ \Rightarrow $ Hence $\left| {\begin{array}{*{20}{c}}
1&a&{bc} \\
1&b&{ca} \\
1&c&{ab}
\end{array}} \right| = \vartriangle = \left( {a - b} \right)\left( {a - c} \right)\left( {c - b} \right)$
Note: Whenever we come up with these types of problems, to make calculations easy, first reduce the determinant by applying row or column operations then expand it. Take care of the signs while carrying out expansions. It is better to expand along a row or column with most 0’s.
Complete step-by-step answer:
$ \Rightarrow $ Let $\vartriangle = $ $\left| {\begin{array}{*{20}{c}}
1&a&{bc} \\
1&b&{ca} \\
1&c&{ab}
\end{array}} \right|$
Now, here we will apply row operations for solving the above determinant.
Now, applying row operation in ${R_2}$
$ \Rightarrow {R_2} \to {R_2} - {R_1}$
\[ \Rightarrow \vartriangle = \left| {\begin{array}{*{20}{c}}
1&a&{bc} \\
0&{b - a}&{ca - bc} \\
1&c&{ab}
\end{array}} \right|\]
Now, applying row operation in R3
$ \Rightarrow {R_3} \to {R_3} - {R_1}$
\[ \Rightarrow \vartriangle = \left| {\begin{array}{*{20}{c}}
1&a&{bc} \\
0&{b - a}&{ca - bc} \\
0&{c - a}&{ab - bc}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
1&a&{bc} \\
0&{b - a}&{c\left( {a - b} \right)} \\
0&{c - a}&{b\left( {a - c} \right)}
\end{array}} \right|\]
Now, for making the expansion easy we can take common terms out from the determinant. So taking (a – b) and (a – c) common from ${R_2}$ and ${R_3}$ respectively, we get,
\[ \Rightarrow \vartriangle = \left( {a - b} \right)\left( {a - c} \right)\left| {\begin{array}{*{20}{c}}
1&a&{bc} \\
0&{ - 1}&c \\
0&{ - 1}&b
\end{array}} \right|\]
Now we can expand the determinant along any row or column.
So, here it is better to expand the above determinant along the first column as two elements of the first column are 0. So, it will make our calculations easier.
So, expanding along first column, we get,
\[ \Rightarrow \vartriangle = \left( {a - b} \right)\left( {a - c} \right)\left| {\begin{array}{*{20}{c}}
{ - 1}&c \\
{ - 1}&b
\end{array}} \right| = \left( {a - b} \right)\left( {a - c} \right)\left( { - b + c} \right)\]
$ \Rightarrow $ Hence $\left| {\begin{array}{*{20}{c}}
1&a&{bc} \\
1&b&{ca} \\
1&c&{ab}
\end{array}} \right| = \vartriangle = \left( {a - b} \right)\left( {a - c} \right)\left( {c - b} \right)$
Note: Whenever we come up with these types of problems, to make calculations easy, first reduce the determinant by applying row or column operations then expand it. Take care of the signs while carrying out expansions. It is better to expand along a row or column with most 0’s.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
State the differences between manure and fertilize class 8 biology CBSE
Why are xylem and phloem called complex tissues aBoth class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
What would happen if plasma membrane ruptures or breaks class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What precautions do you take while observing the nucleus class 11 biology CBSE
What would happen to the life of a cell if there was class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE