Determinants And Matrices Set-1

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Example -1

Find the value of

\[\Delta  = \left| {\begin{align}&(a_1-b_1)^2\quad(a_1-b_2)^2\quad(a_1-b_3)^2\quad(a_1-b_4)^2\\
&(a_2-b_1)^2\quad(a_2-b_2)^2\quad(a_2-b_3)^2\quad(a_2-b_4)^2\\&(a_3-b_1)^2\quad(a_3-b_2)^2\quad(a_3-b_3)^2\quad(a_3-b_4)^2\\&(a_4-b_1)^2\quad(a_4-b_2)^2\quad(a_4-b_3)^2\quad(a_4-b_4)^2\end{align}\;} \right|\]

Solution: The given determinant can be expressed as product of two simpler determinants:

\[\begin{align}\Delta &= \left| {\;\begin{gathered}{a_1^2}&{ - 2{a_1}}&1&0\\{a_2^2}&{ - 2{a_2}}&1&0\\{a_3^2}&{ - 2{a_3}}&1&0\\{a_4^2}&{ - 2{a_4}}&1&0\end{gathered}\;} \right|\;\; \times \;\;\left| {\;\begin{gathered}1&{{b_1}}&{b_1^2}&0\\1&{{b_2}}&{b_2^2}&0\\1&{{b_3}}&{b_3^2}&0\\1&{{b_4}}&{b_4^2}&0\end{gathered}\;} \right|\\&=0\end{align}\]

Example -2

Prove that

\[\left| {\;\begin{array}{*{20}{c}}{bc - {a^2}}&{ca - {b^2}}&{ab - {c^2}\;}\\{ca - {b^2}}&{ab - {c^2}}&{bc - {a^2}}\\{ab - {c^2}}&{bc - {a^2}}&{ca - {b^2}}\end{array}} \right| = \left| {\;\begin{array}{*{20}{c}}p&q&q\\q&p&q\\q&q&p\end{array}\;} \right|,\]

where \(p = {a^2} + {b^2} + {c^2}\;{\rm{and}}\;\;q = ab + .bc + ca.\)

Solution: If you consider the determinant on the LHS say \({\Delta _L}\) carefully, you’ll see it is the determinant of the co-factors of

\[\Delta  = \left| {\;\begin{gathered}a&b&c\\b&c&a\\c&a&b\end{gathered}\;} \right|\]

Also, \(\Delta {\Delta _L} = {\Delta ^3}\left( {{\rm{why}}?} \right),\;\;{\rm{so}}\;{\Delta _L} = {\Delta ^2}\)

Thus,

\[{\Delta _2} = \left| {\;\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\;} \right|\;\; \times \;\;\left| {\;\begin{array}{*{20}{c}}a&b&c\\b&c&a\\
c&a&b\end{array}\;} \right| = {\Delta _R}\;\;\;\left( {{\text{the determinant on the RHS}}} \right)\]

Example -3

Let \(A,B,C\) be the angles of a \(\Delta ABC.\) Find the value of

\[\Delta  = \left| {\;\begin{array}{*{20}{c}}{{e^{i2A}}}&{{e^{ - iC}}}&{{e^{ - iB}}\;}\\{{e^{ - iC}}}&{{e^{i2B}}}&{{e^{ - iA}}}\\{{e^{ - iB}}}&{{e^{ - iA}}}&{{e^{i2C}}}\end{array}} \right|\]

Solution: We’ll make use of the fact that \(A + B + C = \pi .\) Taking out factors of \({e^{i2A}},{e^{ - iC}},{e^{ - iB}}\) from \({C_1},{C_2},{C_3}\) respectively, we have:

\[\Delta  = {e^{i\left( {2A - B - C} \right)}}\left| {\;\begin{array}{*{20}{c}}1&1&1\\{{e^{ - i\left( {2A + C} \right)}}}&{{e^{i\left( {2B + C} \right)}}}&{{e^{i\left( {B - A} \right)}}}\\{{e^{ - i\left( {2A + B} \right)}}}&{{e^{i\left( {C - A} \right)}}}&{{e^{i\left( {B + 2C} \right)}}}\end{array}\;} \right|\]

Now, we note that \(2A + C = \pi  - \left( {B - A} \right),2B + C = \pi  - \left( {A - B} \right),2C + B = \pi  - \left( {A - C} \right)\), \(2A + B = \pi  + \left( {A - C} \right).\)Therefore,

\[\begin{align}\Delta  &= {e^{i\left( {2A - B - C} \right)}}\left| {\begin{gathered}1&1&1\\{ - {e^{i\left( {B - A} \right)}}}&{ - {e^{i\left( {B - A} \right)}}}&{{e^{i\left( {B - A} \right)}}}\\{\; - {e^{i\left( {C - A} \right)}}}&{{e^{i\left( {C - A} \right)}}}&{ - {e^{i\left( {C - A} \right)}}}\end{gathered}\;} \right|\\ &= \left| {\;\begin{gathered}{\;\;1}&{\;\;1}&{\;\;\;1}\\{ - 1}&{ - 1}&{\;\;1}\\{ - 1}&{\;\;1}&{ - 1}
\end{gathered}\;} \right|\\&=4 \end{align}\]

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