Question

Show that $\left| {\begin{array}{*{20}{c}}
  {{a_1}{l_1} + {b_1}{m_1}}&{{a_1}{l_2} + {b_1}{m_2}}&{{a_1}{l_3} + {b_1}{m_3}} \\
  {{a_2}{l_1} + {b_2}{m_1}}&{{a_2}{l_2} + {b_2}{m_2}}&{{a_2}{l_3} + {b_2}{m_3}} \\
  {{a_3}{l_1} + {b_3}{m_1}}&{{a_3}{l_2} + {b_3}{m_2}}&{{a_3}{l_3} + {b_3}{m_3}}
\end{array}} \right| = 0$.

Answer 1

Hint: Firstly, Try to apply some row and column operations and use properties of the determinant to break the given determinant. Use Simple manipulations like $\left| {A + B} \right| = \left| A \right| + \left| B \right|$, $\left| {kA} \right| = k\left| A \right|$, ${R_1} \to {R_1} - {R_2}$.

Complete step by step solution:
Given,
L.H.S. = $\left| {\begin{array}{*{20}{c}}
  {{a_1}{l_1} + {b_1}{m_1}}&{{a_1}{l_2} + {b_1}{m_2}}&{{a_1}{l_3} + {b_1}{m_3}} \\
  {{a_2}{l_1} + {b_2}{m_1}}&{{a_2}{l_2} + {b_2}{m_2}}&{{a_2}{l_3} + {b_2}{m_3}} \\
  {{a_3}{l_1} + {b_3}{m_1}}&{{a_3}{l_2} + {b_3}{m_2}}&{{a_3}{l_3} + {b_3}{m_3}}
\end{array}} \right|$
Let $\Delta = \left| {\begin{array}{*{20}{c}}
  {{a_1}{l_1} + {b_1}{m_1}}&{{a_1}{l_2} + {b_1}{m_2}}&{{a_1}{l_3} + {b_1}{m_3}} \\
  {{a_2}{l_1} + {b_2}{m_1}}&{{a_2}{l_2} + {b_2}{m_2}}&{{a_2}{l_3} + {b_2}{m_3}} \\
  {{a_3}{l_1} + {b_3}{m_1}}&{{a_3}{l_2} + {b_3}{m_2}}&{{a_3}{l_3} + {b_3}{m_3}}
\end{array}} \right|$
Now by Applying Properties of Determinant, $\left| {A + B} \right| = \left| A \right| + \left| B \right|$
$ \Rightarrow \Delta = \left| {\begin{array}{*{20}{c}}
  {{a_1}{l_1}}&{{a_1}{l_2}}&{{a_1}{l_3}} \\
  {{a_2}{l_1}}&{{a_2}{l_2}}&{{a_2}{l_3}} \\
  {{a_3}{l_1}}&{{a_3}{l_2}}&{{a_3}{l_3}}
\end{array}} \right| + \left| {\begin{array}{*{20}{c}}
  {{b_1}{m_1}}&{{b_1}{m_2}}&{{b_1}{m_3}} \\
  {{b_2}{m_1}}&{{b_2}{m_2}}&{{b_2}{m_3}} \\
  {{b_3}{m_1}}&{{b_3}{m_2}}&{{b_3}{m_3}}
\end{array}} \right|$ ………………….. (1)
Now Use The Property of Determinant, $\left| {kA} \right| = k\left| A \right|$
Taking ${a_1}$, ${a_2}$, ${a_3}$common from ${C_1}$, ${C_2}$, ${C_3}$ Respectively, We get,
$\therefore \left| {\begin{array}{*{20}{c}}
  {{a_1}{l_1}}&{{a_1}{l_2}}&{{a_1}{l_3}} \\
  {{a_2}{l_1}}&{{a_2}{l_2}}&{{a_2}{l_3}} \\
  {{a_3}{l_1}}&{{a_3}{l_2}}&{{a_3}{l_3}}
\end{array}} \right| = {a_1}{a_2}{a_3}\left| {\begin{array}{*{20}{c}}
  {{l_1}}&{{l_2}}&{{l_3}} \\
  {{l_1}}&{{l_2}}&{{l_3}} \\
  {{l_1}}&{{l_2}}&{{l_3}}
\end{array}} \right|$ ………………………. (2)
Similarly,
 Use The Same Property of Determinant, $\left| {kA} \right| = k\left| A \right|$
Taking ${b_1}$, ${b_2}$, ${b_3}$common from ${C_1}$, ${C_2}$, ${C_3}$ Respectively, We get,
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {{b_1}{m_1}}&{{b_1}{m_2}}&{{b_1}{m_3}} \\
  {{b_2}{m_1}}&{{b_2}{m_2}}&{{b_2}{m_3}} \\
  {{b_3}{m_1}}&{{b_3}{m_2}}&{{b_3}{m_3}}
\end{array}} \right| = {b_1}{b_2}{b_3}\left| {\begin{array}{*{20}{c}}
  {{m_1}}&{{m_2}}&{{m_3}} \\
  {{m_1}}&{{m_2}}&{{m_3}} \\
  {{m_1}}&{{m_2}}&{{m_3}}
\end{array}} \right|$ …………………… (3)
Put results obtained in equations (2) and (3) in equation (1), We get:
$ \Rightarrow \Delta = {a_1}{a_2}{a_3}\left| {\begin{array}{*{20}{c}}
  {{l_1}}&{{l_2}}&{{l_3}} \\
  {{l_1}}&{{l_2}}&{{l_3}} \\
  {{l_1}}&{{l_2}}&{{l_3}}
\end{array}} \right| + {b_1}{b_2}{b_3}\left| {\begin{array}{*{20}{c}}
  {{m_1}}&{{m_2}}&{{m_3}} \\
  {{m_1}}&{{m_2}}&{{m_3}} \\
  {{m_1}}&{{m_2}}&{{m_3}}
\end{array}} \right|$Now apply the row transformation in both the determinants, ${R_1} \to {R_1} - {R_2}$ , We get:
$ \Rightarrow \Delta = {a_1}{a_2}{a_3}\left| {\begin{array}{*{20}{c}}
  0&0&0 \\
  {{l_1}}&{{l_2}}&{{l_3}} \\
  {{l_1}}&{{l_2}}&{{l_3}}
\end{array}} \right| + {b_1}{b_2}{b_3}\left| {\begin{array}{*{20}{c}}
  0&0&0 \\
  {{m_1}}&{{m_2}}&{{m_3}} \\
  {{m_1}}&{{m_2}}&{{m_3}}
\end{array}} \right|$
Now by using the result that if the Row of a determinant is zero then the value of the determinant is Zero.
$\therefore \Delta = {a_1}{a_2}{a_3}(0) + {b_1}{b_2}{b_3}(0)$
$ \Rightarrow \Delta = 0$= R.H.S.

Hence proved that the value of the determinant $\left| {\begin{array}{*{20}{c}}
  {{a_1}{l_1} + {b_1}{m_1}}&{{a_1}{l_2} + {b_1}{m_2}}&{{a_1}{l_3} + {b_1}{m_3}} \\
  {{a_2}{l_1} + {b_2}{m_1}}&{{a_2}{l_2} + {b_2}{m_2}}&{{a_2}{l_3} + {b_2}{m_3}} \\
  {{a_3}{l_1} + {b_3}{m_1}}&{{a_3}{l_2} + {b_3}{m_2}}&{{a_3}{l_3} + {b_3}{m_3}}
\end{array}} \right|$ is Zero.

Note:
In the above solution, we use a result that if the Row of a determinant is zero then the value of the determinant is Zero.