Multiplication of Two Determinants - Practice Questions & MCQ

Multiplication of Determinant

Let two determinant of third-order be

$
\Delta_1=\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right| \text { and } \Delta_2=\left|\begin{array}{ccc}
\alpha_1 & \beta_1 & \gamma_1 \\
\alpha_2 & \beta_2 & \gamma_2 \\
\alpha_3 & \beta_3 & \gamma_3
\end{array}\right|
$
We can multiply these row-by-row or column-by-column or row-by-column or column-by-row
Row-by-row multiplication of these two determinants is given by

$
\Delta_1 \times \Delta_2=\left|\begin{array}{lll}
a_1 \alpha_1+b_1 \beta_1+c_1 \gamma_1 & a_1 \alpha_2+b_1 \beta_2+c_1 \gamma_2 & a_1 \alpha_3+b_1 \beta_3+c_1 \gamma_3 \\
a_2 \alpha_1+b_2 \beta_1+c_2 \gamma_1 & a_2 \alpha_2+b_2 \beta_2+c_2 \gamma_2 & a_2 \alpha_3+b_2 \beta_3+c_2 \gamma_3 \\
a_3 \alpha_1+b_3 \beta_1+c_3 \gamma_1 & a_3 \alpha_2+b_3 \beta_2+c_3 \gamma_2 & a_3 \alpha_3+b_3 \beta_3+c_3 \gamma_3
\end{array}\right|
$
Multiplication can also be performed row by column; column by row or column by column as required in the problem.

To express a determinant as a product of two determinants, one requires lots of practice and this can be done only by inspection and trial.

Property:

If $\mathrm{A}_1, \mathrm{~B}_1, \mathrm{C}_1, \ldots$...are respectively the cofactors of the elements $\mathrm{a}_1, \mathrm{~b}_1, \mathrm{c}_1$ $\qquad$ of the determinant $\Delta=\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|, \Delta \neq 0$, then $\left|\begin{array}{lll}A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3\end{array}\right|=\Delta^2$

Proof:
given, $\quad \Delta=\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|$ and, $\mathrm{A}_1, \mathrm{~B}_1, \mathrm{C}_1, \ldots$. are respectively the cofactors of the elements $\mathrm{a}_1, \mathrm{~b}_1, \mathrm{c}_1 \ldots \ldots \ldots$. Hence,

$
\begin{aligned}
& \left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|\left|\begin{array}{lll}
A_1 & B_1 & C_1 \\
A_2 & B_2 & C_2 \\
A_3 & B_3 & C_3
\end{array}\right| \\
& =\left|\begin{array}{ccc}
a_1 A_1+b_1 B_1+c_1 C_1 & a_1 A_2+b_1 B_2+c_1 C_2 & a_1 A_3+b_1 B_3+c_1 C_3 \\
a_2 A_1+b_2 B_1+c_2 C_1 & a_2 A_2+b_2 B_2+c_2 C_2 & a_2 A_3+b_2 B_3+c_2 C_3 \\
a_3 A_1+b_3 B_1+c_3 C_1 & a_3 A_2+b_3 B_2+c_3 C_2 & a_3 A_3+b_3 B_3+c_3 C_3
\end{array}\right|
\end{aligned}
$

[row by row multiplication]

$
\begin{aligned}
& =\left|\begin{array}{ccc}
\Delta & 0 & 0 \\
0 & \Delta & 0 \\
0 & 0 & \Delta
\end{array}\right|=\Delta^3 \\
& \because \mathrm{a}_{\mathrm{i}} A_{\mathrm{j}}+\mathrm{b}_{\mathrm{i}} \mathrm{~B}_{\mathrm{j}}+\mathrm{c}_{\mathrm{i}} \mathrm{C}_{\mathrm{j}}=\left\{\begin{array}{cc}
\Delta, & i=j \\
0, & i \neq j
\end{array}\right. \\
& \Rightarrow \Delta\left|\begin{array}{lll}
A_1 & B_1 & C_1 \\
A_2 & B_2 & C_2 \\
A_3 & B_3 & C_3
\end{array}\right|=\Delta^3 \\
& \Rightarrow\left|\begin{array}{ccc}
A_1 & B_1 & C_1 \\
A_2 & B_2 & C_2 \\
A_3 & B_3 & C_3
\end{array}\right|=\Delta^2
\end{aligned}
$