Multiplication of Two Determinants

Determinant of matrix

The determinant of a matrix A is a number that is calculated from the matrix. For a determinant to exist, matrix A must be a square matrix. The determinant of a matrix is denoted by det A or $|\mathrm{A}|$.

For $2 \times 2$ matrices

$
\mathrm{A}=\left[\begin{array}{ll}
a_1 & a_2 \\
b_1 & b_2
\end{array}\right]
$

then $\operatorname{det} \mathrm{A}$ is :

$
|\mathrm{A}|=\left|\begin{array}{ll}
a_1 & a_2 \\
b_1 & b_2
\end{array}\right|=\mathrm{a}_1 \times \mathrm{b}_2-\mathrm{a}_2 \times \mathrm{b}_1
$

For a $3 \times 3$ matrix determinant can be calculated in the following way :

$
\text { let } \mathrm{A}=\left[\begin{array}{lll}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right]
$

then we find $\operatorname{det} \mathrm{A}$ in following way

$
|A|=a_1\left(b_2 \cdot c_3-b_3 \cdot c_2\right)-a_2\left(b_1 \cdot c_3-c_1 b_3\right)+a_3\left(b_1 c_2-b_2 c_1\right)
$
This same process we follow to evaluate the determinant of the matrix of any order. Notice that we start the first term with the +ve sign then 2nd with the -ve sign and 3rd again with the +ve sign, this sign sequence is followed for any order of matrix.

This whole process is row-dependent, the same process can be done using columns, which means we can select an element along a column delete their row and column compute the determinant of the out matrix, and then multiply it with the element that we select. And we will get the same result as we get while doing the whole process along the row.

Multiplication of Determinant

There are two types of matrix multiplication :

1) Multiplication of determinant by scalar quantity

2) Multiplication of determinant by another determinant

Multiplication of determinant by scalar quantity

If A is a square matrix and k is a scalar quantity then, $|\mathrm{kA}|=\mathrm{k}^{\mathrm{n}}|\mathrm{A}|$, where n is the order of A

Multiplication of determinant by another determinant

Determinant multiplication is a binary operation that produces a determinant from two determinants. For determinant multiplication, the order of both the determinants should be the same.

$\begin{equation}
\begin{aligned}
&\text { Let two determinants of third-order be }\\
&\Delta_1=\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right| \text { and } \Delta_2=\left|\begin{array}{ccc}
\alpha_1 & \beta_1 & \gamma_1 \\
\alpha_2 & \beta_2 & \gamma_2 \\
\alpha_3 & \beta_3 & \gamma_3
\end{array}\right|
\end{aligned}
\end{equation}$

We can multiply these row-by-row or column-by-column or row-by-column or column-by-row

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

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

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

Important points

For n-order determinant, $\Delta_c=\Delta^{n-1}$, where $\Delta_c$ is the determinant formed by the cofactors of $\Delta$ and $n$ is the order of determinant. This property is useful in studying the adjoint of a matrix.

Recommended Video Based on Multiplication of Determinant:

Solved Examples Based on Multiplication of Determinant

Example 1:
If $\alpha, \beta \neq 0$, and $f(n)=\alpha^n+\beta^n$ and $\left|\begin{array}{ccc}3 & 1+f(1) & 1+f(2) \\ 1+f(1) & 1+f(2) & 1+f(3) \\ 1+f(2) & 1+f(3) & 1+f(4)\end{array}\right|$
$=K(1-\alpha)^2(1-\beta)^2(\alpha-\beta)^2$, then K is equal to:

Solution: The given determinant can be written as

$
\left|\begin{array}{ccc}
1+1+1 & 1+\alpha+\beta & 1+\alpha^2+\beta^2 \\
1+\alpha+\beta & 1+\alpha^2+\beta^2 & 1+\alpha^3+\beta^3 \\
1+\alpha^2+\beta^2 & 1+\alpha^3+\beta^3 & 1+\alpha^4+\beta^4
\end{array}\right|
$

Expressing it as a product of two determinants

$
=\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & \alpha & \beta \\
1 & \alpha^2 & \beta^2
\end{array}\right|\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & \alpha & \alpha^2 \\
1 & \beta & \beta^2
\end{array}\right|
$

Now we know that each of these determinants equal

$
(1-\alpha)(1-\beta)(\alpha-\beta)
$

Hence, given expression

$
\begin{aligned}
& =(1-\alpha)^2(1-\beta)^2(\alpha-\beta)^2 \\
& \therefore K=1
\end{aligned}
$

Hence, the answer is 1 .