Properties of Determinants - Practice Questions & MCQ

Properties of Determinants

Property 1: The value of the determinant remains unchanged if its rows and columns are interchanged.

For example,
Let, $\Delta=\left|\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right|$
Expanding along first row, we get

$
\begin{aligned}
& \Delta=\mathrm{a}_1\left|\begin{array}{ll}
b_2 & b_3 \\
c_2 & c_3
\end{array}\right|-\mathrm{a}_2\left|\begin{array}{cc}
b_1 & b_3 \\
c_1 & c_3
\end{array}\right|+\mathrm{a}_3\left|\begin{array}{ll}
b_1 & b_2 \\
c_1 & c_2
\end{array}\right| \\
& \Delta=\mathrm{a}_1\left(\mathrm{~b}_2 \mathrm{c}_3-\mathrm{b}_3 \mathrm{c}_2\right)-\mathrm{a}_2\left(\mathrm{~b}_1 \mathrm{c}_3-\mathrm{b}_3 \mathrm{c}_1\right)+\mathrm{a}_3\left(\mathrm{~b}_1 \mathrm{c}_2-\mathrm{b}_2 \mathrm{c}_1\right)
\end{aligned}
$
By interchanging the rows and columns of $\Delta$, we get the determinant

$
\Delta^{\prime}=\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|
$
Expanding $\Delta^{\prime}$ along first column, we get

$
\begin{aligned}
& \Delta^{\prime}=a_1\left(b_2 \mathrm{c}_3-\mathrm{b}_3 \mathrm{c}_2\right)-\mathrm{a}_2\left(\mathrm{~b}_1 \mathrm{c}_3-\mathrm{b}_3 \mathrm{c}_1\right)+\mathrm{a}_3\left(\mathrm{~b}_1 \mathrm{c}_2-\mathrm{b}_2 \mathrm{c}_1\right) \\
& \Delta=\Delta^{\prime}
\end{aligned}
$

Property 2: If any two rows or two columns of a determinant are interchanged, then sign of determinant changes but the numerical value remains unaltered.

For example

Let, $\Delta=\left|\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right|$
Expanding along first row, we get

$
\begin{aligned}
\Delta & =\mathrm{a}_1\left|\begin{array}{ll}
b_2 & b_3 \\
c_2 & c_3
\end{array}\right|-\mathrm{a}_2\left|\begin{array}{ll}
b_1 & b_3 \\
c_1 & c_3
\end{array}\right|+\mathrm{a}_3\left|\begin{array}{ll}
b_1 & b_2 \\
c_1 & c_2
\end{array}\right| \\
\Delta & =\mathrm{a}_1\left(\mathrm{~b}_2 \mathrm{c}_3-\mathrm{b}_3 \mathrm{c}_2\right)-\mathrm{a}_2\left(\mathrm{~b}_1 \mathrm{c}_3-\mathrm{b}_3 \mathrm{c}_1\right)+\mathrm{a}_3\left(\mathrm{~b}_1 \mathrm{c}_2-\mathrm{b}_2 \mathrm{c}_1\right)
\end{aligned}
$
Interchanging first and third rows, the new determinant obtained is given by

$
\Delta^{\prime}=\left|\begin{array}{lll}
c_1 & c_2 & c_3 \\
b_1 & b_2 & b_3 \\
a_1 & a_2 & a_3
\end{array}\right|
$
Expanding along third row, we get

$
\begin{aligned}
\Delta^{\prime} & =\mathrm{a}_1\left(\mathrm{c}_2 \mathrm{~b}_3-\mathrm{c}_3 \mathrm{~b}_2\right)-\mathrm{a}_2\left(\mathrm{c}_1 \mathrm{~b}_3-\mathrm{c}_3 \mathrm{~b}_1\right)+\mathrm{a}_3\left(\mathrm{~b}_2 \mathrm{c}_1-\mathrm{b}_1 \mathrm{c}_2\right) \\
& =-\left[\mathrm{a}_1\left(\mathrm{~b}_2 \mathrm{c}_3-\mathrm{b}_3 \mathrm{c}_2\right)-\mathrm{a}_2\left(\mathrm{~b}_1 \mathrm{c}_3-\mathrm{b}_3 \mathrm{c}_1\right)+\mathrm{a}_3\left(\mathrm{~b}_1 \mathrm{c}_2-\mathrm{b}_2 \mathrm{c}_1\right)\right] \\
\Delta & =-\Delta^{\prime}
\end{aligned}
$
Property 3 If there is an interchange of rows or coloumns twice, then the value of the determinant remains the same.

If $\Delta_n$ is the determinant obtained by n such successive operations, then

$
\Delta_n=\left\{\begin{array}{cc}
-\Delta, & \text { if } \mathrm{n} \text { is odd } \\
\Delta, & \text { if } \mathrm{n} \text { is even }
\end{array}\right.
$
 

Property 4

If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then the value of the determinant is zero.

For Example,

If we interchange the identical rows (or columns) of the determinant $\Delta$, then by property $2, \Delta$ changes its sign
Let, $\Delta=\left|\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ a_1 & a_2 & a_3\end{array}\right|=-\left|\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ a_1 & a_2 & a_3\end{array}\right| \quad$ (interchanging row 1 and row 3) $=-\Delta \quad$ [By property 2]

$
\begin{aligned}
2 \Delta & =0 \\
\Delta & =0
\end{aligned}
$

Property 5

If each element of a row (or a column) of a determinant is multiplied by a constant k , then the value of the determinant is multiplied by k .
For example
Let, $\Delta=\left|\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right|$
and $\Delta^{\prime}$ be the determinant obtained by multiplying the elements of the first row by k .

$
\Delta^{\prime}=\left|\begin{array}{ccc}
k a_1 & k a_2 & k a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right|
$
Expanding along first row, we get

$
\begin{aligned}
& \Delta^{\prime}=k \mathrm{ka}_1\left(\mathrm{~b}_2 \mathrm{c}_3-\mathrm{b}_3 \mathrm{c}_2\right)-\mathrm{ka}_2\left(\mathrm{~b}_1 \mathrm{c}_3-\mathrm{b}_3 \mathrm{c}_1\right)+\mathrm{ka}_3\left(\mathrm{~b}_1 \mathrm{c}_2-\mathrm{b}_2 \mathrm{c}_1\right) \\
& \quad=\mathrm{k}\left[\mathrm{a}_1\left(\mathrm{~b}_2 \mathrm{c}_3-\mathrm{b}_3 \mathrm{c}_2\right)-\mathrm{a}_2\left(\mathrm{~b}_1 \mathrm{c}_3-\mathrm{b}_3 \mathrm{c}_1\right)+\mathrm{a}_3\left(\mathrm{~b}_1 \mathrm{c}_2-\mathrm{b}_2 \mathrm{c}_1\right)\right] \\
& \Delta^{\prime}=\mathrm{k} \Delta
\end{aligned}
$
Hence,

$
\left|\begin{array}{ccc}
k a_1 & k a_2 & k a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right|=\mathrm{k}\left|\begin{array}{ccc}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right|
$

Note:

By this property, we can take out any common factor from any one row or any one column of a given determinant.  

If corresponding elements of any two rows (or columns) of a determinant are proportional (in the same ratio), then the determinant value is zero. 

Property 6

If every element of some row or column of a determinant are expressed as sum of two (or more) terms, then the determinant can be expressed as sum of two (or more) determinants

For example

$
\left|\begin{array}{ccc}
a_1+x & a_2+y & a_3+z \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right|=\left|\begin{array}{ccc}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right|+\left|\begin{array}{ccc}
x & y & z \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right|
$
Proof:

$
\mathrm{LHS}=\left|\begin{array}{ccc}
a_1+x & a_2+y & a_3+z \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right|
$
Expanding along first row, we get

$
\begin{aligned}
& \Delta=\left(\mathrm{a}_1+\mathrm{x}\right)\left(\mathrm{b}_2 \mathrm{c}_3-\mathrm{b}_3 \mathrm{c}_2\right)-\left(\mathrm{a}_2+\mathrm{y}\right)\left(\mathrm{b}_1 \mathrm{c}_3-\mathrm{b}_3 \mathrm{c}_1\right)+\left(\mathrm{a}_3+\mathrm{z}\right)\left(\mathrm{b}_1 \mathrm{c}_2-\mathrm{b}_2 \mathrm{c}_1\right) \\
&= \mathrm{a}_1\left(\mathrm{~b}_2 \mathrm{c}_3-\mathrm{b}_3 \mathrm{c}_2\right)-\mathrm{a}_2\left(\mathrm{~b}_1 \mathrm{c}_3-\mathrm{b}_3 \mathrm{c}_1\right)+\mathrm{a}_3\left(\mathrm{~b}_1 \mathrm{c}_2-\mathrm{b}_2 \mathrm{c}_1\right) \\
&+\mathrm{x}\left(\mathrm{~b}_2 \mathrm{c}_3-\mathrm{b}_3 \mathrm{c}_2\right)-\mathrm{y}\left(\mathrm{~b}_1 \mathrm{c}_3-\mathrm{b}_3 \mathrm{c}_1\right)+\mathrm{z}\left(\mathrm{~b}_1 \mathrm{c}_2-\mathrm{b}_2 \mathrm{c}_1\right) \\
&=\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|+\left|\begin{array}{ccc}
x & y & z \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right|
\end{aligned}
$
 

Property 7

If to each element of any row or column of a determinant, the equimultiples of corresponding elements of other rows (or column) are added, then the value of determinant remains the same, i.e., the value of determinant remains same if we apply the operation

$
\mathrm{R}_{\mathrm{i}} \rightarrow \mathrm{R}_{\mathrm{i}}+\mathrm{kR}_{\mathrm{j}} \quad \text { or } \quad \mathrm{C}_{\mathrm{i}} \rightarrow \mathrm{C}_{\mathrm{i}}+\mathrm{kC}_{\mathrm{j}}
$
Explanation,
Let, $\Delta=\left|\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right|$ and $\Delta^{\prime}=\left|\begin{array}{ccc}a_1+k c_1 & a_2+k c_2 & a_3+k c_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right|$
Here, $\Delta^{\prime}$ is obtained by $R_1 \rightarrow R_1+k R_3$
we can write $\Delta^{\prime}$ as

$
\begin{aligned}
\Delta^{\prime} & =\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|+\left|\begin{array}{ccc}
k c_1 & k c_2 & k c_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right| \\
& =\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|+\mathrm{k}\left|\begin{array}{ccc}
c_1 & c_2 & c_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right| \\
& =\Delta+\mathrm{k} \cdot 0
\end{aligned}
$

hence, $\Delta^{\prime}=\Delta$
Property 8

If each element of one side or the other side or both sides of the principal diagonal of a determinant is zero, then the value of the determinant is the product of the diagonal elements.

I.e.

$
\left|\begin{array}{lll}
a & f & g \\
0 & b & h \\
0 & 0 & c
\end{array}\right|=\left|\begin{array}{lll}
a & 0 & 0 \\
f & b & 0 \\
g & h & c
\end{array}\right|=\left|\begin{array}{lll}
a & 0 & 0 \\
0 & b & 0 \\
0 & 0 & c
\end{array}\right|=\mathrm{abc}
$
Property 9
If a determinant D become 0 for $\mathrm{x}=\mathrm{a}$, then $(x-\alpha)$ is a factor of $\Delta$.
For example,
If $\Delta=\left|\begin{array}{ccc}x & x^2 & x^3 \\ 4 & 16 & 64 \\ 5 & 9 & 11\end{array}\right|$
When, $x=4$ the value of $\Delta$ becomes 0
$\because$ at $\mathrm{x}=4, \mathrm{R}_1$ and $\mathrm{R}_2$ are identical.
and at $\mathrm{x}=0, \Delta=0$, because all element of $\mathrm{R}_1$ becomes 0 hence, $(x-0)$ and $(x-4)$ are the factors of $\Delta$.