Property 4:

If all the elements of a row (or a column) of a square matrix are multiplied by a number, k, then the value of the determinant of the matrix obtained is k times the determinant of the given matrix.

Verification:

Let A = $\left|\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{a}}_{\text{2}}& {\text{a}}_{\text{3}}\\ {\text{b}}_{\text{1}}& {\text{b}}_{\text{2}}& {\text{b}}_{\text{3}}\\ {\text{c}}_{\text{1}}& {\text{c}}_{\text{2}}& {\text{c}}_{\text{3}}\end{array}\right|$

|A| = a₁(b₂c₃ - b₃c₂) - a₂(b₁c₃ - b₃c₁) + a₃(b₁c₂ - b₂c₁)

|A| = a₁b₂c₃ - a₁b₃c₂ - a₂b₁c₃ + a₂b₃c₁ + a₃b₁c₂ - a₃b₂c₁

Let the elements of row 1 be multiplied by a constant, k.

Then B = $\left[\begin{array}{ccc}{\text{ka}}_{\text{1}}& {\text{ka}}_{\text{2}}& {\text{ka}}_{\text{3}}\\ {\text{b}}_{\text{1}}& {\text{b}}_{\text{2}}& {\text{b}}_{\text{3}}\\ {\text{c}}_{\text{1}}& {\text{c}}_{\text{2}}& {\text{c}}_{\text{3}}\end{array}\right]$

Expanding along the first row, we get

|B| = ka₁(b₂c₃ - b₃c₂) - ka₂(b₁c₃ - b₃c₁) + ka₃(b₁c₂ - b₂c₁)

⇒|B| = k|A|

i.e. the determinant of matrix B is k times the determinant of matrix A.

Note:

If the elements of a row (or a column) of a square matrix are k times the elements of another row (or column), then the value of its determinant is zero.

If M and N are two square matrices such that M = kN, where k is a real number, then |M| = k|N|, where n is the order of the matrices M and N.

Property 5:

If the elements of a row (or a column) of the determinant of a matrix are expressed as the sum of two or more terms, then the determinant can be expressed as the sum of two or more determinants.

The elements in the first column of a determinant are expressed as the sum of two terms.

$\left|\begin{array}{ccc}{{\text{a}}_{\text{1}}\text{+l}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {{\text{a}}_{\text{2}}\text{+l}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {{\text{a}}_{\text{3}}\text{+l}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|\text{=}\left|\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{a}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{a}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|\text{+}\left|\begin{array}{ccc}{\text{l}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{l}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{l}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|$    

Consider $\left|\begin{array}{ccc}{{\text{a}}_{\text{1}}\text{+l}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {{\text{a}}_{\text{2}}\text{+l}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {{\text{a}}_{\text{3}}\text{+l}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|$.

Expanding the determinant along the first column, we get

=(a₁ + l₁)(b₂c₃ - b₃c₂) - (a₂ + l₂)(b₁c₃ - b₃c₁) + (a₃ + l₃)(b₁c₂ - b₂c₁)

=a₁(b₂c₃ - b₃c₂) + l₁(b₂c₃ - b₃c₂) - a₂(b₁c₃ - b₃c₁) - l₂(b₁c₃ - b₃c₁) + a₃(b₁c₂ - b₂c₁) + l₃(b₁c₂ - b₂c₁)

$\text{=}\left|\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{a}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{a}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|\text{+}\left|\begin{array}{ccc}{\text{l}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{l}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{l}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|$ 

Note: Similar results can be obtained by considering the other rows and columns.

Property 6:

If the elements of a row (or column) of a square matrix are added with k times (where k is a constant) the corresponding elements of another row (or column), then the value of the determinant of the given matrix does not change.

Or

The value of a determinant remains unchanged if we apply the operation, R_i → R_i + kR_j or C_i → C_i + kC_j.

Let A = $\left[\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{a}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{a}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right]$      Applying operation C₁ → C₁ + kC₂, we have

           $\left[\begin{array}{ccc}{{\text{a}}_{\text{1}}\text{+kb}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {{\text{a}}_{\text{2}}\text{+kb}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {{\text{a}}_{\text{3}}\text{+kb}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right]$ = (B)say

|B| = $\left|\begin{array}{ccc}{{\text{a}}_{\text{1}}\text{+kb}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {{\text{a}}_{\text{2}}\text{+kb}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {{\text{a}}_{\text{3}}\text{+kb}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|$

$\text{=}\left|\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{a}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{a}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|\text{+}\left|\begin{array}{ccc}{\text{kb}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{kb}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{kb}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|$   [By property 5]

The second determinant, each element of the first column is multiplied by k.

$\text{=}\left|\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{a}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{a}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|\text{+k}\left|\begin{array}{ccc}{\text{b}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{b}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{b}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|$   [By property 4]

Here the second determinant whose first and second columns are identical.

= |A| + k(0)  [By property 3]

= |A|

Hence, $\left|\begin{array}{ccc}{{\text{a}}_{\text{1}}\text{+kb}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {{\text{a}}_{\text{2}}\text{+kb}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {{\text{a}}_{\text{3}}\text{+kb}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|$$\text{=}$$\left|\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{a}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{a}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|$

$$
Note: Similar results can be obtained by considering the other rows and columns.