`color{red}{"The value of the determinant remains unchanged if its rows and columns are interchanged."}`

Verification : Let `Delta = |(a_1,a_2,a_3),(b_1,b_2,b_3),(c_1,c_2,c_3)|`

Expanding along first row, we get `Delta = a_1|(b_2,b_3),(c_2,c_3)| - a_2|(b_1,b_3),(c_1,c_3)| + a_3|(b_1,b_2),(c_1,c_2)|`

`=a_1(b_2 c_3– b_3 c_2) - a_2(b_1 c_3– b_3 c_1) + a_3(b_1 c_2 – b_2 c_1)`

● By interchanging the rows and columns of Δ, we get the determinant

` Δ_1 = | (a_1 ,b_1 ,c_1), ( a_2,b_2, c_2), ( a_3,b_3,c_3) |`

● Expanding `Δ_1` along first column, we get

`Δ_1 = a_1 (b_2 c_3 – c_2 b_3) – a_2 (b_1 c_3 – b_3 c_1) + a_3 (b_1 c_2 – b_2 c_1)`

Hence `Δ = Δ_1`


`color{blue}{"Remark :"}` If A is a square matrix, then `color{orange}{det (A) = det (A′),}` where A′ = transpose of A.


`color{blue}{"Note :"}` If `R_i = i^(th)` row and `C_i = i^(th)` column, then for interchange of row and columns, we will symbolically write `C_i↔ R_i`

`color{red}{"If any two rows (or columns) of a determinant are interchanged,"}`
`color{red}{"then sign of determinant changes. "}`

`"Verification :"` Let Δ = ` | (a_1, a_2, a_3 ), ( b_1, b_2 , b_3 ), ( c_1, c_2, c_3) |`

Expanding along first row, we get

`Δ = a_1 (b_2 c_3 – b_3 c_2) – a_2 (b_1 c_3 – b_3 c_1) + a_3 (b_1 c_2 – b_2 c_1)`

`=>` Interchanging first and third rows, the new determinant obtained is given by

`Δ_1 = | (c_1, c_2, c_3 ), ( b_1, b_2 , b_3 ), ( a_1, a_2 , a_3) |`

`=>` Expanding along third row, we get

`Δ_1 = a_1 (c_2 b_3 – b_2 c_3) – a_2 (c_1 b_3 – c_3 b_1) + a_3 (b_2 c_1 – b_1 c_2)`

`= – [a_1 (b_2 c_3 – b_3 c_2) – a_2 (b_1 c_3 – b_3 c_1) + a_3 (b_1 c_2 – b_2 c_1)]`

Clearly `Δ_1 = – Δ`

`"Similarly, we can verify the result by interchanging any two columns."`

`color{orange}{" We can denote the interchange of rows by"\ \ R_i ↔ R_j \ \ "and interchange of
columns by" C_i ↔ C_j}`.

`color{red}{" If any two rows (or columns) of a determinant are identical "}`
`color{red}{"all corresponding elements are same), then value of determinant is zero."}`

`color{blue}{"Proof"}` If we interchange the identical rows (or columns) of the determinant Δ, then Δ
does not change. However, by Property 2, it follows that Δ has changed its sign

Therefore `Δ = – Δ`

or `Δ = 0`

`color{red}{" If each element of a row (or a column) of a determinant is"}`
`color{green}{"multiplied by a constant" k, "then its value gets multiplied by" k}`

`"Verification :"` Let `Δ = | (a_1, b_1, c_1), ( a_2,b_2, c_2), (a_3,b_3,c_3) |`

and `Δ_1` be the determinant obtained by multiplying the elements of the first row by `k.`

Then

`Δ_1 = | (k a_1, k b_1, k c_1 ), ( a_2, b_2, c_2 ), ( a_3, b_3, c_3) |`

`=>` Expanding along first row, we get

`Δ_1 = k a_1 (b_2 c_3 – b_3 c_2) – k b_1 (a_2 c_3 – c_2 a_3) + k c_1 (a_2 b_3 – b_2 a_3)`

`= k [ a_1 (b_2 c_3 – b_3 c_2) – b_1 (a_2 c_3 – c_2 a_3) + c_1 (a_2 b_3 – b_2 a_3)]`

`= k Δ`


`color{orange}{ | (k a_1, k b_1, k c_1 ), ( a_2, b_2, c_2), ( a_3, b_3, c_3) | = k | (a_1, b_1, c_1), ( a_2,b_2 ,c_2), (a_3, b_3, c_3) |}`


`color{blue}{"Remarks :"}`

(i) By this property, we can take out any common factor from any one row or any one column of a given determinant.
(ii) If corresponding elements of any two rows (or columns) of a determinant are proportional (in the same ratio), then its value is zero. For example
`Δ = | (a_1, a_2, a_3 ), (b_1, b_2, b_3 ), ( k a_1, k a_2, k a_3) |=0 ` (rows `R_1` and `R_2` are proportional)

`color{red}{"If some or all elements of a row or column of a determinant are expressed as "}`
`color{red}{"sum of two (or more) terms, then the determinant can be expressed as sum of two (or more) determinants."}`

For example, ` | (a_1+ lambda_1 , a_2 + lambda_2, a_3 + lambda_3 ), ( b_1, b_2, b_3 ), ( c_1, c_2, c_3) | `

`= | (a_1, a_2, a_3 ), ( b_1, b_2 , b_3), (c_1, c_2 , c_3) | + | ( lambda_1, lambda_2, lambda_3 ) , ( b_1, b_2, b_3), (c_!, c_2, c_3) |`

Verification L.H.S.` = | (a_1 + lambda_1 , a_2 +lambda_2 , a_3+lambda_3 ), ( b_1, b_2, b_3), ( c_1, c_2, c_3) |`


`=>` Expanding the determinants along the first row, we get

`Δ = (a_1 + λ_1) (b_2 c_3 – c_2 b_3) – (a_2 + λ_2) (b_1 c_3 – b_3 c_1)`

`+ (a_3 + λ_3) (b_1 c_2 – b_2 c_1)`

`= a_1 (b_2 c_3 – c_2 b_3) – a_2 (b_1 c_3 – b_3 c_1) + a_3 (b_1 c_2 – b_2 c_1)`

`+ λ_1 (b_2 c_3 – c_2 b_3) – λ_2 (b_1 c_3 – b_3 c_1) + λ_3 (b_1 c_2 – b_2 c_1)`
(by rearranging terms)

`= | (a_1, a_2, a_3), ( b_1, b_2, b_3), ( c_1, c_2, c_3) | + | (λ_1 , λ_2, λ_3 ), ( b_1 ,b_2, b_3 ), ( c_1, c_2, c_3) | = R.H.S.`


`=>` Similarly, we may verify Property 5 for other rows or columns.

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

`color{red}{R_i → R_i + k R_j` or `C_i → C_i + k C_j}` .

`"Verification : "` Let ` Δ = | (a_1, a_2, a_3), ( b_1, b_2, b_3), (c_1, c_2, c_3) | `
and ` Δ_1 = | (a_1+k c_1 , a_2 + kc_2 , a_3 + k c_3), ( b_1, b_2, b_3 ), ( c_1, c_2, c_3) |` ,

`=>` where `Δ_1` is obtained by the operation `R_1 → R_1 + k R_3` .

`=>` Here, we have multiplied the elements of the third row `(R_3)` by a constant k and added them to the corresponding elements of the first row `(R_1)`.

Now, again

` Δ_1 = | (a_1 , a_2, a_3), ( b_1, b_2, b_3), ( c_!, c_2, c_3) | + | (k c_1, k c_2, k c_3), (b_1, b_2, b_3), ( c_1, c_2, c_3) | ` (Using Property 5)

`= Δ + 0` (since `R_1` and `R_3` are proportional)

Hence `Δ = Δ_1`

Remarks

(i) If `Δ_1` is the determinant obtained by applying `R_i → k R_i` or `C_i → k C_i` to the determinant Δ, then `Δ_1 = kΔ`.

(ii) If more than one operation like `R_i→ R_i + k R_j` is done in one step, care should be taken to see that a row that is affected in one operation should not be used in another operation. `A` similar remark applies to column operations.