Mathematics Properties of Determinants For CBSE-NCERT
Click for Only Video

### Topic covered

star Properties of Determinants

### Property 1 :

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
Q 3154180954

Verify Property 1 for Δ = | ( 2,-3, 5), ( 6,0,4), ( 1,5, -7) |
Class 12 Chapter 4 Example 6
Solution:

Expanding the determinant along first row, we have

 Δ = 2 | ( 0 ,4) , ( 5,-7) | - (-3) | (6,4), ( 1,-7) | + 5 | ( 6,0 ), (1,5) |

= 2 (0 – 20) + 3 (– 42 – 4) + 5 (30 – 0)
= – 40 – 138 + 150 = – 28
By interchanging rows and columns, we get

Δ_1 = | (2,6,1), (-3,0,5), (5,4 , -7) | (Expanding along first column)

=2 | ( 0 ,5 ), (4,-7) | - (-3) | ( 6,1), (4 , -7) | + 5 | ( 6,1), ( 0 ,5) |

= 2 (0 – 20) + 3 (– 42 – 4) + 5 (30 – 0)
= – 40 – 138 + 150 = – 28
Clearly Δ = Δ_1

Hence, Property 1 is verified.

### Property 2 :

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}.

Q 3184180957

Verify Property 2 for Δ = | (2,-3,5), ( 6,0,4), ( 1,5, -7) |
Class 12 Chapter 4 Example 7
Solution:

Δ = | (2,-3,5), ( 6,0,4), ( 1,5, -7) | = -28 (See Example 6)

Interchanging rows R_2 and R_3 i.e., R_2 ↔ R_3, we have

Δ_1 = | ( 2,-3 ,5), ( 1,5, -7) , ( 6,0,4) |

Expanding the determinant Δ_1 along first row, we have

Δ_1 = 2 | ( 5,-7), (0,4) | - ( -3) | (1,-7), (6,4) | + 5 | ( 1,5), (6 ,0) |

= 2 (20 – 0) + 3 (4 + 42) + 5 (0 – 30)
= 40 + 138 – 150 = 28

Clearly Δ_1 = – Δ

Hence, Property 2 is verified

### Property 3 :

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

Q 3154191054

Evaluate Δ = | ( 3,2,3), ( 2,2,3), ( 3,2,3) |
Class 12 Chapter 4 Example 8
Solution:

Expanding along first row, we get

Δ = 3 (6 – 6) – 2 (6 – 9) + 3 (4 – 6)
= 0 – 2 (–3) + 3 (–2) = 6 – 6 = 0
Here R_1 and R_3 are identical.

### Property 4 :

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)
Q 3104191058

Evaluate  | (102, 18 ,36) , (1,3,4), ( 17,3,6) |
Class 12 Chapter 4 Example 9
Solution:

Note that  | (102,18 , 36), (1,3, 4 ), (17 ,3,6 ) | = | ( 6 (17) , 6 (3) , 6(6) ), ( 1,3,4), ( 17,3,6) | = 6 | (17 ,3,6 ), ( 1,3,4), ( 17,3,6) | = 0