Question

Ex. 1 Show that
|101 202 303
505 606 707 = 0
2 :
3​

Answer 1

Step-by-step explanation:

let∆= (101 202 303)

let∆= (101 202 303) (505 606 707)

let∆= (101 202 303) (505 606 707) (1. 2. 3)

let∆= (101 202 303) (505 606 707) (1. 2. 3)∆=(101 1 2)

let∆= (101 202 303) (505 606 707) (1. 2. 3)∆=(101 1 2) (505 1. 2) applying C2→C2 - C1 and C3 - C1)

let∆= (101 202 303) (505 606 707) (1. 2. 3)∆=(101 1 2) (505 1. 2) applying C2→C2 - C1 and C3 - C1) (1 1. 2)

let∆= (101 202 303) (505 606 707) (1. 2. 3)∆=(101 1 2) (505 1. 2) applying C2→C2 - C1 and C3 - C1) (1 1. 2)= 2 (101 1 1)

let∆= (101 202 303) (505 606 707) (1. 2. 3)∆=(101 1 2) (505 1. 2) applying C2→C2 - C1 and C3 - C1) (1 1. 2)= 2 (101 1 1) (505. 1 1)

let∆= (101 202 303) (505 606 707) (1. 2. 3)∆=(101 1 2) (505 1. 2) applying C2→C2 - C1 and C3 - C1) (1 1. 2)= 2 (101 1 1) (505. 1 1) (1 1 1)

let∆= (101 202 303) (505 606 707) (1. 2. 3)∆=(101 1 2) (505 1. 2) applying C2→C2 - C1 and C3 - C1) (1 1. 2)= 2 (101 1 1) (505. 1 1) (1 1 1)= 0

let∆= (101 202 303) (505 606 707) (1. 2. 3)∆=(101 1 2) (505 1. 2) applying C2→C2 - C1 and C3 - C1) (1 1. 2)= 2 (101 1 1) (505. 1 1) (1 1 1)= 0since two elements columns are identical, the value of the determinant is zero.

let∆= (101 202 303) (505 606 707) (1. 2. 3)∆=(101 1 2) (505 1. 2) applying C2→C2 - C1 and C3 - C1) (1 1. 2)= 2 (101 1 1) (505. 1 1) (1 1 1)= 0since two elements columns are identical, the value of the determinant is zero. =∆= (101 202 303)

let∆= (101 202 303) (505 606 707) (1. 2. 3)∆=(101 1 2) (505 1. 2) applying C2→C2 - C1 and C3 - C1) (1 1. 2)= 2 (101 1 1) (505. 1 1) (1 1 1)= 0since two elements columns are identical, the value of the determinant is zero. =∆= (101 202 303) (505 606 707)

let∆= (101 202 303) (505 606 707) (1. 2. 3)∆=(101 1 2) (505 1. 2) applying C2→C2 - C1 and C3 - C1) (1 1. 2)= 2 (101 1 1) (505. 1 1) (1 1 1)= 0since two elements columns are identical, the value of the determinant is zero. =∆= (101 202 303) (505 606 707) (1 2 3)

let∆= (101 202 303) (505 606 707) (1. 2. 3)∆=(101 1 2) (505 1. 2) applying C2→C2 - C1 and C3 - C1) (1 1. 2)= 2 (101 1 1) (505. 1 1) (1 1 1)= 0since two elements columns are identical, the value of the determinant is zero. =∆= (101 202 303) (505 606 707) (1 2 3) = 0