Step-by-Step Solution

Topic: System of Linear Equations & Matrices

(A) SYSTEM OF EQUATIONS

Let x, y, and z be the number of students in the 1st, 2nd, and 3rd groups respectively.

1. Total Students:

x + y + z = 10

2. Second Condition: “Double the number of 1st group added to 2nd group gives 13”

2x + y = 13

3. Third Condition: “Combined strength of 1st and 2nd is four times the 3rd”

(x + y) = 4z ⇒ x + y – 4z = 0

(B) COEFFICIENT MATRIX A

Extracting the coefficients from the equations above:

(C)(I) MATRIX OF COFACTORS

We calculate the cofactor C_ij for each element A_ij:

• C₁₁ = 1(-4) – 0(1) = -4
• C₁₂ = -[2(-4) – 0(1)] = -[-8] = 8
• C₁₃ = 2(1) – 1(1) = 1
• C₂₁ = -[1(-4) – 1(1)] = -[-5] = 5
• C₂₂ = 1(-4) – 1(1) = -5
• C₂₃ = -[1(1) – 1(1)] = 0
• C₃₁ = 1(0) – 1(1) = -1
• C₃₂ = -[1(0) – 1(2)] = -[-2] = 2
• C₃₃ = 1(1) – 1(2) = -1

(C)(II) SOLVE FOR x, y, z

We can solve the system using the equations derived in Part (a):

1. x + y + z = 10
2. 2x + y = 13
3. x + y – 4z = 0

From (2): y = 13 – 2x.

Substitute y into (1):

x + (13 – 2x) + z = 10 ⇒ -x + z = -3 ⇒ z = x – 3.

Substitute y and z into (3):

x + (13 – 2x) – 4(x – 3) = 0

x + 13 – 2x – 4x + 12 = 0

-5x + 25 = 0 ⇒ x = 5.

Now find y and z:

y = 13 – 2(5) = 3

z = 5 – 3 = 2

⚠️ Silly Mistake Audit: Cofactor Signs

Check the Checkerboard Pattern!
The most common error in finding the Matrix of Cofactors is forgetting to apply the negative sign to the positions (1,2), (2,1), (2,3), and (3,2). Remember that Cofactor C_ij = (-1)^i+j M_ij. Simply calculating the minor is not enough!