We are given the matrices A, B and C, such that

We need to verify that, A(B + C) = AB + AC.

Take L.H.S: A(B + C)

Solving (B + C).

These matrices can be added as they have same order.

Now, multiply A by (B + C).

Let (B + C) = D.

We have,

AD = A(B + C)

Order of A = 1 × 2

Order of D = 2 × 3

Then, order of resulting matrix = 1 × 3

Multiply 1^st row of matrix A by matching members of 1^st column of matrix D, then sum them up.

(2, 1)(4, 9) = (2 × 4) + (1 × 9)

⇒ (2, 1)(4, 9) = 8 + 9

⇒ (2, 1)(4, 9) = 17

Multiply 1^st row of matrix A by matching members of 2^nd column of matrix D, then sum them up.

(2, 1)(5, 7) = (2 × 5) + (1 × 7)

⇒ (2, 1)(5, 7) = 10 + 7

⇒ (2, 1)(5, 7) = 17

Multiply 1^st row of matrix A by matching members of 3^rd column of matrix D, then sum them up.

(2, 1)(5, 8) = (2 × 5) + (1 × 8)

⇒ (2, 1)(5, 8) = 10 + 8

⇒ (2, 1)(5, 8) = 18

So,

Now, take R.H.S: AB + AC

Let us compute AB.

Order of A = 1 × 2

Order of B = 2 × 3

Then, order of AB = 1 × 3

Multiply 1^st row of matrix A by matching members of 1^st column of matrix B, then sum them up.

(2, 1)(5, 8) = (2 × 5) + (1 × 8)

⇒ (2, 1)(5, 8) = 10 + 8

⇒ (2, 1)(5, 8) = 18

Similarly, repeat steps to find the rest of the elements.

Now, let us compute AC.

Order of AC = 1 × 3

Multiply 1^st row of matrix A by matching members of 1^st column of matrix C, then sum them up.

(2, 1)(-1, 1) = (2 × -1) + (1 × 1)

⇒ (2, 1)(-1, 1) = -2 + 1

⇒ (2, 1)(-1, 1) = -1

Similarly, repeat steps to find the rest of the elements.

Add, AB + AC.

Thus,

A(B + C) = AB + AC.