We have the matrices A and B, such that

We need to find matric C, such that AC = BC.

Let C be a non-zero matrix of order 2 × 1, such that

But order of C can be 2 × 1, 2 × 2, 2 × 3, 2 × 4, …

[∵ In order to multiply two matrices, A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.

∴, number of columns in matrix A = number of rows in matrix C = 2]

Take AC.

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

(3, 5)(x, y) = (3 × x) + (5 × y)

⇒ (3, 5)(x, y) = 3x + 5y

Now, take BC.

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

(7, 3)(x, y) = (7 × x) + (3 × y)

⇒ (7, 3)(x, y) = 7x + 3y

And,

AC = BC

⇒ [3x + 5y] = [7x + 3y]

⇒ 3x + 5y = 7x + 3y

⇒ 7x – 3x = 5y – 3y

⇒ 4x = 2y

⇒ y = 2x

Then,

Since, C is of orders, 2 × 1, 2 × 2, 2 × 3, …

In general,

Where, k is any real number.