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
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,
Where, k is any real number.