We are given that,

We need to find the value of x.

Let matrices be,

Then,

Order of A = 1 × 2 [∵, Matrix A has 1 row and 2 columns]

Order of B = 2 × 2 [∵, Matrix B has 2 rows and 2 columns]

Since,

Number of columns in A = Number of rows in B = 2

∴, Order of resulting matrix AB will be 1 × 2.

Resulting matrix = O

O is zero-matrix, where every element of the matrix is zero.

Order of O = 1 × 2

That is,

So,

…(i)

Let,

Let us solve the left hand side of the matrix equation.

In multiplication of matrices,

For z₁₁: Dot multiply 1^st row of first matrix and 1^st column of second matrix, and then sum up.

(x 1)(1 -2) = x × 1 + 1 × -2

⇒ (x 1)(1 -2) = x – 2

So,

For z₁₂: Dot multiply 1^st row of first matrix and 2^nd column of second matrix, and then sum up.

(x 1)(0 0) = x × 0 + 1 × 0

⇒ (x 1)(0 0) = 0 + 0

⇒ (x 1)(0 0) = 0

So,

Substituting the resulting matrix in left hand side of (i),

We know by the property of matrices,

This implies,

a₁₁ = b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂ = b₂₂

Therefore,

x – 2 = 0

⇒ x = 2

Thus, the value of x is 2.