Now we will multiply the two first matrices on LHS using the formula c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj, we get

Again multiply these two LHS matrices, we get

⇒ x² – 2x – 15 = 0. This is form of quadratic equation, we will solve this by splitting the middle term, we get

⇒ x² – 5x + 3x – 15 = 0

⇒ x(x – 5) + 3(x – 5) = 0

⇒ (x – 5)(x + 3) = 0

⇒ x – 5 = 0 or x + 3 = 0

This gives, x = 5 or x = – 3 is the required solution of the matrices.