Now we will multiply the two first matrices on LHS using the formula cij = ai1b1j + ai2b2j + … + ainbnj, we get
Again multiply these two LHS matrices, we get
⇒ 4 + 4x = 0 we will solve this linear equation, we get
⇒ 4x = – 4
This gives, x = – 1 is the required solution of the matrices.
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