Q. 40 A5.0( 2 Votes )

# Solve the matrix 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 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.

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