Cheapest way to chain three products

MediumLinear Algebra~4m

You must compute the product $ABC$ where $A$ is $2\times3$ , $B$ is $3\times4$ , and $C$ is $4\times2$ , using the standard algorithm (an $m\times n$ times $n\times p$ product costs $m\cdot n\cdot p$ scalar multiplications). Matrix multiplication is associative, so you may bracket it as $(AB)C$ or $A(BC)$ . What is the minimum total number of scalar multiplications?

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Cost out both bracketings $(AB)C$ and $A(BC)$ separately.
Each stage costs $m\cdot n\cdot p$ ; add the two stages, then take the smaller total.

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Cheapest way to chain three products

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