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Given $A \in\mathbb{R^{7\times8}}$, $B \in\mathbb{R^{8\times5}}$ and $C \in\mathbb{R^{5\times7}}$

How can one check whether $$det(ABC) = 0$$ is true?

Given their spaces, the multiplications are "valid", but no specific matrices are given.

LyK
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1 Answers1

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Note that $B\in\Bbb R^{8\times 5}$, so $B$ has $5$ columns and the rank of $B$ is at most $5$. ($C$ has $5$ rows so the rank of $C$ is also at most $5$.) Therefore the rank of $ABC$ is at most $5$. But $ABC$ is a $7\times 7$ matrix, so its rank is less than the number of its rows or columns.

Therefore the determinant of $ABC$ is zero.

Note that we need no further information on matrices $A,B,C$.

Rory Daulton
  • 32,288
  • I understand the logic behind the ranks of C and B. But why this means that ABC is also at most the same rank? EDIT: found the answer here http://math.stackexchange.com/questions/978/how-to-prove-and-interpret-operatornamerankab-leq-operatornamemin-ope

    Thank you!

     – LyK Jun 11 '15 at 09:43