# Identifying matrix vector multiplication

I have the following question in a book:

According to the book, the answer is (D). But I don't understand how. Isn't this just scalar multiplication? The solution in the book says that I have to use determinants. How do I work this one out? Is the book wrong? (No?).

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Hint: what if you were doing co-factors to find the determinant, what would it look like? The book is right! –  Amzoti Apr 30 '13 at 2:35

Note that $B$ is equal to the determinant of the second matrix; $A$ is equal to the determinant of the first matrix. Have you tried computing the determinant of each to see that we get B = 27A?

A is not a matrix, B is not the matrix.

We do have that the matrix for which the determinant equals B, call it $B_M = 3 A_M$, where $A_M$ is the matrix whose determinant is equal to A. But, $\det B_M = 3^3 \det A_M$

Recall, that when finding the determinant of a matrix, if a row of one matrix is multiplied by a scalar $c$, then the determinant $B$ of the resulting matrix $3\det A$: for EACH row. Here, you have three rows in the second matrix that are each a multiple of a row in the first matrix, hence we have $$B = 3\cdot 3\cdot 3 A = 3^3 A = 27 A$$

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Yeah, I get that, but why isn't this scalar multiplication? –  Hele Apr 30 '13 at 2:36
$\det? \cdot? what?? – Hele Apr 30 '13 at 2:38 Note that$B$is equal to the determinant of the second matrix;$A$is equal to the determinant of the first matrix. Have you tried computing the determinant of each to see that we get B = 27A? A is not a matrix, B is not the matrix. We do have that the Matrix for which the determinant equals B, call it$B_M = 3 A_M$, where$A_M$is the matrix whose determinant is equal to A. But,$\det B_M = 3^3 \det A_M\$ –  amWhy Apr 30 '13 at 2:43
Thanks you, but my real question was "Why isn't this scalar multiplication?". Vadim answered that question. Thanks anyway :) ! –  Hele Apr 30 '13 at 2:45
I'm just trying to help! ;-) –  amWhy Apr 30 '13 at 2:47

The vertical bars denote determinant. They are not just placeholders to hold the matrix, like square brackets or parentheses.

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Problem solved :) ! Thanks! –  Hele Apr 30 '13 at 2:41