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Finding the determinant

Problem Statement
A=[a b c; d e f; g h i]

Suppose that det(A) = −5. Let B be another 3 × 3 matrix (not given here) with det(B) = 6. Find the determinant of each of the following matrices.

I found it for the other ones in the question except for:

What is the determinant of D=3(A^-1)(B^T)?
Relevant Equations
D=3(A^-1)(B^T)
I assumed that my calculation would be 3(-5^-1)(6) and I got the answer -18/5, however this is incorrect, Im unsure of where I am going wrong. I thought the determinant of a matrix is equal to the determinant of the transpose of the matrix so det(B)=6 would also be det(B^T)=6?

Thank you.
 

Orodruin

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What is the determinant of kA if A is a matrix and k s a constant?
 
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ver_mathstats said:
Relevant Equations D=3(A^-1)(B^T)
What you wrote for a relevant equation isn't really relevant -- it's part of the problem statement.
Some equations that are relevant are shown in this wiki article, https://en.wikipedia.org/wiki/Determinant, in the Properties of the determinant section. The 2nd, 3rd, and 5th properties are especially relevant to your question.
 
What is the determinant of kA if A is a matrix and k s a constant?
Thank you for the reply, det(cA)=(c^n)det(A), I didn't realize this was the case, I ended up with the answer -162/5.
 
What you wrote for a relevant equation isn't really relevant -- it's part of the problem statement.
Some equations that are relevant are shown in this wiki article, https://en.wikipedia.org/wiki/Determinant, in the Properties of the determinant section. The 2nd, 3rd, and 5th properties are especially relevant to your question.
Thank you for the reply, I used the properties, and realized where I went wrong, my answer is -162/5.
 

StoneTemplePython

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Thank you for the reply, det(cA)=(c^n)det(A), I didn't realize this was the case, I ended up with the answer -162/5.
It isn't needed but another way to remember this is we are dealing with n x n matrices here and determinants multiply so

##\det\big(c\mathbf A\big) = \det\big(c\mathbf I\mathbf A\big) = \det\big(c\mathbf I\big) \det\big(\mathbf A\big) = c^n \cdot \det\big(\mathbf A\big)##

where ##\det\big(c\mathbf I\big) ## is easy because it is a diagonal matrix with c on the diagonal -- the determinant of a diagonal matrix is always just the product of the diagonal components.
 

Orodruin

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It isn't needed but another way to remember this is we are dealing with n x n matrices here and determinants multiply so

##\det\big(c\mathbf A\big) = \det\big(c\mathbf I\mathbf A\big) = \det\big(c\mathbf I\big) \det\big(\mathbf A\big) = c^n \cdot \det\big(\mathbf A\big)##

where ##\det\big(c\mathbf I\big) ## is easy because it is a diagonal matrix with c on the diagonal -- the determinant of a diagonal matrix is always just the product of the diagonal components.
In fact, I often think of constants multiplying matrices as a diagonal matrix with all diagonal elements equal to that constant.
 

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