## Linear Algebra - Determinants

1. The problem statement, all variables and given/known data

If det\left[ \begin {array}{ccc} a&1&d\\ \noalign{\medskip} b&1&e\\ \noalign{\medskip} c&1&f \end {array} \right]=-4 and det\left[ \begin {array}{ccc} a&1&d\\ \noalign{\medskip} b&2&e\\ \noalign{\medskip} c&3&f \end {array} \right]=-1,

then det\left[ \begin {array}{ccc} a&8&d\\ \noalign{\medskip} b&8&e\\ \noalign{\medskip} c&8&f \end {array} \right]=___

and det\left[ \begin {array}{ccc} a&-1&d\\ \noalign{\medskip} b&-4&e\\ \noalign{\medskip} c&-7&f \end {array} \right]=___

3. The attempt at a solution

For the first question, I'm pretty sure that I can factor out an 8 as it is a scalar multiple of the second column. One of the properties of determinants is that if a row or column is multiplied by a scalar, then we can factor the scalar out and then multiply the determinant by that scalar. Thus, the answer would be -32.
However, I am really stumped by the second question. I am sure that the scalar -1 is somehow multiplied into the matrix, but I am not sure how the numbers were obtained. I'm thinking that each row must have resulted from the scalar multiple of another row. However, if that's the case, then why are the unknowns unaffected?
I am seriously at my wit's end and any direction would be helpful.
Thank you!
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 Recognitions: Homework Help Science Advisor (-1,-4,-7) is (-3)*(1,2,3)+2*(1,1,1). A determinant is a linear function of each of it's columns.
 Recognitions: Homework Help Note that detA = detA^T. And also the determinant of A+B where A,B differ by a single row is det(A+B). Now multiply the column 1,2,3 of that matrix by 3. What do you have to multiply the 2nd column of the 1,1,1 column matrix such that adding up both gives you -1,-4,-7?

## Linear Algebra - Determinants

Wow! It makes so much more sense now that you've mentioned detA=detA^T to me! Thank you so much!
 Now that you have already seen the algebraic explanation, here is an image tutorial showing the geometric intuition behind it: http://img137.imageshack.us/img137/6...trowopsjx2.png The image explains in 2D, but in higher dimensions everything is the same except parallelograms become paralellipipeds and the determinant measures volume not area. Can you see why the determinant of a singular matrix has to be zero?
 Recognitions: Homework Help Uh, what has that got to do with the OP's question?
 It is a visual proof that 1) Multiplying a column by a constant scales the determinant by that constant, and 2) The determinant is linear in any particular column.

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